Question

Commute times in the U.S. are heavily skewed to the right. We select a random sample of 500 people from the 2000 U.S. Census who reported a non-zero commute time. In this sample the mean commute time is 27.6 minutes with a standard deviation of 19.6 minutes. Are researchers able to conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance. What is the p-value for this hypothesis test?

Answer 1

Answer:

We conclude that the mean commute time in the U.S. is less than half an hour.

Step-by-step explanation:

We are given that a random sample of 500 people from the 2000 U.S. Census is selected who reported a non-zero commute time.

In this sample the mean commute time is 27.6 minutes with a standard deviation of 19.6 minutes.

Let $\mu$ = mean commute time in the U.S..

So, Null Hypothesis, $H_0$ : $\mu \geq$ 30 minutes      {means that the mean commute time in the U.S. is more than or equal to half an hour}

Alternate Hypothesis, $H_A$ : $\mu$ < 30 minutes     {means that the mean commute time in the U.S. is less than half an hour}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

                           T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean commute time = 27.6 minutes

            s = sample standard deviation = 19.6 minutes

            n = sample of people from the 2000 U.S. Census = 500

So, the test statistics  =  $\frac{27.6 -30}{\frac{19.6}{\sqrt{500} } }$  ~ $t_4_9_9$

                                       =  -2.738

The value of t test statistic is -2.738.

Also, P-value of test statistics is given by the following formula;

           P-value = P( $t_4_9_9$ < -1.645)

Since, we know that at large sample size, the t distribution follows like normal distribution, that means;

              P( $t_4_9_9$ < -1.645)  =  P(Z < -1.645) = 1 - P(Z $\leq$ 1.645)

                                         =  1 - 0.95002 = 0.04998

Now, at 5% significance level the t table gives critical values of -1.645 at 499 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -2.378 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean commute time in the U.S. is less than half an hour.