Answer to Question #339698 in Statistics and Probability for Agl

Question

2. A light-bulb manufacturer regularly advertises that his bulbs last 900 hours

with a standard deviation of 75 hours. A random sample is chosen before each

campaign to make sure that the claim is correct. If one such sample of 20 bulbs

show a mean of 925 hours, can the advertising claim be considered an

underestimate at the 0.05 level of significance?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 900$

$H_a:\mu>900$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

The rejection region for thisright-tailed test is $R = \{z: z >1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{925-900}{75/\sqrt{20}}\approx1.4907$$

Since it is observed that $z = 1.4907 <1.6449= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z>1.4907)=0.06802,$ and since $p=0.06802>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is greater than 900, at the $\alpha = 0.05$ significance level.