Chapter 10, Problem 43CE

a.

To determine

State the hypotheses.

Answer to Problem 43CE

Null hypothesis:

$\underset{\_}{H_0:{\pi }_1-{\pi }_2\le 0}$

Alternative hypothesis:

$\underset{\_}{H_1:{\pi }_1-{\pi }_2>0}$

Explanation of Solution

Calculation:

The given information is that, the summary statistic for college student is $x_1=833$ and $n_1=850$. The summary statistic for young children is $x_2=692$ and $n_2=740$.

Here, the claim is that the college students are more likely than young children to eat cereal.

Assume ${\pi }_1$ is the population proportion of the college students and ${\pi }_2$ is the population proportion of the young children.

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

State the hypotheses:

Null hypothesis:

$\underset{\_}{H_0:{\pi }_1-{\pi }_2\le 0}$

That is, the college students are not more likely than young children to eat cereal.

Alternative hypothesis:

$\underset{\_}{H_1:{\pi }_1-{\pi }_2>0}$

That is, the college students are more likely than young children to eat cereal.

b.

To determine

Find the decision rule and sketch it.

Answer to Problem 43CE

The decision rule is, “reject the null hypothesis if  $z_{\text{calc}}>+1.645$, otherwise do not reject the null hypothesis”.

Explanation of Solution

Calculation:

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 Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the critical value for right-tailed test is 1.645.

Decision rule:

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

Otherwise, do not reject the null hypothesis.

c.

To determine

Find the sample proportions and z statistic.

Answer to Problem 43CE

The sample proportion for college students is 0.98 and the sample proportion for young children is 0.9351.

The test statistic is 4.51.

Explanation of Solution

Calculation:

Sample proportion for college students:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{833}{850}\\ =0.98\end{array}$

Thus, the sample proportion for college students is 0.98.

Sample proportion for young children:

$\begin{array}{c}{p}_2=\frac{x_2}{n_2}\\ =\frac{692}{740}\\ =0.9351\end{array}$

Thus, the sample proportion for young children is 0.9351.

Pooled proportion:

$\begin{array}{c}{p}_c=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{833+692}{850+740}\\ =\frac{1,525}{1,590}\\ =0.9591\end{array}$

Thus, the pooled proportion is 0.9591.

Test statistic:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{p_c\left(1-p_c\right)\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}}\\ =\frac{0.98-0.9351}{\sqrt{0.9591\left(1-0.9591\right)\left[\frac{1}{850}+\frac{1}{740}\right]}}\\ =\frac{0.98-0.9351}{\sqrt{0.9591\left(0.0409\right)\left[0.0012+0.0014\right]}}\\ =\frac{0.0449}{\sqrt{\text{0}\text{.0392}\left(0.0026\right)}}\\ =4.51\end{array}$

Thus, the test statistic is 4.51.

d.

To determine

Make a decision.

Explanation of Solution

Here, the test statistic is greater than the critical value.

That is, $z_{\text{calc}}\left(=4.51\right)>+1.645$.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to infer that the college students are more likely than young children to eat cereal.

d.

To determine

Find the p-value and interpret it.

Answer to Problem 43CE

The p-value is 0.

There is evidence to infer that the college students are more likely than young children to eat cereal.

Explanation of Solution

Calculation:

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 4.51.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the p-value is approximately 0.

Decision rule:

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

Conclusion:

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

That is, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to infer that the college students are more likely than young children to eat cereal.

e.

To determine

Check whether the normality assumption is fulfilled or not.

Answer to Problem 43CE

The normality assumption is 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: $n_1{p}_1\ge 10$

$\begin{array}{c}{n}_1{p}_1=850\left(0.98\right)\\ =833\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_1\left(1-p_1\right)=850\left(1-0.98\right)\\ =17\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_2{p}_2=740\left(0.9351\right)\\ =691.974\left(>10\right)\end{array}$

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

$\begin{array}{c}{n}_2\left(1-p_2\right)=740\left(1-0.9351\right)\\ =48.026\left(>10\right)\end{array}$

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