Answer to Question #343682 in Statistics and Probability for Nipuli

Question

A manufacturer of light bulbs claims that its light bulbs have a mean life of 2000 hours with a standard deviation of 2. A random sample of 200 such bulbs is selected for testing. If the sample produces a mean value of 1992 hours and a sample standard deviation of 81, is there sufficient evidence to claim that the mean life is significantly less than the manufacturer claimed at 5% significant level

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge2000$

$H_1:\mu<2000$

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 $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1992-2000}{2/\sqrt{200}}\approx-56.5685$$

Since it is observed that $z=-56.5685<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z<-56.5685)=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 2000, at the $\alpha = 0.05$ significance level.