Question

A statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.80 points, and their standard deviation is 1.06 points. The two-tailed students average 7.64 points, and their standard deviation is 1.32 points. Can you conclude that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions? State the appropriate null and alternate hypotheses, and then compute the P-value. Check all that are true.

Answer 1

Answer:

The null and alternative hypothesis are:

$H_0: \mu_X-\mu_Y=0\\\\H_a:\mu_X-\mu_Y> 0$

P-value=0.1996

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .

Then, the null and alternative hypothesis are:

$H_0: \mu_X-\mu_Y=0\\\\H_a:\mu_X-\mu_Y> 0$

The significance level is 0.05.

The sample X (one-tailed), of size nX=80 has a mean of 7.8 and a standard deviation of 1.06.

The sample Y (two-tailed), of size nY=80 has a mean of 7.64 and a standard deviation of 1.32.

The difference between sample means is Md=0.16.

$M_d=M_X-M_Y=7.8-7.64=0.16$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_X^2+\sigma_Y^2}{n}}=\sqrt{\dfrac{1.06^2+1.32^2}{80}}\\\\\\s_{M_d}=\sqrt{\dfrac{2.866}{80}}=\sqrt{0.036}=0.189$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_X-\mu_Y)}{s_{M_d}}=\dfrac{0.16-0}{0.189}=\dfrac{0.16}{0.189}=0.8453$

The degrees of freedom for this test are:

$df=n_1+n_2-1=80+80-2=158$

This test is a right-tailed test, with 158 degrees of freedom and t=0.8453, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>0.8453)=0.1996$

As the P-value (0.1996) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .