Answer :
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 [tex]\mu[/tex] = mean commute time in the U.S..
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 30 minutes {means that the mean commute time in the U.S. is more than or equal to half an hour}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = 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 = [tex]\frac{27.6 -30}{\frac{19.6}{\sqrt{500} } }[/tex] ~ [tex]t_4_9_9[/tex]
= -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( [tex]t_4_9_9[/tex] < -1.645)
Since, we know that at large sample size, the t distribution follows like normal distribution, that means;
P( [tex]t_4_9_9[/tex] < -1.645) = P(Z < -1.645) = 1 - P(Z [tex]\leq[/tex] 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.
The test statistics are less than the critical value, then sufficient evidence to reject the null hypothesis. Then the mean commute time in the U.S. is less than half-hour.
What are null hypotheses and alternative hypotheses?
In null hypotheses, there is no relationship between the two phenomena under the assumption that it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.
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.
For the null hypothesis
H₀: μ ≥ 30 minutes
For the alternative hypothesis
Hₐ: μ < 30 minutes
The test statistics will be
[tex]TS = \dfrac{\overline{x} - \mu}{s/\sqrt{n}}\\\\\\TS = \dfrac{27.6 - 30}{19.6/\sqrt{500}}\\\\\\TS = -2.738[/tex]
The test statistics = -2.738
Then the p-value of the test statistics will be
p-value = P(t₄₉₉ < -1.645)
Since, we know that at large sample size, the t-distribution follows like a normal distribution that is
P(t₄₉₉ < -1.645) = P(z < -1.645)
P(t₄₉₉ < -1.645) = 1 - P(z ≥ - 1.645)
P(t₄₉₉ < -1.645) = 0.04998
Now, at a 5% significance level, the t-table gives a critical value of -1.645 qat 499 degree of freedom for a left tailed test.
Since, our test statistics are less than the critical value. Then sufficient evidence to reject the null hypothesis as it will fail in the rejection region due to which we reject the null hypothesis.
Thus, the mean commute time in the U.S. is less than half-hour.
More about the null hypotheses and alternative hypotheses link is given below.
https://brainly.com/question/9504281