Chapter 11.3, Problem 40E

a.

To determine

Check whether there is a convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day.

Answer to Problem 40E

The conclusion is that there is a convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day.

Explanation of Solution

Calculation:

$n_1=1,000$, $n_2=1,000$, $x_1=220$, and $x_2=40$.

Step 1:

In this context, $p_1$ denotes the proportion of teens who report that they check a social networking site more than 10 times a day and $p_2$ denotes the proportion of parents who think their teen checks a social networking site more than 10 times a day.

Step 2:

Null hypothesis:

$H_0:p_1-p_2=0$

Step 3:

Alternative hypothesis:

$H_a:p_1-p_2>0$

Step 4:

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.01$.

Step 5:

Test statistic:

$z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\frac{{\widehat{p}}_c\left(1-{\widehat{p}}_c\right)}{n_1}+\frac{{\widehat{p}}_c\left(1-{\widehat{p}}_c\right)}{n_2}}}$

Step 6:

Assumption for the two-proportion z test:

• The response variable should be categorical data.
• The samples should be collected independently.
• The sample sizes should be large. That is, in each group the number of success points and the number of failure points should be at least 10.

Assumption in this particular problem:

• From the given samples, it can be observed that the response variable is categorical, that is, teens and parents who report that they check a social networking site more than 10 times a day.
• The two samples are collected independently.
• The sample sizes are large enough. From the proportion, it is clear that the number of success points and the number of failure points are more than 10 for both groups.

The number of success points and the number of failure points for the first sample is obtained as given below:

$\begin{array}{c}{n}_1{\widehat{p}}_1=1,000\times \left(\frac{220}{1,000}\right)\\ =220\\ >10\\ n_1{\widehat{q}}_1=1,000\times \left(1-\frac{220}{1,000}\right)\\ =1,000\times \left(\frac{780}{1,000}\right)\\ =780\\ >10\end{array}$

The number of success points and the number of failure points for the second sample is obtained as given below:

$\begin{array}{c}{n}_2{\widehat{p}}_2=1,000\times \frac{40}{1,000}\\ =40\\ >10\\ n_2{\widehat{q}}_2=1,000\times \left(1-\frac{40}{1,000}\right)\\ =960\\ >10\end{array}$

Therefore, the success and failure proportions are more than 10.

Therefore, the assumptions are satisfied.

Step 7:

Test statistic:

Software procedure:

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

• Choose Stat > Basic Statistics > 2 Proportions.
• Choose Summarized data.
• In the First sample, enter Trials (n) as 1,000 and Events(x) as 220.
• In the Second sample, enter Trials as 1,000 and Events as 40.
• Choose confidence level as 99.
• Write hypothesized difference as 0.
• Choose Difference $>$ hypothesized value in alternative hypothesis.
• Choose use pooled estimate of the proportions in the test method.
• Click OK.

Output obtained using the MINITAB software is given below:

From the given MINITAB output, the value of test statistic is 11.97.

Step 8:

P-value:

From the MINITAB output, the P-value is 0.

Step 9:

Decision rule:

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

Here, the $P\text{-value}$ of 0 is less than the significance level 0.01.

That is, $P\text{-value}\left(=0\right)<\alpha \left(=0.01\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, there is a convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day.

b.

To determine

Check whether the two-sample z test is used to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted.

Answer to Problem 40E

No, the two-sample z test cannot be used to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted.

Explanation of Solution

Justification:

The given hypotheses compare the single proportion. Hence, one-sample test is used to test the hypotheses. Thus, the two-sample z test cannot be used to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted.

c.

To determine

Check whether there is convincing evidence that more than one-third of all teens have posted something on a social networking site that they later regretted.

Answer to Problem 40E

The conclusion is that there is a convincing evidence that more than one-third of all teens have posted something on a social networking site that they later regretted.

Explanation of Solution

Calculation:

$n=1,000$, and $x=390$.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes population proportion of all teens who say they have posted something on a social networking site that they later regretted.

Step 2:

Null hypothesis:

$H_0:p=\frac{1}{3}$

Step 3:

Alternative hypothesis:

$H_a:p>\frac{1}{3}$

Step 4:

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.05$.

Step 5:

Test statistic, z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$,

Where, $\widehat{p}$ is the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=\frac{1}{3}$ in the test statistic.

$z=\frac{\widehat{p}-\frac{1}{3}}{\sqrt{\frac{\frac{1}{3}\left(1-\frac{1}{3}\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion, which is selected from a random sample.
• The large sample z test can be used if the sample size n satisfies the below conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is assumed that the sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =1,000\left(\frac{1}{3}\right)\\ =\text{333}\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =1,000\left(1-\frac{1}{3}\right)\\ =1,000\left(\frac{2}{3}\right)\\ =\text{667}\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although, the population size is not known, it is reasonable to assume that the sample size of 1,000 teens as the representative for all teens are using social networking sites. It is also definite that the sample size is less than the 10% of the population of the teens are using social networking sites.

Step 7:

Software procedure:

Step-by-step procedure to obtain the test statistic using MINITAB:

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 390. In Number of trials, enter 1,000.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.5.
• Click Options. Under Alternative, and choose greater than.
• Click OK in each dialog box.

Output using MINITAB software is given below:

From the given MINITAB output, the value of test statistic is 3.82.

Step 8:

P-value:

From the MINITAB output, the P-value is 0.

Step 9:

Decision rule:

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

Conclusion:

Here, the $P\text{-value}$ of 0 is less than the significance level 0.05.

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

The decision is that the null hypothesis is rejected.

Hence, there is convincing evidence that more than one-third of all teens have posted something on a social networking site that they later regretted.