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Unformatted text preview: STAT 350 – Spring 2009 Homework #6 – Solution ReCORRECTED through Lecture J Text Exercises Chapter 7, Problems #38, 46*, 48*, 52 * do not assume equal variance! s 3.1 ⇒ 30.2 ± 2.365 7.38 x ± t0.05 / 2,8−1 n 8 ⇒ ( 27.608,32.792 ) 7.46 s12 n1 0.662 20 c= 2 = = 0.741194487 2 s1 s2 0.662 0.392 + + n1 n2 20 20 ( n1 − 1)( n2 − 1) df = 2 ( n2 − 1) c 2 + ( n1 − 1)(1 − c ) = ( 20 − 1)( 20 − 1) 2 ( 20 − 1) 0.7411944872 + ( 20 − 1)(1 − 0.741194487 )
2 s12 s2 + ⇒ n1 n2 = 30.83 df is rounded down to 30. thus t* = 2.042
* x1 − x2 ± tα / 2 8.74 − 4.96 ± 2.042 0.662 0.392 + 20 20 ⇒ ( 3.43, 4.13) Homework #6 – Solution Page 1 of 1 7.48 a.
Boxplot of middle, high
470 460 450 Data 440 430 420 middle high b.
Variable middle high Total Count 17 11 Mean 438.29 437.45 StDev 15.14 6.83 s12 n1 15.142 17 = = 0.760734951 2 s12 s2 15.142 6.832 + + n1 n2 17 11 ( n1 − 1)( n2 − 1) df = 2 ( n2 − 1) c 2 + ( n1 − 1)(1 − c ) c= = (17 − 1)(11 − 1) 2 (11 − 1) 0.7607349512 + (17 − 1)(1 − 0.760734951)
2 s12 s2 + ⇒ n1 n2 = 23.869 df is rounded down to 23 thus t* = 2.069
* x1 − x2 ± tα / 2 438.29 − 437.45 ± 2.069 15.142 6.832 + 17 11 ⇒ ( −7.87,9.55) The fact that the confidence interval for the difference in the means includes zero suggests that we cannot say with confidence that there is a true difference in the means of these two populations. Homework #6 – Solution Page 2 of 2 7.52
House: Indoor: Outdoor: difference 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 0.07 0.08 0.09 0.12 0.12 0.12 0.13 0.14 0.15 0.15 0.17 0.17 0.18 0.18 0.18 0.18 0.19 0.20 0.22 0.22 0.23 0.23 0.25 0.26 0.28 0.28 0.29 0.34 0.39 0.40 0.45 0.54 0.62 0.29 0.68 0.47 0.54 0.97 0.35 0.49 0.84 0.86 0.28 0.32 0.32 1.55 0.66 0.29 0.21 1.02 1.59 0.90 0.52 0.12 0.54 0.88 0.49 1.24 0.48 0.27 0.37 1.26 0.70 0.76 0.99 0.36 0.22 0.60 0.38 0.42 0.85 0.23 0.36 0.70 0.71 0.13 0.15 0.15 1.37 0.48 0.11 0.03 0.83 1.39 0.68 0.30 0.11 0.31 0.63 0.23 0.96 0.20 0.02 0.03 0.87 0.30 0.31 0.45 0.26 Average: 0.42394 Stdev: 0.386773 a. b. Because n > 30 (n = 33), I will use the z critical value for my confidence interval, using a t critical value of 2.042 (or slightly less) would also be appropriate. s 0.386773 ⇒ −0.42394 ± 1.96 ⇒ d ± zα /2 d ( −0.5559, −0.29198 ) n 33 A prediction interval for d34: 1 1 d ± zα /2 sd 1 + ⇒ −0.42394 ± 1.96 ( 0.386773) 1 + n 33 ⇒ ( −1.19342, 0.345535) Homework #6 – Solution Page 3 of 3 Then answer the following problems 1. You are part of a team that has designed a new filter to be used with pitchers for drinking water. You need to study how effective it is at removing various contaminants. You have 12 1gallon pitchers fitted with new filters. You have a supply of water spiked with a high concentration of copper. You add 1 gallon of water to each pitcher, then measure the amount of copper removed. The average copper removed was 72.8 ppb (parts per billion) with a standard deviation of 18.6 ppb. Because of the small sample size (n = 12), we must use a t critical value (from Appendix Table IV) instead of a z critical value. The degrees of freedom (df) for the t critical value are n – 1 = 11. a. Give a 90% CI for the true mean copper removed by these filters from the first gallon of water. s 18.6 x ± tcrit ⇒ 72.8 ± 1.796 ⇒ (63.1566, 82.4434) n 12 b. Give a 95% CI for the true mean copper removed by these filters from the first gallon of water. 18.6 72.8 ± 2.201 ⇒ (60.9820, 84.6180) 12 c. Give a 99% CI for the true mean copper removed by these filters from the first gallon of water. 18.6 72.8 ± 3.106 ⇒ (56.1228, 89.4772) 12 d. Assume you have a 13th filter (not included in the previous tests). i. give a 90% prediction interval for the amount of copper that would be removed from the first gallon of water by this 13th filter. 