Answer to Question #309580 in Statistics and Probability for cleojj

Question

A survey of the National Capital Region finds the average commute time of employees on one way is 45 minutes. The Makati Chamber of Commerce feels that in their city it is greater and want to publicize this. They randomly selected 28 commuters and found the average is 50 minutes with a standard deviation of 6 minutes. At 0.05 level of significance.

Answer 1

Hypotheses

Ho:$\mu$=45

Ha:$\mu$>45

$\alpha$=0.05

Test statistic.

The test statistic is the z score statistic given by the formula:

z= $\frac{x-\mu}{s/\sqrt{n}}$

Rejection region.

From the normal distribution table, using 0.05 level of significance, the critical region z=1.645(for a one tailed test). Therefore, we reject the null hypothesis if the computed test statistic Z>1.645

Computed test statistic.

Given the sample mean x=50, s=6, n=28 and μ=45

Z= $\frac{50-45}{6/\sqrt{28}}$ = 4.41

Decision.

Since the computed test statistic Z=4.41>1.645, we reject the null hypothesis.

Conclusion.

At 0.05 level of significance, there is sufficient evidence to conclude that the average commute time of employees on one way is greater than 45. We may therefore recommend the Makati chamber of commerce to proceed to publicize their city.

$\mu$