Question

An electrical firm manufactures light bulb that has a length of life that is approximately normally distributed with a mean of 800 hours and a standard deviation of 40 hours. The supervising electrical Engineer took a random sample of 30 bulbs with an average life of 788 hours, test the hypothesis that µ=800 hours against the alternative hypothesis µ is greater than 800.

Use confidence level of 96%.

Is it a two tailed or one tailed test?

Null hypothesis?

Alternative hypothesis?

Size of the test?

Test statistics?

Critical region?

Decision ( accept or reject null hypothesis)?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=800$

$H_1:\mu>800$

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

Size of the test is $n=30.$

Based on the information provided, the significance level is $\alpha = 0.04,$ and the critical value for a right-tailed test is $z_c = 1.7507.$

The rejection region for this right-tailed test is $R = \{z:z>1.7507\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{788-800}{40/\sqrt{30}}\approx-1.6432$$

Since it is observed that $z=-1.6432<1.7507=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z>-1.6432)=0.949829,$ and since $p=0.949829>0.04=\alpha,$ 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 800, at the $\alpha = 0.04$ significance level.