Ex Estimation

Examples: Estimation Estimation Problems

CHAPTER 9 One- and Two-Sample Estimation Problems 4/252) An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean mean of all bulbs produced produced by this firm. Soln: Given: n = 30, σ = 40 hours, x = 780 hours 96% confidence interval From Table A.3, zα/2 = z.02 = 2.05 __ __  x - zα/2 σ/√n < µ <  x + zα/2 σ/√n __ __ 780 – 2.05(40)/√30 < µ < 780 + 2.05(40)/√30

765 < µ < 795 6/252) The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. a) Construct a 98% confidence interval interval for the mean height of all college students. b) What can we assert assert with 98% confidence confidence about the possible size size of our error if we estimate the mean height of all college students to be 174.5 centimeters? Soln: Given: n = 50, σ = 6.9 cm, x = 174.5 cm a) 98% confidence interval From Table A.3, zα/2 = z.01 = 2.33 __ __  x - zα/2 σ/√n < µ <  x + zα/2 σ/√n __ __ 174.5 – 2.33(6.9)/√50 < µ < 174.5 + 2.33(6.9)/√50

172.23 < µ < 176.77 ENGSTAT Notes Notes of AM Fillone Fillone


__ b) Error ≤ 2.33(6.9)/√50 ≤ 2.27 cm 8/253) How large a sample is needed in Exercise 4 if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Soln: Given: zα/2 = 2.05, σ = 40 hours, e = 10 hours n = (zα/2σ / e)

2

n = (2.05(40) / 10)

2

n ≅ 68 samples 10/253) An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 95% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds.

Soln: Given: zα/2 = z.025 = 1.96, σ = 40 seconds, e = 15 seconds seconds n = (zα/2σ / e)

2

n = (1.96(40) / 15) n ≅ 28 samples

ENGSTAT Notes Notes of AM Fillone Fillone

2


__ b) Error ≤ 2.33(6.9)/√50 ≤ 2.27 cm 8/253) How large a sample is needed in Exercise 4 if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Soln: Given: zα/2 = 2.05, σ = 40 hours, e = 10 hours n = (zα/2σ / e)

2

n = (2.05(40) / 10)

2

n ≅ 68 samples 10/253) An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 95% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds.

Soln: Given: zα/2 = z.025 = 1.96, σ = 40 seconds, e = 15 seconds seconds n = (zα/2σ / e)

2

n = (1.96(40) / 15) n ≅ 28 samples

ENGSTAT Notes Notes of AM Fillone Fillone

2


13/253) A machine is producing metal pieces that are cylindrical in shape. A sample of pieces is taken and the diameters are 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Find a 99% confidence interval for the mean diameter of pieces from this machine, assuming an approximate normal distribution. Soln: 1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03 x = ------------------------------------------------------------------------- 9  x = 1.006 9

s = Σ (xi - x) /(n-1) = 0.0048 / 8 : Therefore, s = 0.024 2

2

i=1

From Table A.4, tα/2 = t .005 = 3.355 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 1.006 – 3.355(.024)/√9 < µ < 1.006 + 3.355(.024)/√9 0.974 < µ < 1.037 16/253) A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school. Soln: Given: n = 12,  x = 79.3 words/minute, s = 7.8 words/minute @ 95% confidence interval From Table A.4, tα/2 = = t .025 = 2.201 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ ENGSTAT Notes Notes of AM Fillone Fillone

Examples: Estimation Problems

79.3 – 2.201(7.8)/√12 < µ < 79.3 + 2.201(7.8)/√12 74.344 < µ < 84.256 18/253) The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint: 3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6 3.7 2.8 4.4 4.0 5.2 3.0 4.8 Assuming that the measurements represent a random sample from a normal population, find the 99% tolerance limits that will contain 95% of the drying times. Soln: 15

x = Σ xi / n = 3.787  i=1

15

s = Σ (xi - x) /(n-1) : Therefore, s = 0.971 2

2

i=1

@ n = 15, (1- γ) = 0.99, (1- α) = 0.95 From Table A.7, k = 3.507  x ± ks 3.787 ± 3.507(0.971) 3.787 ± 3.405 Therefore, the tolerance limit is from 0.382 to 7.192 hours. 2/263) Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A has an average tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means. Soln: Given: Brand A: nA = 50, σA = 5.6 kg, xA = 78.3 kg Brand B: nB = 50, σB = 6.3 kg, xB = 87.2 kg @ 95% confidence interval From Table A.3, zα/2 = z.025 = 1.96 ___________ ___________ 2 2 2 2 ( xB- xA) - zα/2√σ A/nA+σ B/nB <µB-µA< ( xB- xA) + zα/2√σ A/nA+σ B/nB,

________________ 2 2 (87.2–78.3)–1.96√(5.6) /50 + (6.3) /50 < µB-µA < ENGSTAT Notes of AM Fillone


________________ 2 2 (87.2–78.3)+1.96√(5.6) /50 + (6.3) /50, 6.56 < µB - µA < 11.24

3/263) A study was made to determine if a certain metal treatment has any effect on the amount of metal removed in a pickling operation. A random sample of 100 pieces was immersed in a bath of 24 hours without treatment, yielding an average of 12.2 millimeters of metal removed and a sample standard deviation of 1.1 millimeters. A second sample of 200 pieces was exposed to the treatment followed by the 24-hour immersion in the bath, resulting in an average removal of 9.1 milliliters of metal with a sample standard deviation of 0.9 millimeter. Compute a 98% confidence interval estimate for the difference between the population means. Does the treatment appear to reduce the mean amount of metal removed?

