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. a) Find the p-value of the test Ho: u = 432 hours vs. H1 : ! #432 hours. b) Test Ho:0= 56 hours vs. H1:0> 56 hours rs rs at a 5% level of significance. c) Find the power of the test in part (b) at o = 84 hours.

Answer 1

a)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 432 Ha: u432 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is te = 2.056 The rejection region for this two-tailed test is R = {t \t| > 2.056} (3) Test Statistics The t-statistic is computed as follows: X - Ho t= 8/ n 410 432 = -1.706 67//27 (4) Decision about the null hypothesis Since it is observed that t= 1.706< tc = 2.056, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p 0.0999, and since p = 0.0999 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 432, at the 0.05 significance level.

b)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Но: 0? — 3136 На: >3136 This corresponds to a right-tailed test, for which a Chi-Square test for one population variance will be used (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the {x2 : 38.885 the rejection region for this right-tailed test is R (3) Test Statistics The Chi-Squared statistic is computed as follows: (п — 1)82 (27 - 1) 4489 37.217 02 3136 (4) Decision about the null hypothesis 37.217 Xi Since it is observed that x2 that the null hypothesis is not rejected. 38.885, it is then concluded (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance o2 is greater than 3136, at the 0.05 significance level.

c)

power = probability to reject the null when null is false

P(TS > 38.885 | sigma^2 = 84^2)

= P(X^2 > (38.885* 56^2/ 84^2)

= P(X^2 > 17.2822 )

=chisq.dist.rt(17.2822,27-1)

=0.90031