Question 1

Question 1: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 13 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.072. This means: Your Answer: Correct There was only a 7.2% chance of observing an increase greater than 13 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true. Question 2: Score 0/2 An agronomist measured the height of 150 Wheat plants. The mean height was 204 cm and the standard deviation was 17 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.388 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result!

Question 3: Score 0/2 A medical researcher is interested in whether patients' left arms or right arms are longer. If 11 patients participate in this study (so that 11 left arms and 11 right arms are measured), how many degrees of freedom should the researcher use in Incorrect her t-test critical value? Your Answer: Correct 10 Answer: Comment: The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 11 - 1 = 10

Question 4: Score 0/2 To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 300 cases were studied. The

results were: True Status Innocent Guilty Examiner's Innocent Decision Guilty

19

29

131

121

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as: Your Answer: Correct 0.873 Answer: Commen Type I error is when a true null hypothesis is rejected. In this case, this is the probability t: of the Examiner judging an innocent suspect to be guilty. The probability is:

Question 5: Score 0/2 Researchers tested patients fitted with a medical telemetry unit to see if use of a cellular telephone interferes with the operation of the device. There were 555 tests conducted for one type of cellular telephone; interference with the device was found in 15% of these tests. Incorrect Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p. Your Answer: Correct Answer: (0.111, 0.189) Comment: The confidence interval would be:

= = (0.111, 0.189) Question 6: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 45 stocks listed in the NASDAQ. The mean value of the stocks are

USD 1.53 (in millions) and the standard deviation is USD 0.19 (millions). Incorrect Assuming normality, a 90% confidence interval is: Your Answer: Correct (1.483 , 1.577) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Question 7: Score 0/2 A random sample of 505 printers discovered that 68 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses. Your Answer: Correct Answer: 0.096 < p < 0.174 Comment:

Incorrect

Question 8: Score 0/2 An agronomist measured the height of 132 Corn plants. The mean height was 239 cm and the standard deviation was 17 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.4797 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 9: Score 0/2 A pharmacist is planning to estimate the mean level of a certain drug in a lab. The pharmacist wanted the estimate to be within 7 mg/dLi or less with 95% confidence. The pharmacist also believes that the standard deviation of the drug level is probably about 41 mg/dLi. How large a sample should the pharmacist need to take? Your Answer: Correct Answer: 132

Incorrect

Comment:

Use the equation:

. For a 95% confidence interval we have c = 1.96,

so:

Question 10: Score 0/2 An agronomist measured the height of 123 Corn plants. The mean height was 235 cm and the standard deviation was 13 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.1722±0.01 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 11: Score 0/2

6 squirrels were found to have an average weight of 390 grams with a sample standard deviation of 5.35. Find the 95% confidence interval of the true Incorrect mean weight (assume the t-student distribution). Your Answer: Correct (384.38,395.62) Answer: Comment: Use the formula . With degrees of freedom = 5 we have t0.025 = 2.571 so our confidence interval is

Question 12: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a

7.6 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.094. This means: Your Answer: Correct There was only a 9.4% chance of observing an increase greater than 7.6 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals (units) while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more different mean weight, so more weight gained) occurrence, given that the null hypothesis is true.

Question 13: Score 0/2 In order to study the harmful effects of DDT poisoning, the pesticide was fed to 6 randomly chosen rats out of a group of 12 rats. The other 6 rats were used as the control group. The following data gives the measurements of the amount of tremor detected in the bodies of each rat after the experiment: The more tremor, the more harmful.

Rat:

1

2

3

4

5

6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6 Control Group:

11.1 12.1 9.3 6.6 9.6 8.2

A computer analysis is done with the output below (the differences are computed as control - poisoned) t-test

Incorrect

Difference t-test DF Prob>|t|

Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean

Answer: Comment:

number of tremors for all potential rats in the poisoned group is larger than that in the control group. The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats.

Question 14: Score 0/2 A study of 95 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 10 m. Which of the following is the 97% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (179.77, 184.23) Answer: Comment: For a 97% CI we need to find the z value for the standard normal for which 98.5% of the graph area lies to the left. This is 2.1701 . The CI is:

Question 15: Score 0/2 In a sample of 530 mice, a biologist found that 65% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less. Your Answer: Correct Answer: 0.609% < p < 0.69% Comment: Question 1: Score 0/2 In hypothesis testing, β is the probability of committing an error of Type II. The power of the test, 1 − β is then: Your Answer: Correct Answer: Comment:

Incorrect

Incorrect

the probability of rejecting H0 when HA is true Type II error is when a false null hypothesis is not rejected. β is the probability of failing to reject null hypothesis H0 given that HA is true. 1-β is the opposite of this, so the probability of rejecting H0 given HA is true.

Question 2: Score 0/2 The Bata Shoe Museum states that 33% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this

proportion is within 0.01 of the true proportion. How large a sample is necessary? Your Answer: Correct Answer: 11,965 Comment:

Incorrect

Question 3: Score 0/2 The formula of the t -test for dependent samples is: Incorrect Your Answer: Correct Answer:

Comment:

Question 4: Score 0/2 A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. She measures one egg from each nest. The data are summarized below. Incorrect

The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 5: Score 0/2 A survey of 519 women shoppers found that 38% of them shop on impulse. What is the 98% confidence interval for the true proportion of women

.

shoppers who shop on impulse? Incorrect Your Answer: Correct 0.3304 < p < 0.4296 Answer: Comment: To find a 98% confidence interval not that we have Z = 2.3263, p = 0.38 and:

0.3304 < p < 0.4296 Question 6: Score 0/2 A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In Incorrect this problem, the researcher will commit a type I error if: Your Answer: Correct she concludes that the new diet is better when in fact the diets are equal in Answer: effectiveness. Comment: Her hypotheses are: Ho: New diet worse or equal to old Ha: New diet better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new diet is better when in fact the diets are equal in effectiveness." Question 7: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 40 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.22 (in millions) and the standard deviation is USD 0.17 (millions). Assuming normality, a 90% confidence interval is: Your

Incorrect

Answer: Correct (1.176 , 1.264) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Question 8: Score 0/2 Your response

Correct response

A sample of 600 racing cars showed A sample of 600 racing cars showed that 300 cars cost over $400,000. that 300 cars cost over $400,000. What What is the 98% confidence interval of is the 98% confidence interval of the true the true proportion of cars costing over proportion of cars costing over $400,000 (3 $400,000 (3 decimals)? decimals)? Incorrect

(0%) < p <

(0%)

0.4525±0.01

< p < 0.5475±0.01

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0% Comment: Our estimated mean is

. Our confidence interval is

.

For a 98% Confidence Interval we have z = 2.3263, so our interval is or Question 9: Score 0/2 It was found that in a sample of 350 grandparents, 10% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of grandparents who have received speeding tickets? Incorrect Your Answer:

Correct Answer: 0.074 < p < 0.126 Comment: Question 10: Score 0/2 A clerk researched the average number of years served by 47 different judges on her court. The average number of years served was 13.11 years with a standard deviation of 7.95 years. What is the 95% confidence Incorrect interval for the average number of years served by all such judges? Your Answer: Correct 10.837 < μ < 15.383 Answer: Comment: To find a 95% CI we need the value of Z such that 97.5% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find this value is 1.959964. The CI then is:

=(10.837, 15.383)

Question 11: Score 0/2 The critical value for a left-tailed t-test for dependent samples when the degrees of freedom = 7 and α = 0.025 is : (4 decimals)

Incorrect

Your Answer: Correct -2.365±0.001 Answer: Comment: Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0. Question 12: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 129 customers surveyed, 21 ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Your Answer: Correct Answer: 0.079 < p < 0.247 Comment: Question 13: Score 0/2

Incorrect

It was found that in a sample of 370 single women, 25% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of single women who have received speeding tickets? Incorrect Your Answer: Correct Answer: 0.213 < p < 0.287 Comment: Question 14: Score 0/2 A recent study of 553 Internet users in Europe found that 19% of Internet users were seniors. What is the 99% confidence interval of the true proportion of seniors in Europe who use the Internet? Incorrect Your Answer: Correct 0.147< p < 0.233 Answer: Comment: For a 99% confidence interval we have Z = 2.5758 and p = 0.19 so the interval is:

0.147 < p < 0.233 Question 15: Score 0/2 A sample of 1,020 was used to estimate a proportion with 99% confidence. If p = 0.55, what was the amount of error? Incorrect Your Answer: Correct 0.0401 Answer: Comment:

where we have used

the fact that for a 99% confidence interval, Z = 2.5758 . Question 1: Score 0/2 In hypothesis testing, β is the probability of committing an error of Type II. The power of the test, 1 − β is then:

Incorrec t Your Answer: Correct Answer: Comment:


Question 2: Score 0/2 The Bata Shoe Museum states that 33% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.01 of the true proportion. How large a sample is necessary? Your Answer: Correct Answer: 11,965 Comment:

Incorrect


Comment:


The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 5: Score 0/2 A survey of 519 women shoppers found that 38% of them shop on impulse. What is the 98% confidence interval for the true proportion of women

.

shoppers who shop on impulse? Incorrect Your Answer: Correct 0.3304 < p < 0.4296 Answer: Comment: To find a 98% confidence interval not that we have Z = 2.3263, p = 0.38 and:

0.3304 < p < 0.4296 Question 6: Score 0/2 A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In Incorrect this problem, the researcher will commit a type I error if: Your Answer: Correct she concludes that the new diet is better when in fact the diets are equal in Answer: effectiveness. Comment: Her hypotheses are: Ho: New diet worse or equal to old Ha: New diet better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new diet is better when in fact the diets are equal in effectiveness." Question 7: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 40 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.22 (in millions) and the standard deviation is USD 0.17 (millions). Assuming normality, a 90% confidence interval is: Your

Incorrect

Answer: Correct (1.176 , 1.264) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:


Correct response


(0%) < p <

(0%)

0.4525±0.01

< p < 0.5475±0.01



.