Soln: Sample1 n1 = 100  x1 = 12.2 ml σ1 = 1.1 ml

Sample 2 n2 = 200  x2 = 9.1 ml σ2 = 0.9 ml

@ 98% confidence interval, From Table A.3, zα/2 = z.01 = 2.33 ___________ 2 2 ( x1- x2) - zα/2√σ 1/n1 + σ 2/n2 < µ1 - µ2 < ___________ 2 2 ( x1- x2) + zα/2√σ 1/n1 + σ 2/n2, __________________ 2 2 (12.2–9.1)–2.33 √(1.1) /100 + (0.9) /200 < µ1-µ2 < __________________ 2 2 ( 12.2–9.1)+2.33√(1.1) /100+(0.9) /200 2.804 < µ1-µ2 < 3.396

ENGSTAT Notes of AM Fillone


8/264) An experiment reported in Popular Science, in 1981, compared fuel economies for two types of similarly equipped diesel mini-trucks. Let us suppose that 12 Volkswagen and 10 Toyota trucks are used in 90-kilometer per hour steady-speed tests. If the 12 Volkswagen trucks average 16 kilometers per liter with a standard deviation of 1.0 kilometer per liter and the10 Toyota trucks average 11 kilometers per liter with a standard deviation of 0.8 kilometer per liter, construct a 90% confidence interval for the difference between the average kilometers per liter of these two mini-trucks. Assume that the distances per liter for each truck model are approximately normally distributed with equal variances.

Soln: Volkswagen n1 = 12  x1 = 16 km/l s1 = 1.0 km/l

Toyota n2 = 10  x2 = 11 km/l s2 = 0.8 km/l

@ 90% confidence interval for the difference µ2 - µ1 From Table A.4 , @ ν = 20, tα/2 = t.05 = 1.725 2

2

(12 –1)1 + (10-1)(0.8) 2 s p = --------------------------------- = 0.838 12 + 10 - 2 s p = 0.915 ________ (16 - 11) – 1.725(0.915)√1/12+1/10 < µ2 - µ1 < ________ (16 - 11) + 1.725(0.915)√1/12+1/10,



4.324 < µ2 - µ1 < 5.676

11/264) The government awarded grants to the agricultural departments of 9 universities to test the yield capabilities of two new varieties of wheat. Each variety was planted on plots of equal area at each university and the yields, in kilograms per plot, recorded as follows:

University Variety 1 2 3 4 5 6 1 38 23 35 41 44 29 2 45 25 31 38 50 33 7 2 -4 -3 6 4  d

7 8 9 37 31 38 36 40 43 -1 9 5

Soln: __________________

 n n 2  n Σ d i – ( Σ di )2 _______________ i=1 i=1   9(237)2 – (25)2 sd =  ------------------------- =  -------------------- = 4.5765 n(n-1) 9(9-1) √ √  d = (7 + 2 – 4 – 3 + 6 + 4 –1 + 9 + 5)/9 = 2.78 @ 95% confidence interval From Table A.4, ν = n –1 = 8, t α/2 = t.025 = 2.306 _ _ d - tα/2 (sd / √n ) < µD <  d + tα/2 (sd / √n ),  __ __ 2.78 – 2.306 (4.5765/ √9 ) < µD < 2.78 + 2.306 (4.5765/√9 ) -0.738 < µD < 6.298



3/270) In a random sample of 1000 homes in a certain city, it is found that 228 are heated by oil. Find the 99% confidence interval for the proportion of homes in this city that are heated by oil. Soln: Given:  p = 228/1000 = 0.228,  q = 772/1000 = 0.772

@99% Confidence interval From Table A.3, zα/2 = z.005 = 2.575 ______ ______ p - zα/2√ p q / n < p < p + zα/2√ p q /n  _______________ 0.228 – 2.575 √0.228(0.772)/1000 < p < _______________ 0.228 + 2.575 √0.228(0.772)/1000 0.1938 < p < 0.2622 5/270) A new rocket-launching system is being considered for deployment of small, short-range rockets. The existing system has p = 0.8 as the probability of a successful launch. A sample of 40 experimental launches is made with the new system and 34 are successful. (a) Construct a 95% confidence interval for p. (b) Would you conclude that the new system is better? Soln: Given:  p = 34/40 = 0.85,  q = 6/40 = 0.15

@95% Confidence interval From Table A.3, zα/2 = z.025 = 1.96 a) ______ ______  p - zα/2√ p q / n < p < p + zα/2√ p q /n ___________ ___________ ENGSTAT Notes of AM Fillone


0.85 – 1.96 √0.85(0.15)/40 < p < 0.85 + 1.96 √0.85(0.15)/40 0.739 < p < 0.961 b) No. Since the existing p = .80 is still within the range. 8/271) In the newspaper article referred to in Exercise 7, 32% of the 1600 adults polled said the U.S. space program should emphasize scientific exploration. How large a sample of adults is needed in the poll if one wishes to be 95% confident that the estimated percentage will be within 2% of the true percentage? Soln: Given :  p = 0.32,  q = 0.68, @ 95% Confidence interval From Table A.3, zα/2 = z.025 = 1.96 n = z α/2 p q/e 2 2 n = (1.96) (.32)(.68)/(.02) ≅ 2,090 2

2

12/271) A study is to be made to estimate the percentage of citizens in a town who favor having their water fluoridated. How large a sample is needed if one wishes to be at least 95% confident that our estimate is within 1% of the true percentage? Soln: Given: e = .01, @ 95% Confidence interval

From Table A.3, zα/2 = z.025 = 1.96 2

2

n = z α/2 / 4e 2 2 n = (1.96) /4(.01) ≅ 9,604 14/271) A study is to be made to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant. How large a sample is needed if one wished to be at least 95% confident that the estimate is within 0.04 of the true proportion of residents in this city and its suburbs that favor the construction of the nuclear power plant? Soln: ENGSTAT Notes of AM Fillone


