Answer to Question #96817 in Statistics and Probability for Ojugbele Daniel

Question

the mean life of a sample of 400 fluorescent bulbs produced by a company is found to be 1570 hours with a standard deviation of 150hours. test the hypothesis that the mean life time of the bulbs produced by the company is 1600 hours at 1 percent level of significant.

Answer 1

Given that $n=400, \bar{X}=1570, \sigma=150, \mu_0=1600$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=1600$

$H_a:\mu\not=1600$

$n=400>30, \sigma$ is known $=>z-test.$

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 $\alpha=0.01,$ and the critical value for a two-tailed test is $z_c=2.58$

The rejection region for this two-tailed test is $R=\lbrace z:|z|>2.58\rbrace.$

The z-statistic is computed as follows:

$$Z={\bar{X}-\mu_0 \over \sigma/\sqrt{n}}={1570-1600 \over 150/\sqrt{400}}=-4$$

Since it is observed that $|z|=4>2.58=z_c$, it is then concluded that the null hypothesis is rejected.  

Using the P-value approach:

$$P(Z\leq-4)=0.00003167$$

Two-tailed test

$$2P(Z\leq -4)=0.00006334$$

The p-value is p = 0.00006334 and since $p=0.00006334<0.01,$ it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis $H_0$ is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 1600, at the 0.01 significance level.

The 99% confidence interval is

$$1550.681<\mu<1589.319$$