For a 98% Confidence Interval we have z = 2.3263, so our interval is or Question 9: Score 0/2 It was found that in a sample of 350 grandparents, 10% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of grandparents who have received speeding tickets? Incorrect Your Answer:

Correct Answer: 0.074 < p < 0.126 Comment: Question 10: Score 0/2 A clerk researched the average number of years served by 47 different judges on her court. The average number of years served was 13.11 years with a standard deviation of 7.95 years. What is the 95% confidence Incorrect interval for the average number of years served by all such judges? Your Answer: Correct 10.837 < μ < 15.383 Answer: Comment: To find a 95% CI we need the value of Z such that 97.5% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find this value is 1.959964. The CI then is:

=(10.837, 15.383)


Incorrect

Your Answer: Correct -2.365±0.001 Answer: Comment: Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0. Question 12: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 129 customers surveyed, 21 ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Your Answer: Correct Answer: 0.079 < p < 0.247 Comment: Question 13: Score 0/2

Incorrect

It was found that in a sample of 370 single women, 25% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of single women who have received speeding tickets? Incorrect Your Answer: Correct Answer: 0.213 < p < 0.287 Comment: Question 14: Score 0/2 A recent study of 553 Internet users in Europe found that 19% of Internet users were seniors. What is the 99% confidence interval of the true proportion of seniors in Europe who use the Internet? Incorrect Your Answer: Correct 0.147< p < 0.233 Answer: Comment: For a 99% confidence interval we have Z = 2.5758 and p = 0.19 so the interval is:


where we have used

the fact that for a 99% confidence interval, Z = 2.5758 . Question 1: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 44 stocks listed in the NASDAQ. The mean value of the stocks are

USD 1.46 (in millions) and the standard deviation is USD 0.15 (millions). Incorrect Assuming normality, a 90% confidence interval is: Your Answer: Correct (1.423 , 1.497) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Question 2: Score 0/2

11 squirrels were found to have an average weight of 400 grams with a sample standard deviation of 6.05. Find the 95% confidence interval of the Incorrect true mean weight (assume the t-student distribution). Your Answer: Correct (395.94,404.06) Answer: Comment: Use the formula . With degrees of freedom = 10 we have t0.025 = 2.228 so our confidence interval is Question 3: Score 0/2 A random sample of 591 printers discovered that 55 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses. Your Answer: Correct Answer: 0.062 < p < 0.124 Comment:

Incorrect

Question 4: Score 0/2 A sample of 1,280 was used to estimate a proportion with 90% confidence. If p = 0.42, what was the amount of error? Incorrect Your

Answer: Correct 0.0227 Answer: Comment:

where we have used

the fact that for a 90% confidence interval, Z = 1.6449 .

Question 5: Score 0/2 A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 13 at 15oC and a second random sample of size 3 kept at 18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be Your Answer: Correct Answer: 14 Comment:

Incorrect


Rat:

1

2

3

4

5

6


11.1 12.1 9.3 6.6 9.6 8.2



Estimate

-8.1167

Std Errorargro

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

Incorrect

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats. Question 7: Score 0/2 A previous analysis of celery stalks showed that the the standard deviation of their lengths is 7 millimeters. A packer wishes to find the 92% confidence interval for the average length of celery stalks. How many celery Incorrect stalks must be measured to be accurate within ±7 millimeters? Your Answer: Correct 3 Answer: Comment: The z value corresponding to this confidence level is 1.7507. We need n such that Solving: Question 8: Score 0/2 An agronomist measured the height of 133 Corn plants. The mean height was 239 cm and the standard deviation was 13 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.1272±0.01 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 9: Score 0/2 A cooking school believes that 54% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05

Incorrect

with 95% confidence? Your Answer: Correct 381 Answer: Comment: Use the fact that

. For a 95% confidence level Z = 1.959964,

so in this case: , solving:

Question 10: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 9 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided Incorrect alternative was done and the resulting p-value was 0.013. This means: Your Answer: Correct There was only a 1.3% chance of observing an increase greater than 9 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true. Question 11: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 68% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.01. How many claims should be included in the study? Your Answer: Correct 2,176 Answer: Comment: Since we can solve for n to get:

Incorrect

Question 12: Score 0/2 During the pre-flight check, Pilot Smith discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Smith decides to check the fuel level by hand, it will delay the flight by 30 minutes. If Smith decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect

i) the appropriate null hypothesis? and; ii) a type I error? Your Answer: Correct Null Hypothesis: assume that the warning can be ignored. Type I error: decide to Answer: check the fuel by hand when there is in fact enough fuel. Comment: Question 13: Score 0/2 A recent poll of 1,110 people who work indoors found that 190 of them smoke. If the researchers want to be 98% confident of their results to within 0.03, how large a sample is necessary? Your Answer: Correct Answer: 853 Comment:

Incorrect

Question 14: Score 0/2 In hypothesis testing, β is the probability of committing an error of Type II. The power of the test, 1 − β is then: Your Answer: Correct Answer: Comment:

Incorrect


Question 15: Score 0/2 A recent poll of 1,010 people who work indoors found that 130 of them smoke. If the researchers want to be 98% confident of their results to within

0.03, how large a sample is necessary? Incorrect Your Answer: Correct Answer: 674 Comment: Question 1: Score 0/2 A recent study of 530 Internet users in Europe found that 33% of Internet users were unsupervised children. What is the 98% confidence interval of the true proportion of unsupervised children in Europe who use the Internet? Incorrect Your Answer: Correct 0.2825< p < 0.3775 Answer: Comment: For a 98% confidence interval we have Z = 2.3263 and p = 0.33 so the interval is:

0.2825 < p < 0.3775 Question 2: Score 0/2 The College of Podiatrists states that 75% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.05 of the true proportion. How large a sample is necessary? Your Answer: Correct Answer: 405 Comment:

Incorrect

Question 3: Score 0/2 Your response A random sample of 511 women found that 14% were going to vote for a certain candidate. Find the 95% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

Correct response A random sample of 511 women found that 14% were going to vote for a certain candidate. Find the 95% Incorrect limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

(0%) < p <

(0%)

0.1099±0.01

< p < 0.1701±0.01

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0% Comment: To find a 95% confidence interval not that we have Z = 1.96, p = 0.14 and:


where we have used

the fact that for a 90% confidence interval, Z = 1.6449 . Question 5: Score 0/2 An agronomist measured the height of 110 Canola plants. The mean height was 220 cm and the standard deviation was 16 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.5255±0.01 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result!


A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. She measures one egg from each nest. The data are summarized below.

Incorrect

A Type I (false positive) error would occur if: Your Answer: Correct We conclude that larger nests had larger eggs (on average) when in fact there is no Answer: difference in the mean. Comment: Type I error occurs when a true null hypothesis is rejected in favour of the alternate hypothesis. In this case, Type I error means that we reject the hypothesis that larger nests have the same size of eggs as smaller nests.