Given: e = 0.04, @ 95% Confidence interval From Table A.3, zα/2 = z.025 = 1.96 2

2

n = z α/2 / 4e 2 2 n = (1.96) /4(0.04) ≅ 601 16/271) A cigarette-manufacturing firm claims that its brand A line of cigarettes outsells its brand B line by 8%. If it is found that 42 of 200 smokers prefer brand A and 18 of 150 smokers prefer brand B, compute a 94% confidence interval for the difference between the proportions of sales of the 2 brands and decide if the 8% difference is a valid claim. Soln: Given: Brand A:  pA = 42/200 = 0.21 qA = 158/200 = 0.79  Brand B:  pB = 18/150 = 0.12 qB = 132/150 = 0.88  @ 94% confidence interval Using Table A.3, zα/2 = z.03 = 1.88 __________________ a) ( pA- pB)- zα/2√ pA qA/nA + pB qB/nB


1/276) A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years, construct a 95% confidence 2 2 interval for σ and decide if the manufacturer’s claim that σ = 1 is valid. Assume the population of battery lives to be approximately normally distributed. Soln: 2 2 2 2 2 (n –1)s / χ α/2 < σ < (n - 1)s / χ 1-α/2, n

n

nΣ x i–( 2

i=1

Σ xi )2

i=1

2

5(48.26) – (15) 2 s = ------------------------------ = ------------------------ = 0.815 n (n-1) 5(5 –1) For a 95% confidence interval, α = .05. From Table A.5 with v = 4 degrees 2 2 of freedom, we find χ 0.025 = 11.143 and χ 0.975 = 0.484. 4(0.815)/11.143 < σ < 4(0.815)/0.484 2 0.293 < σ < 6.736 2

Therefore, the claim is valid. 7/276) Construct a 98% confidence interval for σ1/σ2 in Exercise 8 on page 264, where σ1 and σ2 are, respectively, the standard deviations for the distances obtained per liter of fuel by the Volkswagen and Toyota minitrucks. Soln: Given: Volkswagen, s1 = 1 n1 = 12 Toyota, s2 = 0.8 n2 = 10

(s 1/s 2)(1/f α/2(ν1, ν2)) < σ 1/σ 2 < (s 1/s 2)f α/2(ν1, ν2), 2

2

2

2

2

2

For a 98% confidence interval, α = 0.02. Interpolating in Table A.6, we find f 0.01(11,9) = 5.185 and f 0.01(9,11) = 4.63.



σ1 (1) (1) (4.63) ----------------- < -------- < ---------------2 2 σ22 (0.8) (5.185) (0.8) 2

2

2

0.549 < σ1/σ2 < 2.690

Θ1

Θ3 Θ2 θ

θ

Figure 9.1 Sampling distribution of different estimators of θ.

1-α

α/2 -zα/2

α/2 0

Figure 9.2 P(-zα/2 < Z < z α/2) = 1 -


z

zα/2

α


Seatwork XMQUAME 1. Molded-rubber expansion joints used in heating and air-conditioning systems are manufactured to withstand high pressure. A rubber company has purchased new machinery to produce joints with 5-inch internal diameters. For a random sample of 50 molded-rubber expansion joints, the mean internal diameter is  x = 4.985 inches and the standard deviation is equal to 0.03 inch. Estimate the mean internal diameter of all expansion joints produced by the new machinery with a 95% confidence interval. 2. Suppose a regional computer center wants to evaluate the performance of its disk memory system. One measure of performance is the average time between failures of a disk drive. Since the computer center operates two disk drives, it wants to estimate (µ1 - µ2), the difference in mean time between failures of the two disk drives. Independent samples of n1 = 10 and n2 = 15 failures produced the following statistics:

Disk Drive 1 x1 = 92 hours   s1 = 16 hours

Disk Drive 2 x2 = 108 hours  s2 = 12 hours

Estimate (µ1 - µ2) with a 95% confidence interval. Which of the t wo disk drives appear to give better performance? 3. One of American industry’s most fundamental problems – the stagnation in productivity – has economic experts seeking methods of “reindustrializing” the United States. One possible answer to the productivity stagnation may be industrial robots. An industrial robot has a control and memory system, often in the form of minicomputer, which enables it to be programmed to carry out a number of work routines faster and more efficiently than a human. Since the Japanese now operate most of the robots in the world, it is decided to estimate (p1 – p2), the difference between the proportions of U.S. and Japanese firms that currently employ at one industrial robot. Suppose random samples of U.S. and Japanese firms are selected, and the number of firms employing at least one industrial robot are recorded, with the sample sizes and results summarized in the accompanying table. Estimate the true difference in proportions (p1 – p2), using a 95% confidence interval. Which country has the higher proportion of firms that employ industrial robots?

Number of firms sampled Number of sampled firms that employ at least one industrial robot


U.S. N1 = 75 16

JAPAN N2 = 50 22


Examples: *4/252) An electrical firm manufactures light bulbs that have a length of li fe that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.

Soln: Given: n = 30, σ = 40 hours,  x = 780 hours 96% confidence interval From Table A.3, z α/2 = z.02 = 2.05 __ __  x - zα/2 σ/√n < µ <  x + zα/2 σ/√n __ __ 780 – 2.05(40)/ √30 < µ < 780 + 2.05(40)/ √30 765 < µ < 795 5/252) A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with a standard deviation equal to 0.15 deciliter. Find a 95% confidence interval for the mean of all drinks dispensed by this machine if a random sample of 36 drinks has an average content of 2.25 deciliters.

