Answer to Question #274860 in Statistics and Probability for Sunny

Question

The manufacturer of television tube know from the past experience that the average life of tube is 2000 hrs with a s.d. of 200 hrs a simple of 100 tubes has an average life of 1950 hrs test at the 0.01 level of significance to see if this sample come from a normal population of mean 2000 hrs.

Answer 1

In this question, we determine whether the population mean$(\mu)$ is equal to 2000hrs as claimed. To do so, we shall test the following hypotheses,

$H_0:\mu=2000$

$Against$

$H_1:\mu\not=2000$

We are given the following information,

$n=100$

$\bar{x}=1950$

$\sigma=200$ and $\alpha=0.01$

To perform this test, we shall use the standard Normal distribution since our sample size is large. We proceed as follows.

The test statistic is given as,

$Z^*=(\bar{x}-\mu)/(\sigma/\sqrt{n})$

$Z^*=(1950-2000)/(200/\sqrt{100})=-50/20=-2.5$

The test statistic $Z^*$ is compare with the table value at $\alpha=0.01$ significance level. This table value is given as,

$Z_{\alpha/2}=Z_{0.01/2}=Z_{0.005}=2.575$ and the null hypothesis is rejected if $|Z^*|\gt Z_{0.005}$

Since $|Z^*|=|-2.5|=2.5\lt Z_{0.005}=2.575$, we fail to reject the null hypothesis and we conclude that there is sufficient evidence to show that this sample came from a normal population of mean 2000 hours at 1% level of significance.