To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 300 cases were studied. The results were: True Status Innocent Guilty Examiner's Innocent Decision Guilty

15

24

135

126

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as: Your Answer: Correct 0.9 Answer: Comment Type I error is when a true null hypothesis is rejected. In this case, this is the probability : of the Examiner judging an innocent suspect to be guilty. The probability is:


Rat:

1

2

3

4

5

6


11.1 12.1 9.3 6.6 9.6 8.2



Incorrect

Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats. Question 9: Score 0/2 The College of Podiatrists states that 39% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.03 of the true proportion. How large a sample is necessary? Your Answer: Correct Answer: 1,430 Comment:

Incorrect

Question 10: Score 0/2 A cooking school believes that 47% of applicants to that school have parents who were alumni. How large a sample is needed to estimate the true proportion of students who have parents who were alumni to within 0.03 with Incorrect 95% confidence? Your Answer: Correct 1,063 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

Question 11: Score 0/2 To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 200 cases were studied. The results were: True Status Innocent Guilty Examiner's Innocent Decision Guilty

10

30

90

70

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as: Your Answer: Correct 0.3 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is . Question 12: Score 0/2 In hypothesis testing, β is the probability of committing an error of Type II. The power of the test, 1 − β is then: Your Answer: Correct Answer: Comment:

Incorrect



A report states that 42% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.03 with 90% confidence? Your Answer: Correct 732 Answer: Comment: For a 90% confidence level we have Z = 1.6449. Use the fact that

Incorrect

by substituting in the known values and solving for n :

,

Question 14: Score 0/2 An agronomist measured the height of 117 Soybean plants. The mean height was 233 cm and the standard deviation was 12 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.1094 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 15: Score 0/2 A quality control engineer wants to estimate the proportion of defective parts that are being manufactured by his company to within 2.5%. A sample of 550 components showed that 15 were defective. How large a sample is needed to estimate the true proportion of defective parts with 99% confidence? Your Answer: Correct Answer: 282 Comment: Use the fact that

Incorrect

Question 1: Score 0/2 A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher use in Incorrect her t-test critical value? Your Answer: Correct 12 Answer: Comment: The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 13 - 1 = 12 Question 2: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 90% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.014. How many claims should be included in the study? Your Answer: Correct 460 Answer: Comment: Since we can solve for n to

Incorrect

get:

Question 3: Score 0/2 To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 300 cases were studied. The results were:

Incorrect

True Status Innocent Guilty Examiner's Innocent Decision Guilty

20

28

130

122


Question 4: Score 0/2 A pharmacist is planning to estimate the mean level of a certain drug in a lab. The pharmacist wanted the estimate to be within 4 mg/dLi or less with 95% confidence. The pharmacist also believes that the standard deviation Incorrect of the drug level is probably about 48 mg/dLi. How large a sample should the pharmacist need to take? Your Answer: Correct Answer: 553 Comment: Use the equation: . For a 95% confidence interval we have c = 1.96, so:

Question 5: Score 0/2 In a sample of 660 mice, a biologist found that 87% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less.

Incorrect

Your Answer: Correct Answer: 0.842% < p < 0.893% Comment: Question 6: Score 0/2 During the pre-flight check, Pilot Jones discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Jones decides to check the fuel level by hand, it will delay the flight by 50 minutes. If Jones decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect

i) the appropriate null hypothesis? and; ii) a type I error? Your Answer: Correct Null Hypothesis: assume that the warning can be ignored. Type I error: decide to Answer: check the fuel by hand when there is in fact enough fuel. Comment:


Rat:

1

2

3

4

5

6


11.1 12.1 9.3 6.6 9.6 8.2



Incorrect

Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats. Question 8: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 92% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.012. How many claims should be included in the study? Your Answer: Correct 512 Answer: Comment: Since we can solve for n to

Incorrect

get: Question 9: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 109 customers surveyed, 40 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Your Answer: Correct Answer: 0.248 < p < 0.486 Comment:

Incorrect

Question 10: Score 0/2 In a sample of 600 mice, a biologist found that 71% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less. Your Answer: Correct Answer: 0.672% < p < 0.745% Comment:

Incorrect

Question 11: Score 0/2 A previous analysis of paper boxes showed that the the standard deviation of their lengths is 7 millimeters. A pallet manufacturer wishes to find the 96% confidence interval for the average length of paper boxes. How many paper Incorrect boxes must be measured to be accurate within ±4 millimeters? Your Answer: Correct 13 Answer: Comment: The z value corresponding to this confidence level is 2.0537. We need n such that Solving: Question 12: Score 0/2 A previous analysis of celery stalks showed that the the standard deviation of their lengths is 10 millimeters. A bag manufacturer wishes to find the 92% confidence interval for the average length of celery stalks. How many Incorrect celery stalks must be measured to be accurate within ±3 millimeters? Your Answer: Correct 34 Answer: Comment: The z value corresponding to this confidence level is 1.7507. We need n such that Solving: Question 13: Score 0/2 A random sample of 506 seniors found that 23% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. Incorrect Your Answer: Correct Answer: 0.1933 < p < 0.2667 Comment: To find a 95% confidence interval not that we have Z = 1.96, p = 0.23 and:

0.1933 < p < 0.2667 Question 14: Score 0/2 An agronomist measured the height of 110 Soybean plants. The mean height was 242 cm and the standard deviation was 18 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.7162 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result!


13

23

87

77

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as: Your Answer: Correct 0.23 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the

given numbers, this is . Question 1: Score 0/2 A clerk researched the average number of years served by 50 different judges on his court. The average number of years served was 13.98 years with a standard deviation of 7.77 years. What is the 96% confidence Incorrect interval for the average number of years served by all such judges? Your Answer: Correct 11.723 < μ < 16.237 Answer: Comment: To find a 96% CI we need the value of Z such that 98% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find this value is 2.053749. The CI then is:

=(11.723, 16.237)

Question 2: Score 0/2 Your response A sample of 360 racing cars showed that 86 cars cost over $300,000. What is the 98% confidence interval of the true proportion of cars costing over $300,000 (3 decimals)?

Correct response A sample of 360 racing cars showed that 86 cars cost over $300,000. What is the 98% confidence interval of the true proportion of cars costing over $300,000 (3 decimals)? Incorrect

(0%) < p <

(0%)

0.1866±0.01

< p < 0.2912±0.01



.

For a 98% Confidence Interval we have z = 2.3263, so our interval is or Question 3: Score 0/2 A pharmacist is planning to estimate the mean level of a certain drug in a lab. The pharmacist wanted the estimate to be within 7 mg/dLi or less with 95% confidence. The pharmacist also believes that the standard deviation Incorrect of the drug level is probably about 30 mg/dLi. How large a sample should the pharmacist need to take? Your Answer: Correct Answer: 71 Comment: Use the equation: . For a 95% confidence interval we have c = 1.96, so:

Question 4: Score 0/2 A shift manager wants to estimate the proportion of defective parts that are being manufactured by her company to within 2.5%. A sample of 500 components showed that 16 were defective. How large a sample is needed to estimate the true proportion of defective parts with 99% confidence? Your Answer: Correct Answer: 329 Comment: Use the fact that


Incorrect

In a study of stock options, a sample of 198 stock options were observed and 54 were discovered to have a final negative payoff. Construct a 98% confidence interval for the relative frequency of those stock options with Incorrect negative payoff. Your Answer: Correct (0.1993 , 0.3467) Answer: Comment: First . Then for a 98% confidence interval we have Z = 2.3263 so the interval is:

Question 6: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 62% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.013. How many claims should be included in the study? Your Answer: Correct 1,395 Answer: Comment: Since we can solve for n to

Incorrect

get:


Incorrect Your Answer: Correct -2.365±0.001 Answer: Comment: Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0.

Question 8: Score 0/2 A report states that 84% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.01 with 98% confidence? Your Answer: Correct 7,273 Answer: Comment: For a 98% confidence level we have Z = 2.3263. Use the fact that

Incorrect


,


The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 10: Score 0/2

.

9 squirrels were found to have an average weight of 380 grams with a sample standard deviation of 4.15. Find the 95% confidence interval of the true Incorrect mean weight (assume the t-student distribution). Your Answer: Correct (376.81,383.19) Answer: Comment: Use the formula . With degrees of freedom = 8 we have t0.025 = 2.306 so our confidence interval is Question 11: Score 0/2 A cooking school believes that 77% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.01 Incorrect with 95% confidence? (4 decimal accuracy) Your Answer: Correct 6,803±1 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

You need to round off to the integer since you cannot take a fraction of a student.

Question 12: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 14 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.088. This means: Your Answer: Correct There was only a 8.8% chance of observing an increase greater than 14 kg (assuming

Answer: Comment:

the null hypothesis was true). The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.


Incorrect



Incorrect

Question 15: Score 0/2 In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Your Answer: Correct Answer: 99% Comment:

Incorrect

Question 1: Score 0/2 A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 13 at 15oC and a second random sample of size 10 kept at 18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be Your Answer: Correct Answer: 21

Incorrect

Comment: Question 2: Score 0/2 A trade school believes that 25% of applicants to that school have parents who were alumni. How large a sample is needed to estimate the true proportion of students who have parents who were alumni to within 0.03 with Incorrect 95% confidence? Your Answer: Correct 800 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

Question 3: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 12 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.087. This means: Your Answer: Correct There was only a 8.7% chance of observing an increase greater than 12 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.