Soln: Given: n = 36, σ = 0.15 deciliter,  x = 2.25 deciliters 95% confidence interval From Table A.3, z α/2 = z.025 = 1.96 __ __ x - zα/2 σ/√n < µ <  x + zα/2 σ/√n  __ __ 2.25 – 1.96(0.15)/ √36 < µ < 2.25 + 1.96(0.15)/√36 2.201 < µ < 2.299 *6/252) The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. c) Construct a 98% confidence interval for the mean height of all college students. d) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centimeters? Soln: ENGSTAT Notes of AM Fillone


Given: n = 50, σ = 6.9 cm,  x = 174.5 cm a) 98% confidence interval From Table A.3, z α/2 = z.01 = 2.33 __ __ x - zα/2 σ/√n < µ <  x + zα/2 σ/√n  __ __ 174.5 – 2.33(6.9)/ √50 < µ < 174.5 + 2.33(6.9)/√50 172.23 < µ < 176.77 __ c) Error ≤ 2.33(6.9)/ √50 ≤ 2.27 cm 7/253) A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3,900 kilometers. a) Construct a 99% confidence interval for the average number of kilometers an automobile is drive annually in Virginia. b) What can se assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

Soln: Given: n = 100, σ = 3,900 km,  x = 23,500 km a) 99% confidence interval From Table A.3, z α/2 = z.005 = 2.575 __ __ x - zα/2 σ/√n < µ <  x + zα/2 σ/√n  ___ ___ 23,500 – 2.575(3,900)/ √100 < µ < 23,500 + 2.575(3,900)/ √100 22,495.75 < µ < 24,504.25 ___ d) Error ≤ 2.575(3,900)/ √100 ≤ 1,004.25 km *8/253) How large a sample is needed in Exercise 4 if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Soln: Given: zα/2 = 2.05, σ = 40 hours, e = 10 hours

n = (zα/2σ / e)

2

n = (2.05(40) / 10) n ≅ 68 samples


2


9/253) How large a sample is needed in Exercise 5 if we wish to be 95% confident that our sample mean will be within 0.09 deciliter of the true mean?

Soln: Given: zα/2 = z.025 = 1.96, σ = 0.15 deciliter, e = 0.09 deciliter n = (zα/2σ / e)

2

n = (1.96(0.15) / 0.09)

2

n ≅ 11 samples *10/253) An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 95% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds.

Soln: Given: zα/2 = z.025 = 1.96, σ = 40 seconds, e = 15 seconds n = (zα/2σ / e)2 n = (1.96(40) / 15)

2

n ≅ 28 samples 11/253) A UCLA researcher claims that the life span of mice can be extended by as much as 25% when the calories in their food are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Assuming that it is known from previous studies that σ = 5.8 months, how many mice should be included in our sample if we wis h to be 99% confident that the mean life span of the sample will be within 2 months of the population mean for all mice subjected to this reduced diet?

Soln: Given: zα/2 = z.005 = 2.575, σ = 5.8 months, e = 2 months n = (zα/2σ / e)

2

n = (2.575(5.8) / 2)

2

n ≅ 56 samples 12/253) Regular consumption of presweetened cereals contributes to tooth decay, heart disease, and other degenerative diseases according to studies conducted by Dr. W. H. Bowen of the National Institute of Health and Dr. J. Yudben, Professor of Nutrition and Dietetics at the University of London. In a random sample of 20 similar single servings of Alpha-Bits the average sugar content was 11.3 grams with a standard deviation of 2.45 ENGSTAT Notes of AM Fillone


grams. Assuming that the sugar contents are normally distributed, construct a 95% confidence interval for the mean sugar content for single servings of Alpha-Bits. Soln: Given: n = 20 servings,  x = 11.3 grams, s = 2.45 grams 95% confidence interval From Table A.4, t α/2 = t.025 = 2.093 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 11.3 – 2.093 (2.45)/ √20 < µ < 11.3 + 2.093 (2.45)/ √20 10.15 < µ < 12.45 *13/253) A machine is producing metal pieces that are cylindrical in shape. A sample of pieces is taken and the diameters are 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Find a 99% confidence interval for the mean diameter of pieces from this machine, assuming an approximate normal distribution.

Soln: 1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03  x = ---------------------------------------------------------------------------9 x = 1.006  9 s = Σ (xi - x)2/(n-1) = 0.0048 / 8 i=1 s = 0.024 2

From Table A.4, t α/2 = = t .005 = 3.355 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 1.006 – 3.355(.024)/ √9 < µ < 1.006 + 3.355(.024)/√9 0/974 < µ < 1.037 14/253) A random sample of 8 cigarettes of a certain brand has an average nicotine content of 2.6 milligrams and a standard deviation of 0.9 milligram. Construct a 99% confidence interval for the true average nicotine content of this particular brand of cigarettes, assuming the distribution of nicotine contents to be approximately normal.

Soln: Given: n = 8,  x = 2.6 milligrams, s = 0.90 @ 99% confidence interval From Table A.4, t α/2 = = t .005 = 3.355 ENGSTAT Notes of AM Fillone


__ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 1.006 – 3.355(.024)/ √9 < µ < 1.006 + 3.355(.024)/√9 0.974 < µ < 1.037

15/253) A random sample of 12 shearing pins is taken in a study of the Rockwell hardness of the head on the pin. Measurements on the Rockwell hardness were made for each of the 12, yielding an average value of 48.50 wit h a sample standard deviation of 1.5. Assuming the measurements to be normally distributed, construct a 90% confidence interval for the mean Rockwell hardness.

Soln: Given: n = 12,  x = 48.5 hardness, s = 1.5 hardness @ 90% confidence interval From Table A.4, t α/2 = = t .05 = 1.796 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 48.5 – 1.796(1.5)/ √12 < µ < 48.5 + 1.796(1.5)/ √12 47.722 < µ < 49.278 *16/253) A random sample of 12 graduates of a certain s ecretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming a normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school.

Soln: Given: n = 12,  x = 79.3 words/minute, s = 7.8 words/minute @ 95% confidence interval From Table A.4, t α/2 = = t .025 = 2.201 __ __ x - tα/2 s/√n < µ <  x + tα/2 s/√n,  __ __ 79.3 – 2.201(7.8)/ √12 < µ < 79.3 + 2.201(7.8)/ √12 74.344 < µ < 84.256 17/253) A random sample of 25 cigarettes of a certain brand has an average nicotine content of 1.3 milligrams and a standard deviation of 0.17 milligram. Find the 95% tolerance limits that will contain 90% of the nicotine contents for this brand of cigarettes, assuming the measurements to be normally distributed.

