MATH 2565 3.0 (2016 winter) Assignment 3 (Due Date: March 15, 2016) Question 1: Carlinville, Carlinville, Illinois, Illinois, has just started an ambitious ambitious recycling program. Special trucks trucks collect recyclable products once a week, week, sorted into barrels of paper, glass, and plastic. plastic. The amount of glass recycled per week by a single household is normally distributed with mean 27 pounds pounds and standard standard deviation deviation 7 pounds. pounds. Suppose Suppose 12 househol households ds are randoml randomly y selected. a. What is the probabil probability ity that that the total glass glass collected collected for the 12 homes homes will will be less than 350 pounds? ¯ be the average Solution: Let T = T = the total glass collected for the 12 homes and let X be X ¯ N (27 glass collected for the 12 homes. T N (324, (324, 588) or X N (27,, 49/ 49/12).
∼ ∼
¯ < 29. P ( P (T < 350) = P = P ((X 29 .17) = P = P ((Z < (29. (29 .17
27)/ − 27)/
∼ ∼
49/ 49/12) = P = P ((Z < 1. 1 .07) = 0. 0.8577
b. If the total glass glass collect collected ed for the 12 homes homes is more more than 400 pounds, pounds, the recyc recycli ling ng plant will make a profit. Find a value of population mean such that the probability of making a profit is 0.10. ¯ ¯ > 33. Solution: T N (12 N (12µ, µ, 588) or X N ( N (µ, 49/ 49/12) P ( P (T > 400) = P ( P (X 33.33) =
∼
0.1 P ( P (Z < (33. (33.33 therefore, µ therefore, µ = = 30. 30.74.
⇒
∼ ∼
− µ)/
49/ 49/12) = 0. 0.9
⇒
(33.33 − µ)/ ⇒ (33.
49/ 49/12 = 1. 1.2816 or 1. 1.28;
Question Question 2: There are three factors that affect the width of a large-sample confidence interval for population proportion p: the confidence confidence level, level, the sample sample size, and the sample sample proportion p. pˆ. In each of the following following problems, determine determine whether the width of the resulting confidence interval for p for p increases or decreases. a. Co Confid nfidenc encee lev level el and p are pˆ are constant, and n increases. Solution: Solution: As n increases, the square root becomes smaller and the resulting width of CI decreases. b. Samp Sample le size size and and p are pˆ are constant, and confidence level decreases. Solution: Solution: As the confidence level decreases, decreases, α increases, the critical values decreases, and the resulting width decreases. c. Sample Sample size size and confiden confidence ce level level are constan constant, t, given given pˆ > 0. 0 .5 and p increases. pˆ increases. Solu Soluti tion: on: As pˆ increases and given pˆ > 0.5, the product p(1 pˆ(1 resulting width of CI decreases.
pˆ) decreases and the − p)
Question 3: The Australian dragon fly, the world’s fastest insect, has been clocked at 36 miles miles per p er hour over over short distances. distances. A biolog biologist ist captured captured 25 Austral Australian ian dragon flies and she recorded recorded the average average wingspan wingspan of these these captured captured dragon flies of 6.6cm. 6.6cm. The biologist biologist assumed that the wingspan is normally distributed with a standard deviation of 0.5cm. 1
a. Find a 98.3% confidence interval for the true mean wingspan for all dragon flies. b. Interpret the interval obtained in part (a) in the context of this question. c. If a 90% confidence interval for the true mean wingspan for all dragon flies is calculated, do you expect this interval be wider or narrower than the interval obtained in part (a)? Why? d. If the biologist assumed that the standard deviation of the wingspan of Australia dragon fly is 0.25cm and a 98.3% confidence interval for the true mean wingspan for all dragon flies is calculated, do you expect this interval be wider or narrower than the interval obtained in part (a)? Why? Solution: Let X be the wingspan of dragon fly. It is given that n = 25; x¯ = 6.6, σ = 0.5 and the wingspan is distributed as normal. a. A 98.3% 2confidence interval for the true mean wingspan for all dragon flies is 6.6 2.39 0.5 /25 (6.361;6.839).
⇒
±
b. Based on the given data, we are 98.3% certain that the mean wingspan for all dragon files is between 6.361cm and 6.839cm. c. Since the required table value will be smaller in the 90% CI than the 98.3% CI, therefore the interval will be narrower for the 90% CI. d. When σ = 0.25, the margin of error will be smaller than the margin of error for σ = 0.5, therefore the 98.3% CI obtained using σ = 0.25 will be narrower than the 98.3% CI obtained
Question 4: Consider a random sample of 11 observations, given in the following, from a normal population with hypothesized mean 57.71. 59.94 58.93 59.41 60.66 59.00 60.98 58.85 55.21 59.02 61.14 59.25 a. Compute the value of the test statistic for H 0 : µ = 57.71 versus H a : µ > 57.71 with α = 0.01. Draw a conclusion about the population mean. b. Write down the formula to get the p value associated with this hypothesis test. ¯
−√ µ −√ 57.71 Solution: (a)T = X , the observed test statistic t = 591..3082 = 3.3053 S/ n 6037/ 11 2.7638. There is evidence to suggest that the mean is greater than 57.71. (b) p = P (T 3.3053), T has t-distribution with 10 df. 0
≥ t
0.01,10
=
≥
Question 5: Kiln-dried solid grade A teak wood should have a moisture content of no more than 12%. A furniture company recently purchased a large shipment of this wood to construct dining room sets. Thirty-seven pieces of teak wood were randomly selected and carefully measured for moisture content. The sample mean moisture content was 12.3%. Assume the population standard deviation is 1.25%. 2
a. Is there any evidence to suggest that the true population mean moisture content is greater than 12%? use α = 0.05. Solution: H 0 : µ = 12% versus H a : µ > 12%. The observed TS = (12.3% 12%)/(1.25%/ 37) = 1.46 < z 0.05 = 1.645 FTRH 0 . There is no evidence to suggest that the population mean moisture content is greater than 12%.
−
√
b. Find the probability of a Type II error if the true population mean moisture content is 12.1%. ¯ −12% ¯ √ Solution: P ( 1.X 1.645) = 0.05 P (X 12.338%) = 0.05 β (12.1%) = 25%/ 37 ¯ −12.1% 12.1% ¯ < 12.338% µ = 12.1%) = P ( X √ < 12.338%−√ P (FTRH 0 µ = 12.1%) = P (X )=
|
P (Z < 1.158) = 87.7%
≥
⇒ |
3
≥
⇒
1.25%/
37
1.25%/
37