1 1 72.8 ± 1.796 (18.6 ) 1 + x ± tcrit ( s ) 1 + ⇒ ⇒ (38.0303, 107.5697) 12 n ii. give a 95% prediction interval for the amount of copper that would be removed from the first gallon of water by this 13th filter. 1 72.8 ± 2.201(18.6 ) 1 + ⇒ (30.1898, 115.4102) 12 iii. give a 99% prediction interval for the amount of copper that would be removed from the first gallon of water by this 13th filter. 1 72.8 ± 3.106 (18.6 ) 1 + ⇒ (12.6694, 132.9306) 12 Homework #6 – Solution Page 4 of 4 e. Explain, conceptually, why the interval in part (d.i) is wider than that obtained in part (a). The prediction interval for a given sample will always be wider than a confidence interval for the mean. To understand why, imagine that we know the values of mu and sigma exactly. The confidence interval for mu would be a single number (mu). But individual observation will vary. Even knowing mu and sigma exactly, I can be 90% confident that the next observed value will be somewhere between μ − 1.645σ and μ + 1.645σ . I cannot know it exactly. 2. The lifetimes of 2 models of specialized batteries are compared for their lifetime in a prototype hearing aid. 20 hearing aids were loaded with Brand 1 batteries and 6 were loaded with Brand 2 batteries. The hearing aids were turned on and monitored until the batteries died. The mean lifetime of the 20 Brand 1 batteries was 108.6 hours with a standard deviation of 14.6 hours. The mean lifetime of the 6 Brand 2 batteries was 137.2 hours with a standard deviation of 20.1 hours. a. Obtain a 95% CI for the true difference in the mean lifetimes of these two brands of batteries (that is, obtain a 95% CI for µ1  µ2), assuming that σ1 = σ2. sp = ( n1 − 1) s12 + ( n2 − 1) s22
n1 + n2 − 2 = ( 20 − 1)14.62 + ( 6 − 1) 20.12
20 + 6 − 2 = 24 = 15.9035 df = n1 + n2 – 2 = 20 + 6 – 2 The t critical value is then 2.064 95% CI: ( x1 − x2 ) ± tcrit s p 11 + n1 n2 ⇒ (108.6 − 137.2 ) ± 2.064 (15.9035) 11 + 20 6 → (43.8791, 13.3209) b. Obtain a 95% CI for the true difference in the mean lifetimes of these two brands of batteries (that is, obtain a 95% CI for µ1  µ2), without assuming that σ1 = σ2. s12 n1 14.62 20 c= 2 = = 0.1366532894 2 s1 s2 14.62 20.12 + + n1 n2 20 6 df = ( n1 − 1)( n2 − 1) 2 ( n2 − 1) c 2 + ( n1 − 1)(1 − c ) = 6.664163 → round down to 6. * x1 − x2 ± tα /2 2 14.62 20.12 s12 s2 + ⇒ + (108.6 − 137.2 ) ± 2.447 20 6 n1 n2 = (50.2103, 6.98965) Homework #6 – Solution Page 5 of 5 3. Shortly before giving a second exam, Dr. Random, Professor of Statistics, was asked if the second exam in his class was usually harder than the first. Being a good statistician, he wanted to test this. He took a random sample of 20 students from the past several semesters that he taught the class (these scores are given below). Obtain a 95% confidence interval for the true mean difference in average scores for the first and second exams in Dr. Random’s class.
Student Exam 1 Score Abby Albert Betty Monica Gus Denise Shannon Radu Charles Matt Lisa Sally Penny John Rebecca Michael Charlie Joe John Yan 68 60 46 31 71 57 80 48 82 45 31 80 78 84 51 56 45 54 31 67 Exam 2 Score 85 73 76 65 49 34 81 58 92 57 73 86 83 83 81 85 82 58 51 74 difference ‐17 ‐13 ‐30 ‐34 22 23 ‐1 ‐10 ‐10 ‐12 ‐42 ‐6 ‐5 1 ‐30 ‐29 ‐37 ‐4 ‐20 ‐7 mean stdev 58.25 71.3 ‐13.05 17.48947 15.44123 17.57233 df = n – 1 = 20 – 1 = 19
* d ± tα /2 sd n ⇒ −13.05 ± 2.093 17.57233 = (21.274, 4.826) 20 Homework #6 – Solution Page 6 of 6 ...
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This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff
 Variance

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