Soln: ENGSTAT Notes of AM Fillone


Given: n = 25,  x = 1.3 milligrams, s = 0.17 milligrams @ n = 25, (1- γ) = 0.95, (1From Table A.7, k = 2.208

α) = 0.90

x ± ks  1.3 ± 2.208(0.17) 1.3 ± (0.375) Therefore, the tolerance limit is from 0.925 to 1.675 milligrams. *18/253) The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint:

3.4 2.8 4.4

2.5 3.3 4.0

4.8 5.6 5.2

2.9 3.7 3.0

3.6 2.8 4.8

Assuming that the measurements represent a random sample from a normal population, find the 99% tolerance limits that will contain 95% of the drying times. Soln: 15 x = Σ xi / n = 3.787  i=1 15 s = Σ (xi - x)2/(n-1) i=1 2

s = 0.971 @ n = 15, (1- γ) = 0.99, (1From Table A.7, k = 3.507

α) = 0.95

x ± 

ks 3.787 ± 3.507(0.971) 3.787 ± 3.405 Therefore, the tolerance limit is from 0.382 to 7.192 hours. 19/253) Referring to Exercise 7, construct a 99% tolerance interval containing 99% of the miles traveled by automobiles annually in Virginia. Assume the distribution of measurements to be approximately normal. Soln: Given: n = 100,  x = 23,500 kilometers, s = 3,900 kilometers

@ n = 100, (1- γ) = 0.99, (1From Table A.7, k = 3.096


α) = 0.99


 x ± ks 23,500 ± 3.096(3,900) Therefore, the tolerance limit is from 11,426 to 35,574 kilometers. 20/253) Referring to Exercise 15, construct a 95% interval containing 905 of the measurements.

Soln: Given: n = 12,  x = 48.5 hardness, s = 1.5 hardness @ n = 12, (1- γ) = 0.95, (1From Table A.7, k = 2.655

α) = 0.90

x ± ks  48.5 ± 2.655(1.5) Therefore, the tolerance limit is from 44.52 to 52.48 hardness. Examples: 1/263) A random sample of size n1 = 25 taken from a normal population with a s tandard deviation σ1 = 5 has a mean x1 = 80. A second random sample size n 2 = 36, taken from a different normal population with a standard deviation σ2 = 3, has a mean x2 = 75. Find a 94% confidence interval µ1 - µ2.

Soln: Given: n1 = 25, σ1 = 5, x1 = 80 n2 = 36, σ2 = 3, x2 = 75 @ 94% confidence interval From Table A.3, z α/2 = z.03 = 1.88 ___________ ___________ 2 2 ( x1- x2) - zα/2√σ 1/n1 + σ 2/n2 < µ1 - µ2 < ( x1- x2) + zα/2√σ21/n1 + σ22/n2, ___________ __________ 2 2 2 2 (80 - 75) – 1.88 √5 /25 + 3 /36 < µ1 - µ2 < (80 - 75) + 1.88 √5 /25 + 3 /36, 2.9 < µ1 - µ2 < 7.1 *2/263) Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A has an average tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means.

Soln: Given: Brand A: nA = 50, σA = 5.6 kg,  xA = 78.3 kg Brand B: n B = 50, σB = 6.3 kg,  xB = 87.2 kg @ 95% confidence interval ENGSTAT Notes of AM Fillone


From Table A.3, z α/2 = z.025 = 1.96 ___________ ___________ 2 2 ( xB- xA) - zα/2√σ A/nA + σ B/nB < µB - µA < ( xB- xA) + zα/2√σ2A/nA + σ2B/nB, ________________ ________________ 2 2 2 2 (87.2–78.3)–1.96√(5.6) /50 + (6.3) /50 < µB-µA <(87.2–78.3)+1.96 √(5.6) /50 + (6.3) /50, 6.56 < µB - µA < 11.24 *3/263) A study was made to determine if a certain metal treatment has any effect on the amount of metal removed in a pickling operation. A random sample of 100 pieces was immersed in a bath of 24 hours without treatment, yielding an average of 12.2 millimeters of metal removed and a sample standard deviation of 1.1 millimeters. A second sample of 200 pieces was exposed to the treatment followed by the 24-hour immersion in the bath, resulting in an average removal of 9.1 milliliters of metal with a sample standard deviation of 0.9 millimeter. Compute a 98% confidence interval es timate for the difference between the population means. Does the treatment appear to reduce the mean amount of metal removed?

Soln: Sample1 n1 = 100  x1 = 12.2 ml σ1 = 1.1 ml

Sample 2 n2 = 200  x2 = 9.1ml σ2 = 0.9 ml

@ 98% confidence interval, From Table A.3, z α/2 = z.01 = 2.33 ___________ ___________ 2 2 2 2 ( x1- x2) - zα/2√σ 1/n1 + σ 2/n2 < µ1 - µ2 < ( x1- x2) + zα/2√σ 1/n1 + σ 2/n2, ___________________ _________________ 2 2 (12.2–9.1)–2.33 √(1.1) /100 + (0.9) /200 < µ1-µ2 < ( 12.2– 2 2 9.1)+2.33√(1.1) /100+(0.9) /200 2.804 < µ1-µ2 < 3.396 5/263) Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average examination grade of 84 with a standard deviation of 4, and 18 students in the section without labs made an average grade of 77 with a standard deviation of 6, find a 99% confidence interval for the difference between the average grades for the two courses. Ass ume the populations to be approximately normally distributed with equal variances.

