Question

2. Suppose the lifetime of a particular brand of light bulbs is normally distributed. A random sample of 27 light bulbs yields an average lifetime of 410 hours and a sample standard deviation of 67 hours. n ave irs a) Find the p-value of the test Ho: u = 432 hours vs. H1: U #432 hours. b) Test Ho:0 = 56 hours vs. H1:0 > 56 hours at a 5% level of significance. Find the power of the test in part (b) at = 84 hours.

Answer 1

n=27 ; a= 410 ; S = 67 Hoi Hi l = 432 ell 7432 Test statistic = T - ll slu 410-432 671/27 z - 1.7062 p-value- Plt & -107062] = 0.0999 6) Ho: s=56 Hii o > 56 2=0.05 or 02 = 3136 L-0.05 Ho 3 H: 82 > 3136

Test statistic is x² - 6x1) 4² xht x² = 26x672 = 37.217 3136

p-value=0.07 so, we fail to reject a since, p-value >c. the will hypothesis- .. we have sbart that enough evidences to r= 56.