Question 4: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 6.7 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.084. This means: Your Answer: Correct There was only a 8.4% chance of observing an increase greater than 6.7 kg (assuming

Answer: Comment:

the null hypothesis was true). The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals (units) while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more different mean weight, so more weight gained) occurrence, given that the null hypothesis is true.

Question 5: Score 0/2 In a sample of 510 mice, a biologist found that 82% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less. Your Answer: Correct Answer: 0.784% < p < 0.85% Comment:

Incorrect

Question 6: Score 0/2 The College of Podiatrists states that 59% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.05 of the true proportion. How large a sample is necessary? Your Answer: Correct Answer: 523 Comment:

Incorrect


11 squirrels were found to have an average weight of 420 grams Incorrect with a sample standard deviation of 4.3. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer: Correct (417.11,422.89) Answer: Comment: Use the formula . With degrees of freedom = 10 we have t0.025 = 2.228 so our confidence interval is Question 8: Score 0/2

A survey of 565 men shoppers found that 41% of them shop on impulse. What is the 99% confidence interval for the true proportion of men shoppers who shop on impulse? Incorrect Your Answer: Correct 0.3567 < p < 0.4633 Answer: Comment: To find a 99% confidence interval not that we have Z = 2.5758, p = 0.41 and:

0.3567 < p < 0.4633 Question 9: Score 0/2 A random sample of 543 printers discovered that 54 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses. Your Answer: Correct Answer: 0.066 < p < 0.133 Comment:

Incorrect

Question 10: Score 0/2 A study of elephants wishes to determine the average weight of a certain subspecies of elephants. The standard deviation of the population is 2,074 kilograms. How many elephants need to be weighed so we can be 95% confident Incorrect to be accurate within 308 kilograms? Your Answer: Correct 174 Answer: Comment: For a 95% CI we need to find z such that 97.5% of the standard normal curve area lies to the left of z. This value is z = 1.96. Since the CI is

we need to find

n such that

. Solving:


18

27

132

123

Incorrect


Question 12: Score 0/2 During the pre-flight check, Pilot Singh discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Singh decides to check the fuel level by hand, it will delay the flight by 55 minutes. If Singh decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect

i) the appropriate null hypothesis? and; ii) a type I error? Your Answer:

Correct Answer: Comment:

Null Hypothesis: assume that the warning can be ignored. Type I error: decide to check the fuel by hand when there is in fact enough fuel.


11 squirrels were found to have an average weight of 480 grams with a sample standard deviation of 4.85. Find the 95% confidence interval of the Incorrect true mean weight (assume the t-student distribution). Your Answer: Correct (476.74,483.26) Answer: Comment: Use the formula . With degrees of freedom = 10 we have t0.025 = 2.228 so our confidence interval is Question 14: Score 0/2 In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Your Answer: Correct Answer: 99% Comment:

Incorrect


Correct response

Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 48% of these tests.

Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 48% of these tests.

Construct a 95% Conficence Interval (4 decimal accuracy). Hint: use the General

Construct a 95% Conficence Interval (4 decimal accuracy). Hint: use the General

Incorrect

confidence interval for p.

confidence interval for p.

(

( 0.4394±0.001 , 0.5206±0.001 )

(0%) ,

(0%) )

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0% Comment: The confidence interval would be:

= = (0.4394, 0.5206) Question 1: Score 0/2 A study of 93 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 11 m. Which of the following is the 99% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (179.06, 184.94) Answer: Comment: For a 99% CI we need to find the z value for the standard normal for which 99.5% of the graph area lies to the left. This is 2.5758 . The CI is:

Question 2: Score 0/2 A researcher is going to conduct an experiment in order to compare two treatments – a new treatment and an old treatment. The researcher would like to see whether there is sufficient evidence to say that the new treatment is better than the old treatment. In this problem, the researcher will commit a type I Incorrect error if: Your Answer: Correct she concludes that the new treatment is better when in fact the treatments are Answer: equal in effectiveness. Comment: Her hypotheses are: Ho: New treatment worse or equal to old Ha: New treatment better than old For a type I error, she needs to reject the null hypothesis when it is actually true.

So the answer is: "she concludes that the new treatment is better when in fact the treatments are equal in effectiveness." Question 3: Score 0/2 A college believes that 60% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05 with Incorrect 95% confidence? (4 decimal accuracy) Your Answer: Correct 369±1 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:



Incorrect


Question 5: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only vegetarian pizza. Out of 127 customers surveyed, 23 ordered vegetarian pizza. What is the 99% confidence interval of the true proportion of

Incorrect

customers who order only vegetarian pizza? Your Answer: Correct Answer: 0.093 < p < 0.269 Comment:


Incorrect


Incorrect


11 squirrels were found to have an average weight of 310 grams with a sample standard deviation of 5.85. Find the 95% confidence interval of the Incorrect true mean weight (assume the t-student distribution). Your Answer: Correct (306.07,313.93) Answer: Comment: Use the formula . With degrees of freedom = 10 we have t0.025 = 2.228 so our confidence interval is Question 9: Score 0/2

In an stock portfolio selection process, a financial consultant observed the value of 47 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.17 (in millions) and the standard deviation is USD 0.15 (millions). Incorrect Assuming normality, a 90% confidence interval is: Your Answer: Correct (1.134 , 1.206) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Question 10: Score 0/2 During the pre-flight check, Pilot Jones discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Jones decides to check the fuel level by hand, it will delay the flight by 50 minutes. If Jones decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect



Rat:

1

2

3

4

5

6

Incorrect


11.1 12.1 9.3 6.6 9.6 8.2



Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats. Question 12: Score 0/2 A clerk researched the average number of years served by 49 different judges on the District Court. The average number of years served was 13.26 years with a standard deviation of 7.95 years. What is the 96% confidence Incorrect interval for the average number of years served by all such judges? Your Answer: Correct 10.928 < μ < 15.592 Answer: Comment: To find a 96% CI we need the value of Z such that 98% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find this value is 2.053749. The CI then is:

=(10.928, 15.592)

Question 13: Score 0/2 A cooking school believes that 17% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot Incorrect help the student financially to within 0.02 with 95% confidence? Your Answer: Correct 1,355 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

Question 14: Score 0/2 In a study of stock options, a sample of 110 stock options were observed and 59 were discovered to have a final negative payoff. Construct a 98% confidence interval for the relative frequency of those stock options with Incorrect negative payoff. Your Answer: Correct (0.4254 , 0.6466) Answer: Comment: First . Then for a 98% confidence interval we have Z = 2.3263 so the interval is:

Question 15: Score 0/2 An agronomist measured the height of 109 Corn plants. The mean height was 241 cm and the standard deviation was 14 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.341±0.01 Comment: The Standard Error is , where s is sample SD.

Notice that the sample mean does not matter for this result! Question 1: Score 0/2 A college believes that 71% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05 with Incorrect 95% confidence? Your Answer: Correct 316 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:


The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 3: Score 0/2 A college believes that 79% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true

.

proportion of students who have parents who have remarried to within 0.04 with Incorrect 95% confidence? (4 decimal accuracy) Your Answer: Correct 398±1 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

You need to round off to the integer since you cannot take a fraction of a student. Question 4: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 97% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.011. How many claims should be included in the study? Your Answer: Correct 241 Answer: Comment: Since we can solve for n to

Incorrect

get: Question 5: Score 0/2 A recent study of 595 Internet users in Europe found that 45% of Internet users were seniors. What is the 98% confidence interval of the true proportion of seniors in Europe who use the Internet? Incorrect Your Answer: Correct 0.4026< p < 0.4974 Answer: Comment: For a 98% confidence interval we have Z = 2.3263 and p = 0.45 so the interval is:

0.4026 < p < 0.4974 Question 6: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 116 customers surveyed, 21 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Your Answer: Correct Answer: 0.089 < p < 0.273 Comment:

Incorrect

Question 7: Score 0/2 A study of 90 bolts of carpet showed that their average length was 183 meters. The standard deviation of the population is 15 m. Which of the following is the 95% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (179.9, 186.1) Answer: Comment: For a 95% CI we need to find the z value for the standard normal for which 97.5% of the graph area lies to the left. This is 1.96 . The CI is:


Incorrect

Rat:

1

2

3

4

5

6


11.1 12.1 9.3 6.6 9.6 8.2



Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

-3.006 10 0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats. Question 9: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 76% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.012. How many claims should be included in the study? Your Answer: Correct 1,267 Answer: Comment: Since we can solve for n to get:

Incorrect

Question 10: Score 0/2 An agronomist measured the height of 138 Canola plants. The mean height was 235 cm and the standard deviation was 15 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.2769±0.01 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result!


