Chapter 10, Problem 41CE

a.

To determine

State the hypotheses and a decision rule for $\alpha =0.10$.

Answer to Problem 41CE

Hypotheses:

Null hypothesis:

$H_0:{\pi }_{Men}-{\pi }_{Women}=0$

Alternative hypothesis:

$H_1:{\pi }_{Men}-{\pi }_{Women}\ne 0$

Decision rule:

• If $z_{\text{calc}}<-1.645$, then reject the null hypothesis.
• If $z_{\text{calc}}>+1.645$, then reject the null hypothesis.

Explanation of Solution

Calculation:

The given information is that, 15 out of 25 men ranked fresh fruit in their top five snack choices and 22 out of 32 women ranked fresh fruit in their top five snack choices. That is, $x_1=15$, $n_1=25$, $x_2=22$ and $n_2=32$.

Here, the claim is that there is a difference in the proportion of men and women who rank fresh fruit in their top five list of snacks.

Here, the direction of the test is two-tailed.

State the hypotheses:

Null hypothesis:

$H_0:{\pi }_{Men}-{\pi }_{Women}=0$

That is, there is no difference in the proportion of men and women who rank fresh fruit in their top five list of snacks.

Alternative hypothesis:

$H_1:{\pi }_{Men}-{\pi }_{Women}\ne 0$

That is, there is a difference in the proportion of men and women who rank fresh fruit in their top five list of snacks.

Critical value:

Software procedure:

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

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 0 and the Standard deviation as 1.
• Click the Shaded Area tab.
• Choose Probability Value and Both Tails for the region of the curve to shade.
• Enter the Probability value as 0.10.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the critical value for two-tailed test is $\pm 1.645$.

Decision rule:

• If $z_{\text{calc}}<-1.645$, then reject the null hypothesis.
• If $z_{\text{calc}}>+1.645$, then reject the null hypothesis.

b.

To determine

Find the sample proportions.

Answer to Problem 41CE

The sample proportion for men is 0.6 and the sample proportion for women is 0.6875.

Explanation of Solution

Calculation:

Sample proportion for men:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{15}{25}\\ =0.6\end{array}$

Thus, the sample proportion for men is 0.6.

Sample proportion for women:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{22}{32}\\ =0.6875\end{array}$

Thus, the sample proportion for women is 0.6875.

c.

To determine

Find the test statistic and its p-value.

State the conclusion.

Answer to Problem 41CE

The test statistic is –0.687.

The p-value is 0.492.

There is no significant difference in the proportion of men and women who rank fresh fruit in their top five list of snacks.

Explanation of Solution

Calculation:

Pooled proportion:

$\begin{array}{c}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{15+22}{25+32}\\ =\frac{37}{57}\\ =0.649\end{array}$

Thus, the pooled proportion is 0.649.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.6-0.6875}{\sqrt{0.649\left(1-0.649\right)\left[\frac{1}{25}+\frac{1}{32}\right]}}\\ =\frac{0-0.0875}{\sqrt{0.649\left(0.351\right)\left[0.04+0.03125\right]}}\\ =\frac{-0.0875}{\sqrt{\text{0}\text{.2278}\left(0.07125\right)}}\\ =-0.687\end{array}$

Thus, the test statistic is –0.687.

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.
• Enter the Mean as 0 and the Standard deviation as 1.
• Click the Shaded Area tab.
• Choose X Value and Both Tails for the region of the curve to shade.
• Enter the X value as –0.687.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the p-value for two-tailed test is $2\left(0.2460\right)=0.492$.

Decision rule:

If $p\text{-value}<\alpha$, then reject the null hypothesis.

Otherwise, do not reject the null hypothesis.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.492\right)>\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is not rejected.

Hence, there is no significant difference in the proportion of men and women who rank fresh fruit in their top five list of snacks.

d.

To determine

Check whether the normality of $p_1-p_2$ can be assumed or not.

Answer to Problem 41CE

The normality of $p_1-p_2$ can be assumed.

Explanation of Solution

Calculation:

Rule for normality:

• Rule 1: $n_1{p}_1\ge 10$
• Rule 2: $n_1\left(1-p_1\right)\ge 10$
• Rule 3: $n_2{p}_2\ge 10$
• Rule 4: $n_2\left(1-p_2\right)\ge 10$

Check the rule:

Rule 1: $n_1{p}_1\ge 10$

$\begin{array}{c}{n}_1{p}_1=25\left(0.6\right)\\ =15\left(>10\right)\end{array}$

Rule 2: $n_1\left(1-p_1\right)\ge 10$

$\begin{array}{c}{n}_1\left(1-p_1\right)=25\left(1-0.6\right)\\ =10\left(=10\right)\end{array}$

Rule 3: $n_2{p}_2\ge 10$

$\begin{array}{c}{n}_2{p}_2=32\left(0.6875\right)\\ =22\left(>10\right)\end{array}$

Rule 4: $n_2\left(1-p_2\right)\ge 10$

$\begin{array}{c}{n}_2\left(1-p_2\right)=32\left(1-0.6875\right)\\ =10\left(=10\right)\end{array}$

Since $n_1{p}_1>10$, $n_1\left(1-p_1\right)\ge 10$, $n_2{p}_2>10$ and $n_2\left(1-p_2\right)\ge 10$, the normality of the sample proportion can be assumed.