Question

(11) Test the claim that the average time is more than 38 minutes if a sample of 36 individuals showed a mean of 41 minutes. Assume time taken is normally distributed with a standard deviation of 10. Use a 2% level of significance.

Ho:

Ha:                                                               Tail

Decision Rule (DR) clearly showing the critical value(s) and rejection region(s).

Test Statistic (TS) or calculated value

Decision   Reject Ho   / Do not reject Ho

Conclusion in the context of the problem statement.

Answer 1

Solution: The null and the alternative hypotheses is:

$H_{a}:\mu>38$

The given test right tailed test.

Decision Rule: The decision rule is that if the test statistic is less than the critical value, then we fail to reject the null hypothesis. And if the test statistic is greater than the critical value, then we reject the null hypothesis.

The right tailed critical value at 0.02 significance level is:

$z_{critical}=2.054$

And the rejection region is:

The test statistic is given below:

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

   $=\frac{41-38}{\frac{10}{\sqrt{36}}}$

$=\frac{3}{1.6667}$

$=1.8$

Therefore, the test statistic is $z=1.8$

Decision: Do not reject H₀

Conclusion in the context of the problem statement.

Fail to reject the null hypothesis, therefore there is insufficient evidence to support the claim that the average time is more than 38 minutes.