Incorrect

Question 12: Score 0/2 A cooking school believes that 35% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot help the student financially to within 0.01 with 95% confidence? (4 decimal Incorrect accuracy) Your Answer: Correct 8,739±1 Answer: Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:


Question 13: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 13 kg bigger gain than the old feed. A two-sample t-test with the proper oneIncorrect sided alternative was done and the resulting p-value was 0.057. This means: Your Answer: Correct There was only a 5.7% chance of observing an increase greater than 13 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true. Question 14: Score 0/2 A study of rhinos wishes to determine the average weight of a certain family group of rhinos. The standard deviation of the population is 2,086 kilograms. How many rhinos need to be weighed so we can be 92% confident to Incorrect be accurate within 340 kilograms? Your Answer: Correct 115 Answer: Comment: For a 92% CI we need to find z such that 96% of the standard normal curve area lies to the left of z. This value is z = 1.7507. Since the CI is such that

we need to find n

. Solving:


11

29

89

71

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we

Incorrect

could estimate the probability of making a type II error as: Your Answer: Correct 0.29 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is . Question 1: Score 0/2 In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Your Answer: Correct Answer: 99% Comment:

Incorrect

Question 2: Score 0/2 During the pre-flight check, Pilot VanDerBoek discovers a minor problem a warning light indicates that the fuel guage may be broken. If VanDerBoek decides to check the fuel level by hand, it will delay the flight by 30 minutes. If VanDerBoek decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect



Incorrect

Your Answer: Correct -2.365±0.001 Answer: Comment: Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0.


15

20

135

130

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as: Your Answer: Correct 0.9 Answer: Comment Type I error is when a true null hypothesis is rejected. In this case, this is the probability : of the Examiner judging an innocent suspect to be guilty. The probability is:

Question 5: Score 0/2 In a study of stock options, a sample of 112 stock options were observed and 53 were discovered to have a final negative payoff. Construct a 95% confidence interval for the relative frequency of those stock options with negative payoff. Your Answer: Correct (0.3805 , 0.5655) Answer:

Incorrect

Comment:

First

. Then for a 95% confidence interval we have Z = 1.96 so

the interval is:

Question 6: Score 0/2 An agronomist measured the height of 136 Corn plants. The mean height was 210 cm and the standard deviation was 11 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 0.9432 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 7: Score 0/2 A report states that 82% of home owners had a gazebo. How large a sample is needed to estimate the true proportion of home owners who have gazebos to within 0.02 with 98% confidence? Your Answer: Correct 1,997 Answer: Comment: For a 98% confidence level we have Z = 2.3263. Use the fact that

Incorrect


,

Question 8: Score 0/2 A study of 95 bolts of carpet showed that their average length was 180 meters. The standard deviation of the population is 14 m. Which of the

following is the 99% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (176.3, 183.7) Answer: Comment: For a 99% CI we need to find the z value for the standard normal for which 99.5% of the graph area lies to the left. This is 2.5758 . The CI is:

Question 9: Score 0/2 An agronomist measured the height of 149 Canola plants. The mean height was 221 cm and the standard deviation was 19 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.5565 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 10: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 62% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.01. How many claims should be included in the study? Your Answer: Correct 2,356 Answer: Comment: Since we can solve for n to get: Question 11: Score 0/2 In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that

Incorrect

was used? Your Answer: Correct Answer: 99% Comment:

Incorrect

Question 12: Score 0/2 A random sample of 530 seniors found that 22% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. Incorrect Your Answer: Correct Answer: 0.1847 < p < 0.2553 Comment: To find a 95% confidence interval not that we have Z = 1.96, p = 0.22 and:

0.1847 < p < 0.2553 Question 13: Score 0/2 A recent study of 538 Internet users in Europe found that 40% of Internet users were unsupervised children. What is the 99% confidence interval of the true proportion of unsupervised children in Europe who use the Internet? Incorrect Your Answer: Correct 0.3456< p < 0.4544 Answer: Comment: For a 99% confidence interval we have Z = 2.5758 and p = 0.4 so the interval is:

0.3456 < p < 0.4544


To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 200 cases were studied. The results were: True Status Innocent Guilty Examiner's Innocent Decision Guilty

15

29

85

71

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as: Your Answer: Correct 0.29 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is .

Question 15: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided Incorrect alternative was done and the resulting p-value was 0.007. This means: Your Answer: Correct There was only a 0.7% chance of observing an increase greater than 8 kg (assuming Answer: the null hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true. Question 1: Score 0/2 11 squirrels were found to have an average weight of 340 grams with a sample standard deviation of 6.4. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer:

Incorrect


(335.7,344.3) Use the formula

. With degrees of freedom = 10 we have t0.025 =

2.228 so our confidence interval is Question 2: Score 0/2 A sample of 1,490 was used to estimate a proportion with 99% confidence. If p = 0.16, what was the amount of error? Incorrect Your Answer: Correct 0.0245 Answer: Comment:

where we have used

the fact that for a 99% confidence interval, Z = 2.5758 . Question 3: Score 0/2 During the pre-flight check, Pilot Smith discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Smith decides to check the fuel level by hand, it will delay the flight by 30 minutes. If Smith decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Incorrect

i) the appropriate null hypothesis? and; ii) a type I error? Your Answer: Correct Null Hypothesis: assume that the warning can be ignored. Type I error: decide to Answer: check the fuel by hand when there is in fact enough fuel. Comment: Question 4: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 105 customers surveyed, 38

ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Your Answer: Correct Answer: 0.241 < p < 0.483 Comment:

Incorrect


Incorrect


A Type I (false positive) error would occur if: Your Answer: Correct We conclude that larger nests had larger eggs (on average) when in fact there is no Answer: difference in the mean. Comment: Type I error occurs when a true null hypothesis is rejected in favour of the alternate hypothesis. In this case, Type I error means that we reject the hypothesis that larger nests have the same size of eggs as smaller nests.

Question 7: Score 0/2 A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value?

Incorrect

Your Answer: Correct Answer: Comment:

12 The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 13 - 1 = 12


Correct response


(0%) < p <

(0%)

0.4624±0.01

< p < 0.5558±0.01



.

For a 95% Confidence Interval we have z = 1.96, so our interval is or


The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 10: Score 0/2 A random sample of 554 printers discovered that 63 of them were being used in small businesses . Find the 99% limit for the population proportion

.

of printers that are used in small businesses. Your Answer: Correct Answer: 0.079 < p < 0.148 Comment:

Incorrect


18

25

82

75

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as: Your Answer: Correct 0.25 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is . Question 12: Score 0/2 A study of 92 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 15 m. Which of the following is the 99% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (177.97, 186.03) Answer: Comment: For a 99% CI we need to find the z value for the standard normal for which 99.5% of the graph area lies to the left. This is 2.5758 . The CI is:

Question 13: Score 0/2 An agronomist measured the height of 111 Canola plants. The mean height was 218 cm and the standard deviation was 18 cm. Calculate the standard error of the mean. (3 decimal accuracy) Incorrect Your Answer: Correct Answer: 1.7085±0.01 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 14: Score 0/2 In a study of stock options, a sample of 179 stock options were observed and 50 were discovered to have a final negative payoff. Construct a 95% confidence interval for the relative frequency of those stock options with Incorrect negative payoff. Your Answer: Correct (0.2133 , 0.3447) Answer: Comment: First . Then for a 95% confidence interval we have Z = 1.96 so the interval is:

Question 15: Score 0/2 A public defender researched the average number of years served by 46 different judges on the District Court. The average number of years served was 13.85 years with a standard deviation of 7.33 years. What is the 95% Incorrect confidence interval for the average number of years served by all such judges? Your Answer: Correct 11.732 < μ < 15.968 Answer: Comment: To find a 95% CI we need the value of Z such that 97.5% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find this

value is 1.959964. The CI then is:

=(11.732, 15.968)

Question 1: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 119 customers surveyed, 39 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Your Answer: Correct Answer: 0.217 < p < 0.439 Comment:

Incorrect


Correct response

Researchers tested patients fitted with a cardiac pacemaker to see if use of a cellular telephone interferes with the operation of the device. There were 518 tests conducted for one type of cellular telephone; interference with the device was found in 49% of these tests.

Researchers tested patients fitted with a cardiac pacemaker to see if use of a cellular telephone interferes with the operation of the device. There were 518 tests conducted for one type of cellular telephone; interference with the device was found in 49% of these tests. Incorrect

Construct a 98% Conficence Interval (4 decimal accuracy). Hint: use the General confidence interval for p.

