Final Answers

9.4 a) European Perspective: H 0 : Genetically modified food is safe H 1 : Genetically modified food is not safe b) U.S. farmer Perspective: H 0 : Genetically modified food is not safe H 1 : Genetically modified food is safe 9.14a Test the hypothesis H0: µ ≤ 100 H1: µ > 100 Using a random sample of n = 25, a probability of Type I error equal to 0.05, and the following sample statistics. x = 106; s = 15 Reject H0 if t = x − µ 0 s √ n > t n-1,α t = 106 − 100 15 √ 25 = 2 > 1.711 t = n-1 = 25 – 1 = 24 (from t distribution table the value is 1.711 with lambda 0.05) Based on this result we rejected the null hypothesis 9.28 A random sample of women is obtained, and each person in the sample is asked if she would purchase a new shoe model. The new shoe model would be successful in meeting corporate profit objective if more than 25% of the women in the population would purchase the shoe more. The following hypothesis test is performed at a level of  = 0.03 using ^ p as the sample proportion of women who said yes. H0: P ≤ 0.25 H1: P > 0.25 What value of the sample proportion, ^ p , is required to reject the null hypothesis given the following sample sizes? For this question we find z at level of α = 0.03 using the table 1 it is 1.88 z α = 1.88 a) n = 400 reject if H0 is ^ p > ^ pc = p 0 + zα ( √ p 0 ( 1 − p 0 )/ n ) 0.25 + 1.88 √ 0.25 ( 1 − 0.25 )/ 400 ) = 0.291 0.291 is the value of sample proportion p hat required to reject null hypothesis

b) n = 225 reject if H0 is ^ p > ^ pc = p 0 + zα ( √ p 0 ( 1 − p 0 )/ n ) 0.25 + 1.88 √ 0.25 ( 1 − 0.25 )/ 225 ) = 0.304 0.304 is the value of sample proportion p hat required to reject null hypothesis c) n = 625 reject if H0 is ^ p > ^ pc = p 0 + zα ( √ p 0 ( 1 − p 0 )/ n ) 0.25 + 1.88 √ 0.25 ( 1 − 0.25 )/ 625 ) = 0.282 0.282 is the value of sample proportion p hat required to reject null hypothesis d) n = 900 reject if H0 is ^ p > ^ pc = p 0 + zα ( √ p 0 ( 1 − p 0 )/ n ) 0.25 + 1.88 √ 0.25 ( 1 − 0.25 )/ 900 ) = 0.277 0.277 is the value of sample proportion p hat required to reject null hypothesis 9.48 At the insistence of a government inspector, a new safety device is installed in an assembly-line operation. After the installation of this device, a random sample of 8 days’ output gave the following results for numbers of finished components produced: 618 660 638 625 571 598 639 582 Management is concerned about the variability of daily output and views any variance above 500 as undesirable. Test, at the 10% significance level, the null hypothesis that the population variance for daily output does not exceed 500 n=8 Testing thenullhypothesis H 0 : σ 2 ≤σ 0 2 = 500 against the alternative H 1 : σ 2 > 500 From the data provided, s 2 = 933.98 χ n − 1 2 = ( n − 1 ) s 2 σ 0 2 = 7 x 933.98 500 = 13.076 χ n − 1 , α 2 = χ 7,0.10 2 = 12.017

t = d S d √ n = − 10 30 5 = 1.67 Since t < t n − 1 ,α / 2 (1.67 < 2.064) and t > - t n − 1 ,α / 2 ( 1.67 >− 2.064 ) Hypothesis H 0 is accepted. We can conclude that these two different production processes have similar mean numbers of units produced per hour. c) S d = 15 t = d S d √ n = − 10 15 5 = 3.33 Since t > t n − 1 ,α / 2 (3.33 > 2.064) => Hypothesis H 0 is rejected. We can conclude that these two different production processes have different mean numbers of units produced per hour. d) S d = 40 t = d S d √ n = − 10 40 5 = 1.25 Since t < t n − 1 ,α / 2 (1.25 < 2.064) and t > - t n − 1 ,α / 2 ( 1.25 >− 2.064 ) Hypothesis H 0 is accepted. We can conclude that these two different production processes have similar mean numbers of units produced per hour. 10.3 Given: n = 145 d = 0.0518 S d = 0.3055 To test null hypothesis: H 0 :  x -  y = 0 H 1 :  x -  y  0 t = d S d √ n = 0.0518 0.3055 √ 145 = 0.0518 0.3055 12.0415 = 0.0518 0.0253 = 2.0474 t n − 1 ,α / 2 = t 144 , 0.025 = 1.960 Since t > t n − 1 ,α / 2 H 0 is rejected.

10.14 Test the hypotheses H 0 : P x - P y = 0 H 1 : P x - P y < 0 using the following statistics from random samples. a. ^ p x = 0.42, n x = 500; . ^ p y = 0.50, n y = 600 ^ p 0 = n x ^ p x + n y ^ p y n x + n y = 500 ( 0.42 ) + 600 ( 0.50 ) 500 + 600 = 510 1100 = 0.464 Z = ( ^ p x − ^ p y ) √ ^ p 0 ( 1 − ^ p 0 ) n x + ^ p 0 ( 1 − ^ p 0 ) n y = 0.42 − 0.5 √ 0.464 ( 1 − 0464 ) 500 + 0.464 ( 1 − 0.464 ) 600 = − 0.08 0.031 =− 2.581 If -2.581 > z α 2 ∨− 2.581 <− z α 2 ¿ > The hypothesiswillbe rejected. If not ,the hypothesiswill beaccepted .