STAT 400 Lecture AL1 Answers for 7.2, 8.2 Spring 2015 Dalpiaz 1. Dr. Statman claims that his new revolutionary study method “Study While You Sleep” is more effective than the traditional study methods. In an experiment, 250 students enrolled in the same section of STAT 100 at UIUC were divided into two groups. One hundred students volunteered to study using SWYS method, and the other 150 students did whatever students usually do. At the end of the semester, the averages of the total number of points (out of 500) were compared for the two groups. Note: This is NOT a good experiment design! SWYS Traditional (sample) average total points 450 410 (sample) standard deviation 20 45 a) Construct a 95% confidence interval for the difference in the average total points for SWYS and traditional study methods. n 1 = 100 and n 2 = 150 are large, t α/2 can be approximated by z α/2. ( )2221212αYX nsnsz +±− ⋅ ( ) 1504510020 96.141045022 +±−⋅ 40 ± 8.2 ( 31.8 , 48.2 ) b) Perform the appropriate test at a 1% level of significance. Claim: µ New > µ Old H 0 : µ New – µ Old = 0 vs. H 1 : µ New – µ Old > 0 Test Statistic: Z = ( )( )15045100200410450YX222221210 ssδ+−−=+−−nn = 9.562. n 1 and n 2 are large Rejection Region: Reject H 0 if Z > z 0.01 = 2.326. Reject H 0 at α = 0.01.c) Test H 0 : µ S – µ T ≤ 30 vs. H 1 : µ S – µ T > 30 at α = 0.05. Test Statistic: Z = ( )( )150451002030410450YX222221210 ssδ+−−=+−−nn = 2.39. Rejection Region: Reject H 0 if Z > z 0.05 = 1.645. Reject H 0 at α = 0.05 p-value ≈ 0.0084. 2. Two work designs are being considered for possible adoption in an assembly plant. A time study is conducted with 10 workers using design A and 12 workers using design B. The sample means and sample standard deviations of their assembly times (in minutes) are Design A Design B Sample Mean 78.3 85.6 Sample Standard deviation 4.8 6.5 Construct a 90% confidence interval for the difference in the mean assembly times between design A and Design B. Use Welch’s T. −+−+⋅⋅ 1111 22222212112222121 nnnnnnssss = −+−+⋅⋅ 125.61121108.41101125.6108.4 2222222 = 76324.19 = 19 degrees of freedom ( )2221212αYX nsnst +±− ⋅ ( ) 125.6108.4 729.16.853.7822 +±− ⋅ – 7.3 ± 4.173 ( – 11.473 , – 3.127 )3. A national equal employment opportunities committee is conducting an investigation to determine if women employees are as well paid as their male counterparts in comparable jobs. Random samples of 14 males and 11 females in junior academic positions are selected, and the following calculations are obtained from their salary data. Male Female Sample Mean $48,530 $47,620 Sample Standard deviation 780 750 Assume that the populations are normally distributed with equal variances. a) Construct a 95% confidence interval for the difference between the mean salaries of males and females in junior academic positions. ( ) ( )21114750111780114222 pooled −+−+−=⋅⋅s = 588,443.47826 pooleds = 767.1 ( )21pooled211YX nnst +α±− ⋅⋅ 14 + 11 – 2 = 23 degrees of freedom ( ) 111141 1.767069.2620,47530,48 +±− ⋅⋅ 910 ± 639.47 ( 270.53 , 1,549.47 ) b) What is the p-value of the test H 0 : µ Male = µ Female vs. H 1 : µ Male ≠ µ Female? Test Statistic: T = ( )( )111141 767.10620,47530,4811 YX21pooled0 sδ+−−=+−−⋅⋅nn = 2.944. t 0.005 ( n 1 + n 2 – 2 = 23 d.f. ) = 2.807. p-value ( 2 – tailed ) < 0.005 × 2 = 0.01 ( p-value ≈ 0.0073 )4. A new revolutionary diet-and-exercise plan is introduced. Eight participants were weighed in the beginning of the program, and then again a week later. The results were as follows: Participant 1 2 3 4 5 6 7 8 Weight Before 213 222 232 201 230 188 218 182 Weight After 207 220 224 198 219 183 220 175 Pounds Lost 6 2 8 3 11 5 – 2 7 a) Construct a 90% confidence interval for the average number of pounds lost during one week on that plan. 8408725113826D =+−+++++==∑ndi = 5 pounds. d d 2 d D−d ( )2 D−d 6 2 8 3 11 5 – 2 7 36 4 64 9 121 25 4 49 OR 6 2 8 3 11 5 – 2 7 1 – 3 3 – 2 6 0 – 7 2 1 9 9 4 36 0 49 4 312 0 112 ( )( )784031212222D nndds−=−−=∑∑ = 16. ( )71121D22D =−−=∑nds = 16. ss 162DD == = 4 pounds. Confidence interval: nst D2D⋅α±. n – 1 = 7 degrees of freedom. 90% confidence level, α = 0.10, α/2 = 0.05, 2tα = 1.895. 84895.15 ⋅± 5 ± 2.68 ( 2.32 , 7.68 )b) Is there enough evidence to conclude that the average weight loss is less than 7 pounds per week? (Use α = 0.05.) What is the p-value of this test? Claim: δ < 7 H 0 : δ ≥ 7 H 1 : δ < 7 Test Statistic: T = 8475D D0δ−=−ns = – 1.4142. Rejection Region: Reject H 0 if T < – t 0.05 ( 8 – 1 = 7 d.f. ) = – 1.895. The value of the test statistic does not fall into the Rejection Region. Do NOT Reject H 0 at α = 0.05. ( p-value ≈ 0.10.
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