Construct a 98% Conficence Interval (4 decimal accuracy). Hint: use the General confidence interval for p.

(

( 0.4389±0.001 , 0.5411±0.001 )

(0%) ,

(0%) )

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0% Comment: The confidence interval would be:

= = (0.4389, 0.5411) Question 3: Score 0/2

A researcher is going to conduct an experiment in order to compare two treatments – a new treatment and an old treatment. The researcher would like to see whether there is sufficient evidence to say that the new treatment is better than the old treatment. In this problem, the researcher will commit a type I Incorrect error if: Your Answer: Correct she concludes that the new treatment is better when in fact the treatments are Answer: equal in effectiveness. Comment: Her hypotheses are: Ho: New treatment worse or equal to old Ha: New treatment better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new treatment is better when in fact the treatments are equal in effectiveness." Question 4: Score 0/2 Your response

Correct response

A random sample of 544 men found that 35% were going to vote for a certain candidate. Find the 98% limit for the population proportion of men who will vote for that candidate. (3 decimal accuracy.)

A random sample of 544 men found that 35% were going to vote for a certain candidate. Find the 98% limit for the population proportion of men who will vote for that candidate. (3 decimal accuracy.)

(0%) < p <

(0%)

0.3024±0.01

< p < 0.3976±0.01


0.3024 < p < 0.3976

Incorrect


Incorrect


Incorrect


Rat:

1

2

3

4

5

6


11.1 12.1 9.3 6.6 9.6 8.2



Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

-3.006 10 0.0132

Incorrect

Upper 95%

-2.1003

(Assuming equal variances.) Which of the following is correct? Your Answer: Correct The confidence interval does not include 0. Hence, there is evidence that the mean Answer: number of tremors for all potential rats in the poisoned group is larger than that in the control group. Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats.


19

22

81

78

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as: Your Answer: Correct 0.22 Answer: Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is . Question 9: Score 0/2 Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 600 tests conducted for one type of cellular telephone; interference with the

Incorrect

device was found in 31% of these tests. Which of the following is a 98% Confidence Interval? Hint: use the General confidence interval for p. Your Answer: Correct Answer: (0.2661, 0.3539) Comment: The confidence interval would be:

= = (0.2661, 0.3539)


The null and alternate hypothesis of interest is: Your Answer: Correct H : μL = μS; A : μL > μS Answer: Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or . The null hypothesis is that the egg size is the same no matter which nest the egg came from from, Question 11: Score 0/2 A study of 90 bolts of carpet showed that their average length was 180 meters. The standard deviation of the population is 10 m. Which of the

.

following is the 97% confidence interval for the mean length per bolt of carpet? Incorrect Your Answer: Correct (177.71, 182.29) Answer: Comment: For a 97% CI we need to find the z value for the standard normal for which 98.5% of the graph area lies to the left. This is 2.1701 . The CI is:

Question 12: Score 0/2 The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 111 customers surveyed, 28 ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Your Answer: Correct Answer: 0.146 < p < 0.358 Comment:

Incorrect


Incorrect

Question 14: Score 0/2 An agronomist measured the height of 146 Soybean plants. The mean height was 247 cm and the standard deviation was 19 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.5725 Comment: The Standard Error is , where s is sample SD. Notice that the sample mean does not matter for this result! Question 15: Score 0/2

A cooking school believes that 28% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot Incorrect help the student financially to within 0.01 with 95% confidence? Your Answer: Correct Answer: 7,744 Comment: Use the fact that . For a 95% confidence level Z = 1.959964, so in this case: , solving:

STAT 202 1121 Introductory Statistics for Scientists : Quiz 10 Welcome Adam Joseph Amador View Details

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Feedback: Details Report [PRINT] STAT 202 1121 Introductory Statistics for Scientists, Quiz 10 Adam Joseph Amador, 3/27/12 at 11:37 AM

Question 1: Score 0/2 A study of 95 bolts of carpet showed that their average length was 185 meters. The standard deviation of the population is 10 m. Which of the following is the 97% confidence interval for the mean length per bolt of carpet? Your Answer: Correct Answer: Comment:

Incorrect

(182.77, 187.23) For a 97% CI we need to find the z value for the standard normal for which 98.5% of the graph area lies to the left. This is 2.1701 . The CI is:


Comment:

Question 3: Score 0/2 A cooking school believes that 81% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.01 with 95% confidence?

Incorrect

Your Answer: Correct 5,912 Answer: Comment: Use the fact that

. For a 95% confidence level Z = 1.959964, so in this

case: , solving:

Question 4: Score 0/2 A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. She measures one egg from each nest. The data are summarized below.

Incorrect

The null and alternate hypothesis of interest is: Your Answer: Correct Answer: Comment:

H : μ L = μ S ; A : μL > μ S The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg size in smaller nests, or

. The null hypothesis is that the egg size is the same no matter

which nest the egg came from from,

.

Question 5: Score 0/2 12 squirrels were found to have an average weight of 490 grams with a sample standard deviation of 6. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer:

Incorrect



. With degrees of freedom = 11 we have t0.025 = 2.201 so our

confidence interval is

Question 6: Score 0/2 Researchers tested patients fitted with a cardiac pacemaker to see if use of a cellular telephone interferes with the operation of the device. There were 549 tests conducted for one type of cellular telephone; interference with the device was found in 39% of these tests. Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect

Your Answer: Correct Answer: (0.3364, 0.4436) Comment: The confidence interval would be:

= =

(0.3364, 0.4436)

Question 7: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 43 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.6 (in millions) and the standard deviation is USD 0.2 (millions). Assuming normality, a 90% confidence Incorrect interval is: Your Answer: Correct Answer: (1.55 , 1.65) Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Question 8: Score 0/2 A retailer wants to estimate with 99% confidence the number of people who buy at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the true proportion. The minimum sample size necessary would be 1,100. Your Answer: Correct Answer: False Comment:

Incorrect

Question 9: Score 0/2 The critical value for a left-tailed t-test for dependent samples when the degrees of freedom = 7 and α = 0.025 is : (4 decimals) Your

Incorrect

Answer: Correct -2.365±0.001 Answer: Comment: Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0.


Incorrect

Question 11: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 50 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.73 (in millions) and the standard deviation is USD 0.2 (millions). Assuming normality, a 90% confidence interval is: Your Answer: Correct (1.683 , 1.777) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Incorrect

Question 12: Score 0/2 A report states that 18% of home owners had a gazebo. How large a sample is needed to estimate the true proportion of home owners who have gazebos to within 0.04 with 98% confidence? Your Answer: Correct Answer: Comment:

Incorrect

499 For a 98% confidence level we have Z = 2.3263. Use the fact that

by

substituting in the known values and solving for n : ,

Question 13: Score 0/2 A recent poll of 1,190 people who work indoors found that 200 of them smoke. If the researchers want to be 98% confident of their results to within 0.04, how large a sample is necessary? Your Answer: Correct Answer: 473 Comment:

Incorrect

Question 14: Score 0/2 A recent study of 518 Internet users in Europe found that 20% of Internet users were unsupervised children. What is the 99% confidence interval of the true proportion of unsupervised children in Europe who use the Internet? Your Answer: Correct Answer: 0.1547< p < 0.2453 Comment: For a 99% confidence interval we have Z

= 2.5758

and p

Incorrect

= 0.2 so the interval is:

0.1547 < p < 0.2453


Incorrect Your Answer: Correct 10 Answer: Comment: The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 11 - 1 = 10


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A study of 95 bolts of carpet showed that their average length was 185 meters. The standard deviation of the population is 10 m. Which of the following is the 97% confidence interval for the mean length per bolt of carpet? Your Answer: Correct Answer: Comment:

Incorrect



Comment:


Incorrect



case: , solving:


Incorrect





.


Incorrect






Incorrect


= =

(0.3364, 0.4436)



Incorrect


Incorrect



Incorrect


Incorrect


Incorrect


by



Incorrect


= 2.5758

and p

Incorrect


0.1547 < p < 0.2453




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Incorrect



Comment:


Incorrect



case: , solving:


Incorrect





.


Incorrect






Incorrect


= =

(0.3364, 0.4436)



Incorrect


Incorrect



Incorrect


Incorrect


Incorrect

499 For a 98% confidence level we have Z = 2.3263. Use the fact that by substituting in the known values and solving for n :

,

Question 13: Score 0/2 A recent poll of 1,190 people who work indoors found that 200 of them smoke. If the researchers want to be 98% confident of their results to within 0.04, how large a sample is necessary? Your Answer:

Incorrect

Correct Answer: 473 Comment:


= 2.5758

and p

Incorrect


0.1547 < p < 0.2453




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Incorrect



Comment:


Incorrect



case: , solving:


Incorrect





.


