Answer to Question #344034 in Statistics and Probability for YUI

Question

the mean lifetime of a sample of 100 light tubes produced by a company is found to be 1,580 hours. Test the hypothesis at 5% level of significance that the mean lifetime of the tubes produced by the company is 1,600 hours with standard deviation of 90 hours

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1600$

$H_1:\mu\not=1600$

This corresponds to a two-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 $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1580-1600}{90/\sqrt{100}}\approx-2.2222$$

Since it is observed that $|z|=2.2222>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-2.2222)=0.02627,$ and since $p= 0.02627<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 1600, at the $\alpha = 0.05$ significance level.