Answer to Question #287766 in Statistics and Probability for Tey

Question

The quality manager of a factory wants to determine the life expectancy for manufactured lamps of his factory and found the mean is 375 hours. The standard deviation of the population is 100 hours. A sample of 64 lamps showed a mean life time of 350 hours. At the 5% level, prove that the mean life expectancy is less than 375 hours. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq375$

$H_1:\mu<375$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{350-375}{100/\sqrt{64}}=-2$$

Since it is observed that $z = -2 <-1.6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-2)=0.02275,$  p and since $p = 0.02275 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than $375,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the mean life expectancy is less than 375 hours, at the $\alpha = 0.05$ significance level.