Soln: Physics (3-hour) n3 = 18  x3 = 77 s3 = 6 ENGSTAT Notes of AM Fillone

Physics (4-hour) n4 = 12  x4 = 84 s4 = 4


@ 99% confidence interval From Table A.4, with v = n 1 + n2 –2 = 28, t α/2 = t.005 = 2.763 (n3 – 1)s 23 + (n4 –1)s24 s2 p = --------------------------------n3 + n4 - 2 (18 –1)6 2 + (12-1)42 2 s p = --------------------------- = 28.143 18 + 12 - 2 s p = 5.305 ________ _________ ( x4 - x3) - tα/2s p√1/n1 + 1/n2 < µ4 - µ3 < ( x4 - x3) + tα/2 s p√1/n1 + 1/n2, _________ _________ (84-77)-2.763(5.305)√1/18 + 1/12 < µ4 - µ3 < (84-77)+2.763(5.305) √1/18 + 1/12 1.537 < µ4 - µ3 < 12.463 7/264) The following data, recorded in days, represent the length of time to recovery for patients randomly treated with one of two medications to clear up severe bladder infections:

Medication 1 n1 = 14 x1 = 17  2 s 1 = 1.5

Medication 2 n2 = 16 x2 = 19  2 s 2 = 1.8

Find a 99% confidence interval for the difference µ2 - µ1 in the mean recovery time for the two medications, assuming normal populations with equal variances. Soln: @ 99% confidence interval for the difference µ2 - µ1 From Table A.4 , @ ν = 28, tα/2 = t.005 = 2.763

(14 –1)(1.5) + (16-1)(1.8) = --------------------------------- = 1.66 14 + 16 - 2 s p = 1.2887 s2 p

________ ________ (19 - 17) – 2.763(1.289) √1/14+1/16 < µ2 - µ1 < (19 - 17) + 2.763(1.289) √1/14+1/16, 0.697 < µ2 - µ1 < 3.303 ENGSTAT Notes of AM Fillone


*8/264) An experiment reported in Popular Science, in 1981, compared fuel economies for two types of similarly equipped diesel mini-trucks. Let us suppose that 12 Volkswagen and 10 Toyota trucks are used in 90-kilometer per hour steady-speed tests. If the 12 Volkswagen trucks average 16 kilometers per liter with a standard deviation of 1.0 kilometer per liter and the10 Toyota trucks average 11 kilometers per liter with a standard deviation of 0.8 kilometer per liter, construct a 90% confidence interval for the difference between the average kilometers per liter of these two mini-trucks. Assume that the distances per liter for each truck model are approximately normally distributed with equal variances.

Soln: Volkswagen n1 = 12 x1 = 16 km/l  s1 = 1.0 km/l

Toyota n2 = 10 x2 = 11 km/l  s2 = 0.8 km/l

@ 90% confidence interval for the difference µ2 - µ1 From Table A.4 , @ ν = 20, tα/2 = t.05 = 1.725 2

2

(12 –1)1 + (10-1)(0.8) 2 s p = --------------------------------- = 0.838 12 + 10 - 2 s p = 0.915 ________ ________ (16 - 11) – 1.725(0.915) √1/12+1/10 < µ2 - µ1 < (16 - 11) + 1.725(0.915) √1/12+1/10, 4.324 < µ2 - µ1 < 5.676 *11/264) The government awarded grants to the agricultural departments of 9 universities to test the yield capabilities of two new varieties of wheat. Each variety w as planted on plots of equal area at each university and the yields, in kilograms per plot, recorded as follows:

Variety 1 2 d 

sd =

1 38 45 7

2 23 25 2

3 35 31 -4

__________________ n  n 2  n Σ d i – ( Σ di )2 i=1  i=1  ------------------------- =


University 4 5 6 41 44 29 38 50 33 -3 6 4

7 37 36 -1

8 31 40 9

9 38 43 5

_______________  9(237)2 – (25)2  -------------------- = 4.5765


√

√

n(n-1)

9(9-1)

d = (7 + 2 – 4 – 3 + 6 + 4 –1 + 9 + 5)/9 = 2.78  @ 95% confidence interval From Table A.4, ν = n –1 = 8, t α/2 = t.025 = 2.306 _ d - tα/2 (sd / √n ) < 

_ d + tα/2 (sd / √n ), µD <  __ __ 2.78 – 2.306 (4.5765/ √9 ) < µD < 2.78 + 2.306 (4.5765/ √9 ) -0.738 < µD < 6.298 Example: 1/270) (a) A random sample of 200 voters is selected and 114 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit. (b) What can we assert with 96% confidence about the possible size of our error if we estimate the fraction of voters favoring the annexation suit to be 0.57? Soln: Given:  p = 114/200 = 0.57,  q = 86/200 = 0.43

@96% Confidence interval From Table A.3, z α/2 = z.02 = 2.05 a)

______ ______ p - zα/2√ p q / n < p < p + zα/2√ p q /n  ____________ ____________ 0.57 – 2.05 √0.57(0.43)/200 < p < 0.57 + 2.05 √0.57(0.43)/200 0.498 < p < 0.642 ____________ b) Error ≤ 2.05 √0.57(0.43)/200

≤ .072

2/270) a) A random sample of 500 cigarette smokers is selected and 86 are found to have a preference for brand X. Find the 90% confidence interval for the fraction of the population of cigarette smokers who prefer brand X. b) What can we assert with 90% confidence about the possible size of our error if we estimate the fraction of cigarette smokers who prefer brand X to be 0.172? Soln: Given:  p = 86/500 = 0.172,  q = 414/500 = 0.828


______ ______  p - zα/2√ p q / n < p < p + zα/2√ p q /n



____________ ____________ 0.172 – 1.645 √0.172(0.828)/500 < p < 0.172 + 1.645 √0.172(0.828)/500 0.1442 < p < 0.1998 ____________ b) Error ≤ 1.645 √0.172(0.828)/500

≤ .0278

*3/270) In a random sample of 1000 homes in a certain city, it is found that 228 are heated by oil. Find the 99% confidence interval for the proportion of homes in this city that are heated by oil. Soln: Given:  p = 228/1000 = 0.228,  q = 772/1000 = 0.772

@99% Confidence interval From Table A.3, z α/2 = z.005 = 2.575 ______ ______  p - zα/2√ p q / n < p < p + zα/2√ p q /n _______________ _______________ 0.228 – 2.575 √0.228(0.772)/1000 < p < 0.228 + 2.575 √0.228(0.772)/1000 0.1938 < p < 0.2622 4/270) Compute a 98% confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives. Soln: Given:  p = 8/100 = 0.08,  q = 92/100 = 0.92

