Chapter 8, Problem 3RE

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 3RE

Null hypothesis: $H_0:p=0.6$.

Alternate hypothesis: $H_1:p\ne 0.6$.

Explanation of Solution

In a poll, of 200 students, 134 reports that they log in to Facebook daily

Denote p as the true population proportion of the students who log in to Facebook daily.

The given test hypotheses are:

Null hypothesis:

 $H_0:p=0.6$.

That is, the proportion of the students who log in to Facebook daily is 0.60.

Alternate hypothesis:

$H_1:p\ne 0.6$.

That is, the proportion of the students who log in to Facebook daily is different from 0.6.

b.

To determine

Find the value of test statistic.

Answer to Problem 3RE

The value of test statistic is 2.02.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• • The samples taken from the population are simple random samples.
• • The population is at least 20 times as large as the sample.
• • The samples in the population are divided into two categories.
• • The values of $n{\widehat{p}}_0$ and $n{\widehat{q}}_0$ should be greater than or equal to 10.

Requirement check:

• • The sample of 200 students is simple random samples.
• • The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• • The samples in the population seemed to be categorized into two parts. That is, students who log in and students who do not log in.
• • Verify the condition: $n{\widehat{p}}_0\ge 10$ and $n{\widehat{q}}_0\ge 10$

  Here,

  $\begin{array}{c}{\widehat{p}}_0=\frac{134}{200}\\ =0.67\end{array}$

  Substitute n as 200 and ${\widehat{p}}_0$ as 0.67,

  $\begin{array}{c}n{\widehat{p}}_0=200\left(0.67\right)\\ =134\\ >10\end{array}$

  Substitute n as 200 and ${\widehat{q}}_0$ as $0.33\left(=1-{\widehat{p}}_0\right)$,

  $\begin{array}{c}n{\widehat{q}}_0=200\left(0.33\right)\\ =66\\ >10\end{array}$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$,

Where, $\widehat{p}$ be the sample proportion, $p_0$ be the population proportion and n be the sample size.

Software procedure:

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

• • Choose Stat > Basic Statistics > 1 Proportion.
• • Choose Summarized data.
• • In Number of events, enter 134. In Number of trials, enter 200.
• • Check Perform hypothesis test. In Hypothesized proportion, enter 0.6.
• • Click Options. Under Alternative, and choose $\mathrm{Pr}\text{oportion}\ne \text{Hypothesized proportion}$.
• • Click OK in each dialog box.

The output using Minitab is given below:

From the MINITAB output, the test statistic, that is, the z-value is 2.02.

Thus, the value of test statistic z is 2.02.

c.

To determine

Decide whether the null hypothesis $H_0$ is rejected at $\alpha =0.05$ level.

Answer to Problem 3RE

The null hypothesis $H_0$ is rejected at $\alpha =0.05$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 2.02.

From the given hypotheses, the alternative hypothesis contains the not equal $(\ne )$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for two-tailed test at $\alpha =0.05$ level are $\pm 1.96$.

Decision based on the critical value method:

For left-tailed test:  If $z\le -z_{\alpha }$, reject $H_0$.

For right-tailed test:  If $z\ge z_{\alpha }$, reject $H_0$.

For two-tailed test:  If $z\le -z_{\frac{\alpha }{2}}\text{or}z\ge z_{\frac{\alpha }{2}}$, reject $H_0$.

Conclusion:

The critical value at $\alpha =0.05$ level is $\pm 1.96$ and the test statistic is 2.02.

Here, the test statistic value of 2.02 lie in the critical region.

That is, $z\left(=2.02\right)>z_{\frac{\alpha }{2}}\left(=1.96\right)$.

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

Answer to Problem 3RE

The conclusion is that, there is evidence that the proportion of the students who log in to Facebook daily is different from 0.6.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who log in to Facebook daily is different from 0.6.