Question

Commute times in the U.S. are heavily skewed to the right. We select a random sample of 45 people from the 2000 U.S. Census who reported a non-zero commute time. In this sample the mean commute time is 25.2 minutes with a standard deviation of 19.1 minutes. Required:a. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour?b. Conduct a hypothesis test at the 5% level of significance. c. What is the p-value for this hypothesis test?

Answer 1

Answer:

The mean commute time in the U.S. is less than half an hour.

Step-by-step explanation:

In this case we need to test whether the mean commute time in the U.S. is less than half an hour.

The information provided is:

 $n=45\\\bar x=25.5\\s=19.1\\\alpha =0.05$

(a)

The hypothesis for the test can be defined as follows:

H₀: The mean commute time in the U.S. is not less than half an hour, i.e. μ ≥ 30.

Hₐ: The mean commute time in the U.S. is less than half an hour, i.e. μ < 30.

(b)

As the population standard deviation is not known we will use a t-test for single mean.

Compute the test statistic value as follows:

 $t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{25.2-30}{19.1/\sqrt{45}}=-1.58$

Thus, the test statistic value is -1.58.

(c)

Compute the p-value of the test as follows:

$p-value=P(t_{(n-1)}$  

*Use a t-table.

The p-value of the test is 0.061.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

p-value = 0.061> α = 0.05

The null hypothesis will not be rejected at 5% level of significance.

Thus, concluding that the mean commute time in the U.S. is less than half an hour.