Question #339673


A company that produces batteries claims that the life expectancy of their batteries is 90 hours with a standard deviation of 10 hours. A consumer interest group believes that the actual battery life expectancy of these batteries is shorter than the claimed battery life. In order to prove this, they test a random sample of 20 batteries which resulted a mean of 87 hours. Conduct a hypothesis test with a significance level of 0.05


1
Expert's answer
2022-05-13T05:00:52-0400

The following null and alternative hypotheses need to be tested:

H0:μ=90H_0:\mu=90

Ha:μ<90H_a:\mu<90

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a left-tailed test is zc=1.6449.z_c = -1.6449.

The rejection region for this left-tailed test is R={z:z<1.6449}.R = \{z: z < -1.6449\}.

The z-statistic is computed as follows:


z=xˉμσ/n=879010/201.34164z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{87-90}{10/\sqrt{20}}\approx-1.34164

Since it is observed that z=1.34164>1.6449=zc,z = -1.34164>-1.6449= z_c , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=P(Z<1.34164)=0.089856,p=P(Z<-1.34164)=0.089856, and since p=0.089856>0.05=α,p=0.089856>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is less than 90, at the α=0.05\alpha = 0.05 significance level.


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