@98% Confidence interval From Table A.3, z α/2 = z.01 = 2.33 ______ ______ p - zα/2√ p q / n < p < p + zα/2√ p q /n  ____________ ____________ 0.08 – 2.33 √0.08(0.92)/100 < p < 0.08 + 2.33 √0.08(0.92)/100 0.0168 < p < 0.1432 *5/270) A new rocket-launching system is being considered for deployment of small, short-range rockets. The existing system has p = 0.8 as the probability of a successful launch. A sample of 40 experimental launches is made with the new sy stem and 34 are successful. (c) Construct a 95% confidence interval for p. (d) Would you conclude that the new system is better? Soln: Given:  p = 34/40 = 0.85,  q = 6/40 = 0.15


______


______


 p - zα/2√ p q/n

< p < p + zα/2√ p q /n ____________ ____________ 0.85 – 1.96 √0.85(0.15)/40 < p < 0.85 + 1.96 √0.85(0.15)/40 0.739 < p < 0.961 b) No. Since the existing p = .80 is still within the range. 6/271) A geneticist is interested in the proportion of African males that have a certain minor blood disorder. In a random sample of 100 African males, 24 are found to be afflicted. a) Compute a 99% confidence interval for the proportion of African males that havet his blood disorder. b) What can we assert with 99% confidence about the possible size of our error if we estimate the proportion of African males with this blood disorder to be 0.24? Soln: Given:  p = 24/100 = 0.24,  q = 76/100 = 0.76


______ ______ p - zα/2√ p q / n < p < p + zα/2√ p q /n  ____________ ____________ 0.24 – 2.575 √0.24(0.76)/100 < p < 0.24 + 2.575 √0.24(0.76)/100 0.130 < p < 0.350 ____________ b) Error ≤ 2.575 √0.24(0.76)/100 ≤ 0.110

7/271) a) According to a report in the Roanoke Times & World-News, August 20, 1981, approximately 2/3 of the 1600 adults polled by telephone said they think the space shuttle program is a good investment for the country. Find a 95% confidence interval for the proportion of American adults who think the space shuttle program is a good investment for the country. b) What can we assert with 95% confidence about the possible size of our error if we estimate the proportion of American adults who think the space shuttle program is a good investment to be 2/3? Soln: Given:  p = 0.667,  q = 0.333


______ ______ p - zα/2√ p q / n < p < p + zα/2√ p q /n 



_______________ _______________ 0.667 – 1.96 √0.667(0.333)/1600 < p < 0.667 + 1.96 √0.667(0.333)/1600 0.6439 < p < 0.6901 _______________ b) Error ≤ 1.96 √0.667(0.333)/1600

≤ 0.023

*8/271) In the newspaper article referred to in Exercise 7, 32% of the 1600 adults polled said the U.S. space program should emphasize scientific exploration. How large a sample of adults is needed in the poll if one wishes to be 95% confident that the estimated percentage will be within 2% of the true percentage? Soln: Given :  p = 0.32,  q = 0.68

n = z α/2 p q/e . 2

2

@ 95% Confidence interval From Table A.3, z α/2 = z.025 = 1.96 2

n = (1.96) (.32)(.68)/(.02)

2

≅ 2,090

9/271) How large a sample is needed in Exercise 1 if we wish to be 96% confident that our sample proportion will be within 0.02 of the true fraction of the voting population? Soln: Given :  p = 0.57,  q = 0.43

n = z α/2 p q/e . 2

2

@ 96% Confidence interval From Table A.3, z α/2 = z.02 = 2.05 n = (2.05) 2(.57)(.43)/(.02)2 ≅ 2,576 10/271) How large a sample is needed in Exercise 3 if we wish to be 99% confident that our sample proportion will be within 0.05 of the true proportion of homes in this city that are heated by oil? Soln: Given :  p = 0.228,  q = 0.772

n = z α/2 p q/e . 2

2


2

n = (2.575) (.228)(.772)/(.05)

≅ 467

11/271) How large a sample is needed in Exercise 4 if we wish to be 98% confident that our sample proportion will be within 0.05 of the true proportion defective? Soln: Given :  p = 0.08,  q = 0.92

n = z α/2 p q/e . 2


2



n = (2.33) (.08)(.92)/(.05)

2

≅ 160

*12/271) A study is to be made to estimate the percentage of citizens in a town who favor having their water fluoridated. How large a sample is needed if one wishes to be at least 95% confident that our estimate is within 1% of the true percentage?

Soln; Given: e = .01 @ 95% Confidence interval From Table A.3, z α/2 = z.025 = 1.96 n = z2α/2 / 4e2 2

n = (1.96) /4(.01)

2

≅ 9,604

13/271) According to Dr. Memory Elvin-Lewis, head of t he microbiology department of Washington University School of Dental Medicine in St. Louis, a couple of cups of either green or oolong tea each day will provide sufficient fluoride to protect your teeth from decay. People who do not like tea and who live in unfluoridated areas should ask t heir local governments to consider having their water fluoridated. How large a sample is needed to estimate the percentage of citizens in a certain town who favor having their water fluoridated if one wishes to be at least 99% confident that the estimate is within 1% of the true percentage.