Incorrect






Incorrect


= =

(0.3364, 0.4436)



Incorrect


Incorrect



Incorrect


Incorrect


Incorrect


by



Incorrect


= 2.5758

and p

Incorrect


0.1547 < p < 0.2453




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Incorrect



Comment:


Incorrect



case: , solving:


Incorrect





.


Incorrect






Incorrect


= =

(0.3364, 0.4436)



Incorrect


Incorrect



Incorrect


Incorrect


Incorrect


by



Incorrect


= 2.5758

and p

Incorrect


0.1547 < p < 0.2453




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Incorrect



Comment:


Incorrect



case: , solving:


Incorrect





.


Incorrect






Incorrect


= =

(0.3364, 0.4436)



Incorrect


Incorrect



Incorrect


Incorrect


Incorrect


by



Incorrect


= 2.5758

and p

Incorrect


0.1547 < p < 0.2453




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A previous analysis of celery stalks showed that the the standard deviation of their lengths is 8 millimeters. A pallet manufacturer wishes to find the 91% confidence interval for the average length of celery stalks. How many celery stalks must be measured to be accurate within ±7 millimeters? Your Answer: Correct 4 Answer: Comment: The z value corresponding to this confidence level is 1.6954. We need n such that

Incorrect

Solving:

Question 2: Score 0/2 A recent poll of 1,230 people who work indoors found that 160 of them smoke. If the researchers want to be 98% confident of their results to within 0.05, how large a sample Your Answer: Correct Answer: 245 Comment:



Question 4: Score 0/2 An agronomist measured the height of 139 Corn plants. The mean height was 228 cm and the standard deviation was 19 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.6116 Comment: The Standard Error is

, where s is sample SD.

Notice that the sample mean does not matter for this result!

Question 5: Score 0/2 Your response A random sample of 514 women found that 10% were going to vote for a certain candidate. Find the 99% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

Correct response A random sample of 514 women found that 10% were going to vote for a certain candidate. Find the 99% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

Incorrect

(0%)

(0%)

0.0659±0.01

< p < 0.1341±0.01


0.0659 < p < 0.1341

Question 6: Score 0/2 A survey of 535 men shoppers found that 42% of them shop on impulse. What is the 98% confidence interval for the true proportion of men shoppers who shop on impulse?

Incorrect

Your Answer: Correct Answer: 0.3704 < p < 0.4696 Comment:

To find a 98% confidence interval not that we have Z = 2.3263, p = 0.42 and:

0.3704 < p < 0.4696

Question 7: Score 0/2 9 squirrels were found to have an average weight of 290 grams with a sample standard deviation of 6.1. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer: Correct (285.31,294.69) Answer: Comment: Use the formula



Incorrect


Your response A random sample of 512 seniors found that 16% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. (3 decimal accuracy.) (0%)

(0%)

Correct response A random sample of 512 seniors found that 16% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. (3 decimal accuracy.)

0.1282±0.01

< p < 0.1918±0.01

Incorrect


0.1282 < p < 0.1918

Question 9: Score 0/2 A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the Incorrect researcher will commit a type I error if: Your Answer: Correct Answer: she concludes that the new diet is better when in fact the diets are equal in effectiveness. Comment: Her hypotheses are: Ho: New diet worse or equal to old Ha: New diet better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new diet is better when in fact the diets are equal in effectiveness."


Comment:

Question 11: Score 0/2 A recent poll of 1,030 people who work indoors found that 190 of them smoke. If the researchers want to be 98% confident of their results to within 0.03, how large a sample is necessary? Your Answer: Correct Answer: 905

Incorrect

Comment:



Question 13: Score 0/2 In a study of stock options, a sample of 180 stock options were observed and 53 were discovered to have a final negative payoff. Construct a 99% confidence interval for the relative frequency of those stock options with negative payoff. Your Answer: Correct Answer: Comment:

Incorrect

(0.2065 , 0.3815) First

. Then for a 99% confidence interval we have Z

= 2.5758 so the

interval is:


Incorrect

(0.3744 , 0.6096) First


= 2.5758 so the

interval is:

Question 15: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting Incorrect p-value was 0.089. This means: Your Answer: Correct There was only a 8.9% chance of observing an increase greater than 8 kg (assuming the null Answer: hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the

new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.


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Question 1: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 15 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting Incorrect p-value was 0.027. This means: Your Answer: Correct There was only a 2.7% chance of observing an increase greater than 15 kg (assuming the null Answer: hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.



Question 3: Score 0/2 A random sample of 594 printers discovered that 53 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses.

Incorrect


Question 4: Score 0/2 Researchers tested patients fitted with a remote blood glucose monitor to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 14% of these tests. Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect


= =

(0.1029, 0.1771)


Incorrect

(0.2493 , 0.3567) First


= 1.6449 so the

interval is:

Question 6: Score 0/2 Suppose a study is being planned to estimate the relative frequency of companies that are labeled as risky. What sample size is needed so that the standard error will be no larger than 0.018. Hint: Find p that maximizes the standard error.

Incorrect

Your Answer: Correct Answer: 771 Comment: Since

, SE is maximized by setting

. Do so and solve for n:

Question 7: Score 0/2 An agronomist measured the height of 108 Canola plants. The mean height was 245 cm and the standard deviation was 20 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.9245 Comment: The Standard Error is




Incorrect

Your Answer: Correct Answer: 1,736 Comment: Since





Question 10: Score 0/2 6 squirrels were found to have an average weight of 490 grams with a sample standard deviation of 5.85. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer: Correct (483.86,496.14) Answer:

Incorrect

Comment: Use the formula




13

28

137

122

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as: Your Answer: Correct 0.913 Answer: Comment: Type I error is when a true null hypothesis is rejected. In this case, this is the probability of the Examiner judging an innocent suspect to be guilty. The probability is:

Question 12: Score 0/2 A survey of 516 depressed shoppers found that 36% of them shop on impulse. What is the 99% confidence interval for the true proportion of depressed shoppers who shop on impulse? Your Answer: Correct Answer: 0.3056 < p < 0.4144 Comment:


0.3056 < p < 0.4144

Question 13: Score 0/2 A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient

Incorrect

evidence to say that the new diet is better than the old diet. In this problem, the researcher will Incorrect commit a type I error if: Your Answer: Correct Answer: she concludes that the new diet is better when in fact the diets are equal in effectiveness. Comment: Her hypotheses are: Ho: New diet worse or equal to old Ha: New diet better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new diet is better when in fact the diets are equal in effectiveness."

Question 14: Score 0/2 A report states that 83% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.03 with 95% confidence? Your Answer: Correct Answer: Comment:

Incorrect

602 For a 95% confidence level we have Z = 1.96. Use the fact that by substituting in the known values and solving for n : ,


Incorrect



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Question 3: Score 0/2 A random sample of 594 printers discovered that 53 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in

small businesses. Your Answer: Correct Answer: 0.059 < p < 0.119 Comment:

Incorrect

Question 4: Score 0/2 Researchers tested patients fitted with a remote blood glucose monitor to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 14% of these tests. Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect


= =

(0.1029, 0.1771)


Incorrect

(0.2493 , 0.3567) First


= 1.6449 so the

interval is:


Incorrect

Your Answer: Correct Answer: 771 Comment: Since



Question 7: Score 0/2 An agronomist measured the height of 108 Canola plants. The mean height was 245 cm and the standard deviation was 20 cm. Calculate the standard error of the mean. Incorrect Your Answer: Correct Answer: 1.9245 Comment: The Standard Error is




Incorrect

Your Answer: Correct Answer: 1,736 Comment: Since





Question 10: Score 0/2 6 squirrels were found to have an average weight of 490 grams with a sample standard deviation of 5.85. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your Answer: Correct (483.86,496.14) Answer: Comment: Use the formula


Incorrect



13

28

137

122

Incorrect

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as: Your Answer: Correct 0.913 Answer: Comment: Type I error is when a true null hypothesis is rejected. In this case, this is the probability of the Examiner judging an innocent suspect to be guilty. The probability is:

Question 12: Score 0/2 A survey of 516 depressed shoppers found that 36% of them shop on impulse. What is the 99% confidence interval for the true proportion of depressed shoppers who shop on impulse?

Incorrect



0.3056 < p < 0.4144

Question 13: Score 0/2 A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the Incorrect researcher will commit a type I error if: Your Answer: Correct Answer: she concludes that the new diet is better when in fact the diets are equal in effectiveness. Comment: Her hypotheses are: Ho: New diet worse or equal to old

Ha: New diet better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new diet is better when in fact the diets are equal in effectiveness."

