Question

The useful lifetime of a particular battery is a normally distributed random variable with mean 40 hours and known standard deviation σ-4 hours. A new chemical composition is introduced aiming to increase the useful lifetime of the battery. It is assumed that the new chemical composition does not affect the standard deviation. After implementing the new process, a random sample of n 100 batteries has a sample mean useful lifetime of 40.9 hours. Has the new process increased the useful lifetime of the battery? What is the p-value? Choose the correct statement out of the following Select one: p-value 0.0122: reject null hypothesis at the 5% significance level but not 1% level p-value 0.0122: reject null hypothesis at both 5% and 1% significance levels p-value = 0.0244: reject null hypothesis at the 5% significance level but not 1% level

Answer 1

H_0: mu = 40 hours H_a: mu > 40 hours X bar = 40.9, n = 100, sigma = 4 Test statistic, Z = X bar - mu/sigma/squareroot n = 40.9 - 40/4/squareroot 100 = 2.25 therefore p - value = 0.0122 (right - tailed) option (A) p - value = 0.0122: reject null hypothesis at the 5% significance level but not 1% level.