Soln: Given: e = .01 @ 99% Confidence interval From Table A.3, z α/2 = z.005 = 2.575 2

2

n = z α/2 / 4e

n = (2.575) 2/4(.01)2 ≅ 16,577 *14/271) A study is to be made to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant. How large a sample is needed if one wished to be at least 95% confident that the estimate is within 0.04 of the true proportion of residents in this city and its suburbs that favor the construction of the nuclear power plant? Soln: Given: e = 0.04

@ 95% Confidence interval From Table A.3, z α/2 = z.025 = 1.96



2

2

n = z α/2 / 4e

n = (1.96) 2/4(0.04)2 ≅ 601 15/271) A certain geneticist is interested in the proportion of males and females in the population that have a certain minor blood disorder. In a random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 f emales tested appear to have the disorder. Compute a 95% confidence interval for the difference between the proportion of males and females that have the blood disorder. Soln: Given: Males:  pm = 250/1000 = 0.25 qm = 750/1000 = 0.75  Females: pf = 275/1000 = 0.275  qf = 725/1000 = 0.725

@ 95% confidence interval Using Table A.3, zα/2 = z.025 = 1.96 a) ________________ __________________ ( p1 - p2) - zα/2√ p1 q1/n1 + p2 q2/n2 < p –p p1 - p2) + zα/2√ p1 q1/n1 + p2 q2 / n2, f m < ( ________________________________ (0.275 – 0.25) – 1.96 √(0.25)(0.75)/1000 + (0.275)(0.725)/1000 < p f – pm < (0.275-0.25) + _________________________________ 1.96 √ (0.25)(0.75)/1000 + (0.275)(0.725)/1000 -0.01355 < p f – pm < 0.06355 b) Yes *16/271) A cigarette-manufacturing firm claims that its brand A line of cigarettes outsells its brand B line by 8%. If it is found that 42 of 200 smokers prefer brand A and 18 of 150 smokers prefer brand B, compute a 94% confidence interval for the difference between the proportions of sales of the 2 brands and decide if the 8% difference is a valid claim. Soln: Given: Brand A: pA = 42/200 = 0.21  qA = 158/200 = 0.79  Brand B: pB = 18/150 = 0.12  qB = 132/150 = 0.88 

@ 94% confidence interval Using Table A.3, zα/2 = z.03 = 1.88 a) _________________ __________________ ( pA- pB)- zα/2√ pA qA/nA + pB qB/nB


0.0164 < p A – pB < 0.1636 b) Yes. It’s a valid claim. 17/271) A clinical trial is conducted to determine if a certain type of inoculation has an effect on the incidence of a certain disease. A sample of 1000 rats was kept in a controlled environment for a period of 1 year and 500 of the rats were given the inoculation. Of the group not given the drug, there were 120 incidences of the disease, while 98 of the inoculated group contracted it. If we call p1 the probability of incidence of the disease in uninoculated rats and p2 the probability of incidence after receiving the drug, compute a 90% confidence interval for p1 – p2.

Soln: Given: Uninoculated:  p1 = 120/500 = 0.24 q1 = 380/500 = 0.76  Inoculated: p2 = 98/500 = 0.196  q2 = 402/500 = 0.804  @ 90% confidence interval Using Table A.3, zα/2 = z.05 = 1.645 ________________ ___________________ ( p1- p2) - zα/2√ p1 q1/n1 + p2 q2/n2 < p1 –p2 < ( p1 - p2) + zα/2√ p1 q1/n1 +  p2 q2 / n2, ______________________________ (0.24 – 0.196) – 1.645 √(0.24)(0.76)/500 + (0.196)(0.804)/500 < p 1 – p2 < (0.24-0.196) + _______________________________ 1.645 √ (0.24)(0.76)/500 + (0.196)(0.804)/500 0.0011 < p 1 – p2 < 0.0869 20/272) A well-known sock company wants t o determine if their medium-size socks satisfy the same proportion of boys and girls who wear their socks. In a survey of 1000 randomly selected teenagers, 462 were boys and 538 were girls. Among the 282 boys who wear the company’s medium-size socks, 267 say they feel comfortable with its size, while 290 girls out of 313 who wear the company’s socks say were satisfied. The rest say that the socks are either too small or too big. Compute a 95% confidence interval for the difference between the proportion of boys and girls who w ere satisfied with the company’s medium-size socks.

Soln: Given: Boys: Girls:

pB = 267/282 = 0.9468  qB = 15/282 = 0.0532  pG = 290/313 = 0.9265  qG = 23/313 = 0.0735 

@ 95% confidence interval Using Table A.3, zα/2 = z.025 = 1.96 _________________ __________________ ( pB- pG)-zα/2√ pB qB/nB + pG qG/nG

_____________________________________ (0.9468–0.9265) –1.96 √(0.9468)(0.0532)/282 + (0.9265)(0.0735)/313 < p B – pG < ______________________________________ (0.9468-0.9265) + 1.645 √ (0.9468)(0.0532)/282 + (0.9265)(0.0735)/313 -0.0187 < p B – pG < 0.0593 Example: *1/276) A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years, construct a 95% confidence interval for σ2 and decide if the manufacturer’s claim that σ2 = 1 is valid. Assume the population of battery lives to be approximately normally distributed.

Soln: (n –1)s / χ α/2 < σ < (n - 1)s / χ 1-α/2, 2

2

n

2

2

2

n

n Σ x i – ( Σ xi ) 2 i=1 i=1 5(48.26) – (15) 2 s = ------------------------------ = ------------------------ = 0.815 n (n-1) 5(5 –1) 2

2

For a 95% confidence interval, α = .05. From Table A.5 with v = 4 degrees of freedom, 2 2 we find χ 0.025 = 11.143 and χ 0.975 = 0.484.

4(0.815)/11.143 <

σ2 < 4(0.815)/0.484

0.293 < σ2 < 6.736 Therefore, the claim is valid. 2/276) A random sample of 20 students obtained a mean of x = 72 and a variance of s = 16 on a college placement test in mathematics. Assuming the scores to be normally 2 distributed, construct a 98% confidence interval for σ . 2

Soln: (n –1)s / χ α/2 < σ < (n - 1)s / χ 1-α/2, 2

2

2

2

2

For a 98% confidence interval, α = .02. From Table A.5 with v = 19 degrees of freedom, 2 2 we find χ 0.01 = 36.191 and χ 0.99 = 7.633.

(20-1)(16)/36.191 <

σ2 < (20-1)(16)/7.633

8.400 < σ < 39.827 2


Ex Estimation

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