Chapter 6.5, Problem 7E

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 7.79 points, and their standard deviation is 1.06 points. The two-tailed students average 7.64 points, and their standard deviation is 1.31 points.

a.    Can you conclude that the mean score μ₁ on one-tailed hypothesis test questions is higher than the mean score μ₂ on two-tailed hypothesis test questions? State the appropriate null and alternate hypotheses, and then compute the P-value.

b.    Can you conclude that the mean score μ₁ on one-tailed hypothesis test questions differs from the mean score μ₂ on two-tailed hypothesis test questions? State the appropriate null and alternate hypotheses, and then compute the P-value.

To determine

Check whether there is evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions is higher than the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Answer to Problem 7E

There is no evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions is higher than the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Explanation of Solution

Given info:

For one tailed hypothesis tests, the number of students is $n_X=80$, the mean is, $\overline{X}=7.79\text{points}$ and ${\sigma }_X=1.06\text{points}$.

For two tailed hypothesis tests, the number of students is $n_Y=80$, the mean is, $\overline{Y}=7.64\text{points}$ and ${\sigma }_Y=1.31\text{points}$.

Calculation:

Consider ${\mu }_{\text{1}}={\mu }_{\text{X}}$ and ${\mu }_{\text{2}}={\mu }_{\text{Y}}$.

State the test hypotheses.

Null hypotheses:

$H_0:{\mu }_{\text{X}}-{\mu }_{\text{Y}}\le 0$

That is, the mean score ${\mu }_1$ on one tailed hypothesis test questions is not higher than the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Alternative hypothesis:

$H_1:{\mu }_{\text{X}}-{\mu }_{\text{Y}}>0$

That is, the mean score ${\mu }_1$ on one tailed hypothesis test questions is higher than the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

The formula for z-score is,

$z=\frac{\left(\overline{X}-\overline{Y}\right)-{\Delta }_0}{\sqrt{\frac{{\sigma }_X^2}{n_X}+\frac{{\sigma }_Y^2}{n_Y}}}$

The z-score is calculated as follows,

$\begin{array}{c}z=\frac{\left(7.79-7.64\right)-0}{\sqrt{\frac{{\left(1.06\right)}^2}{80}+\frac{{\left(1.31\right)}^2}{80}}}\\ =\frac{0.15}{\sqrt{0.0355}}\\ =\frac{0.15}{0.1884}\\ =0.80\end{array}$

Thus, the z-score is 0.80.

P-value:

Software Procedure:

Step-by-step procedure to obtain the P- value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the data value as 0.80.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value for right tail test is 0.2119.

Decision rule:

If $P\text{-value}\le \alpha \text{,}$ then reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}>\alpha \text{,}$ then fail to reject the null hypothesis $\left(H_0\right)$.

Conclusion:

Here, the P-value is 0.2119, which is not small.

That is, the P-value is larger than any level of significance.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions is higher than the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

To determine

Check whether there is evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions differs from the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Answer to Problem 7E

There is no evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions differs from the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypotheses:

$H_0:{\mu }_{\text{X}}-{\mu }_{\text{Y}}=0$

That is, the mean score ${\mu }_1$ on one tailed hypothesis test questions do not differs from the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

Alternative hypothesis:

$H_1:{\mu }_{\text{X}}-{\mu }_{\text{Y}}\ne 0$

That is, the mean score ${\mu }_1$ on one tailed hypothesis test questions differs from the mean score ${\mu }_2$ on two-tailed hypothesis test questions.

From part a., the z-score is 0.80.

P-value:

Software Procedure:

Step-by-step procedure to obtain the P- value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tails for the region of the curve to shade.
• Enter the data value as 0.80.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value for two tail test is, $0.2119+0.2119=\text{0}\text{.4238}$.

Conclusion:

Here, the P-value is 0.4238, which is not small.

That is, the P-value is larger than any level of significance.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean score ${\mu }_1$ on one tailed hypothesis test questions differs from the mean score ${\mu }_2$ on two-tailed hypothesis test questions.