Question 14: Score 0/2 A report states that 83% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.03 with 95% confidence? Your Answer: Correct Answer: Comment:

Incorrect


by



Incorrect



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STAT 202 1121 Introductory Statistics for Scientists, Quiz 10 Adam Joseph Amador, 3/27/12 at 11:55 AM


Incorrect

(181.86, 188.14) For a 98% CI we need to find the z value for the standard normal for which 99% of the graph area lies to the left. This is 2.3263 . The CI is:

Question 2: Score 0/2 A pharmacist is planning to estimate the mean level of a certain drug in a lab. The pharmacist wanted the estimate to be within 6 mg/dLi or less with 95% confidence. The pharmacist also believes that the standard deviation of the drug level is probably about 43 mg/dLi. How large a sample should the pharmacist need to take? Your Answer: Correct Answer: 197 Comment: Use the equation:

Incorrect

. For a 95% confidence interval we have c = 1.96, so:

Question 3: Score 0/2 A previous analysis of cam shafts showed that the the standard deviation of their lengths is 9 millimeters. A bag manufacturer wishes to find the 98% confidence interval for the average length of cam shafts. How many cam shafts must be measured to be accurate within ±5 millimeters? Your Answer: Correct 18 Answer: Comment: The z value corresponding to this confidence level is 2.3263. We need n such that

Solving:


Incorrect

The Pizza Shop wanted to determine what proportion of its customers ordered cheese pizza. Out of 134 customers surveyed, 21 ordered cheese pizza. What is 99% confidence interval of the true proportion of customers who order only cheese pizza? Your Answer: Correct Answer: 0.076 < p < 0.238 Comment:

only the Incorrect

Question 5: Score 0/2 A researcher is going to conduct an experiment in order to compare two drugs – a new drug and an old drug. The researcher would like to see whether there is sufficient evidence to say that the new drug is better than the old drug. In this problem, the Incorrect researcher will commit a type I error if: Your Answer: Correct Answer: she concludes that the new drug is better when in fact the drugs are equal in effectiveness. Comment: Her hypotheses are: Ho: New drug worse or equal to old Ha: New drug better than old For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is: "she concludes that the new drug is better when in fact the drugs are equal in effectiveness."


Question 7: Score 0/2 Researchers tested patients fitted with a cardiac pacemaker to see if use of a cellular telephone interferes with the operation of the device. There were 582 tests conducted for one type of cellular telephone; interference with the device was found in 38% of these tests. Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p. Your Answer: Correct Answer: (0.3282, 0.4318) Comment: The confidence interval would be:

= =

(0.3282, 0.4318)


Incorrect

In a sample of 640 mice, a biologist found that 64% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less.

Incorrect

Your Answer: Correct Answer: 0.603% < p < 0.677% Comment:



Question 10: Score 0/2 An agronomist measured the height of 144 Wheat plants. The mean height was 209 cm and the standard deviation was 14 cm. Calculate the standard error of the mean. (3 decimal accuracy)

Incorrect

Your Answer: Correct Answer: 1.1667±0.01 Comment: The Standard Error is



Question 11: Score 0/2 It was found that in a sample of 390 teenage boys, 90% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets? Incorrect Your Answer: Correct Answer: 0.875 < p < 0.925 Comment:

Question 12: Score 0/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 5 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting Incorrect p-value was 0.04. This means: Your Answer: Correct There was only a 4% chance of observing an increase greater than 5 kg (assuming the null hypothesis Answer: was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals (units) while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more different mean weight, so more weight gained) occurrence, given that the null hypothesis is true.


Comment:

Question 14: Score 0/2 In a study of human mortality rate, an Actuary estimated that in US and Canada, about 65% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.018. How many claims should be included in the study? Your Answer: Correct 703 Answer: Comment: Since

Incorrect

we can solve for n to

get:

Question 15: Score 0/2 In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Incorrect Your Answer: Correct Answer: 99% Comment:


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Question 1: Score 2/2 In a sample of 640 mice, a biologist found that 79% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less.

Correct

Your Answer: 0.757% < p < 0.82% Comment:

Question 2: Score 0/2 The College of Podiatrists states that 66% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.03 of the true proportion. How large a sample is necessary?

Incorrect

Your Answer: 1,483 Correct Answer: 1,349 Comment:

Question 3: Score 2/2 A previous analysis of cam shafts showed that the the standard deviation of their lengths is 8 millimeters. A bag manufacturer wishes to find the 97% confidence interval for the average length of cam shafts. How many cam shafts must be measured to be accurate within ±7 millimeters? Your 6 Answer: Comment: The z value corresponding to this confidence level is 2.1701. We need n such that

Solving:

Correct

Question 4: Score 2/2 An agronomist measured the height of 120 Wheat plants. The mean height was 228 cm and the standard deviation was 20 cm. Calculate the standard error of the mean. Correct Your Answer: 1.8257 Comment: The Standard Error is



Question 5: Score 2/2 In order to study the harmful effects of DDT poisoning, the pesticide was fed to 6 randomly chosen rats out of a group of 12 rats. The other 6 rats were used as the control group. The following data gives the measurements of the amount of tremor detected in the bodies of each rat after the experiment: The more tremor, the more harmful. Rat:

1

2

3

4

5

6


11.1 12.1 9.3

6.6 9.6 8.2

A computer analysis is done with the output below (the differences are computed as control poisoned) t-test

Difference

Estimate

-8.1167

Std Error

2.7002

Lower 95%

-14.1331

Upper 95%

-2.1003

t-test

Correct

DF Prob>|t|

-3.006 10

0.0132

(Assuming equal variances.) Which of the following is correct? Your Answer: Comment:

The confidence interval does not include 0. Hence, there is evidence that the mean number of tremors for all potential rats in the poisoned group is larger than that in the control group. The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats.

Question 6: Score 2/2 An agronomist measured the height of 121 Corn plants. The mean height was 210 cm and the standard deviation was 14 cm. Calculate the standard error of the mean. Correct Your Answer: 1.2727 Comment: The Standard Error is Notice that the sample mean does not matter for this result!



We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting Correct p-value was 0.006. This means: Your There was only a 0.6% chance of observing an increase greater than 8 kg (assuming the null Answer: hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.

Question 8: Score 2/2 We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8.4 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting Correct p-value was 0.067. This means: Your There was only a 6.7% chance of observing an increase greater than 8.4 kg (assuming the null Answer: hypothesis was true). Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals (units) while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more different mean weight, so more weight gained) occurrence, given that the null hypothesis is true.

Question 9: Score 2/2 During the pre-flight check, Pilot VanDerBoek discovers a minor problem - a warning light indicates that the fuel guage may be broken. If VanDerBoek decides to check the fuel level by hand, it will delay the flight by 40 minutes. If VanDerBoek decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be: Correct i) the appropriate null hypothesis? and; ii) a type I error? Your Null Hypothesis: assume that the warning can be ignored. Type I error: decide to check the fuel by Answer: hand when there is in fact enough fuel. Comment:

Question 10: Score 2/2 Your response Researchers tested patients fitted with a medical telemetry unit to see if use of a cellular telephone interferes with the operation of the device. There were 535 tests conducted for one type of cellular telephone; interference with the device was found in 37% of these tests. Construct a 99% Conficence Interval (4 decimal accuracy). Hint: use the General confidence interval for p. ( 0.316 (50%) , 0.424 (50%) ) Comment: The confidence interval would be:

=

Correct

=

(0.3162, 0.4238)

Question 11: Score 2/2 A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value? Your Answer: Comment:

Correct

12 The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 13 - 1 = 12

Question 12: Score 2/2 6 squirrels were found to have an average weight of 350 grams with a sample standard deviation of 4.95. Find the 95% confidence interval of the true mean weight (assume the t-student distribution). Your (344.8,355.2) Answer: Comment: Use the formula

Correct



Question 13: Score 0/2 In an stock portfolio selection process, a financial consultant observed the value of 40 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.06 (in millions) and the standard deviation is USD 0.15 (millions). Assuming normality, a 90% confidence interval is: Your Answer: (1.037, 1.099) Correct (1.021 , 1.099) Answer: Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:

Incorrect

Question 14: Score 0/2 A recent poll of 1,430 people who work indoors found that 160 of them smoke. If the researchers want to be 98% confident of their results to within 0.05, how large a sample is necessary?

Incorrect

Your Answer: 628 Correct Answer: 215 Comment:

Question 15: Score 2/2 An agronomist measured the height of 143 Wheat plants. The mean height was 218 cm and the standard deviation was 10 cm. Calculate the standard error of the mean. (3 decimal accuracy) Your Answer: 0.836

Correct

Comment: The Standard Error is Notice that the sample mean does not matter for this result!


Question 1

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