Answer to Question #258260 in Statistics and Probability for Vijay

Question

an electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hours and the standard deviation of 40 hours. test the hypothesis at .05 level of significance that mean= 800 hours against the alternative mean=/ 800 hours if the random sample of 30 bulbs has an average life of 788 hours. Use 5% and 1% level of significance ; and critical region approach and p-value approach

Answer 1

$H_0: \mu = 800 \\ H_1: \mu ≠ 800 \\ n=30 \\ \bar{x} = 788 \\ \sigma= 40$

Test-statistic

$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \\ Z = \frac{788-800}{40 / \sqrt{30}} \\ Z = \frac{-12}{7.302967} \\ Z = -1.6433$

The test is two-tailed

α=0.01

Reject the null hypothesis if Z ≤ - 2.575 or Z ≥ 2.575

$Z= -1.6433 >Z_{crit} = -2.575$

We accept the null hypothesis at 0.01 level of significance.

α=0.05

Reject the null hypothesis if $Z_{crit} ≤ -1.96$ or $Z_{crit} ≥ 1.96$

$Z= -1.6433 > Z_{crit}$

We accept the null hypothesis at 0.05 level of significance.

P-value:

$P = 2P(Z≥|Z|) \\ = 2P(Z≥ |-1.64|) \\ = 2P(Z< -1.64)$

Using normal table

$P = 2P(Z< -1.64) \\ = 2(0.0505) \\ = 0.1002$

Decision:

P>0.05 and P>0.01

P-value of the test is 0.1 which is greater than the assumed level of significance 0.05 or 0.01

We will accept the null hypothesis.

Therefore, we can conclude that the average lifetime of each electric bulb is 800 hours.