Answer to Question #258572 in Statistics and Probability for deev

Question

the manufacture of television TUBes knows from the past experience thatthe average life of tubes is 2000 hrs with a s.d. of 200hrs . a sample of 100 tube has an average life of 1950 hrs. test at the 0.01 level of significance to see if this sample came from a normal population of mean 2000hrs. step by step explanation

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu=2000$

$H_1:\mu\not=2000$

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 $z_c=2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{1950-2000}{200/\sqrt{100}}=-2.5$$

Since it is observed that $|z| = 2.5 \le 2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=2P(Z<-2.5)\approx0.012419,$ and since $p=0.012419>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $2000,$ at the $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that this sample came from a normal population of mean 2000 hrs, at the $\alpha = 0.01$ significance level.