Chapter 8.4, Problem 11E

a.

To determine

Compute the sample proportion $\widehat{p}$.

Answer to Problem 11E

The value of sample proportion $\widehat{p}$ is 0.56.

Explanation of Solution

In a simple random sample, of 75 samples, only 42 individuals are in the group of interest.

The sample proportion $\widehat{p}$ is:

$\widehat{p}=\frac{x}{n}$

Where, x be the number of observations, n be the sample size.

Substitute the corresponding values to get the proportion,

$\begin{array}{c}\widehat{p}=\frac{42}{75}\\ =0.56\end{array}$

Thus, the sample proportion $\widehat{p}$ is 0.56.

b.

To determine

Check whether the assumptions for the hypothesis test are satisfied.

Answer to Problem 11E

Yes, the assumptions for the hypothesis test are satisfied.

Explanation of Solution

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 samples.
• 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 75 is simple random samples.
• However, the information about the population is not known. 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, individuals under category of interest and individuals not under the category of interest.
• Verify the condition: $n{\widehat{p}}_0\ge 10$ and $n{\widehat{q}}_0\ge 10$

  Substitute n as 75and ${\widehat{p}}_0$ as 0.56,

  $\begin{array}{c}n{\widehat{p}}_0=75\left(0.56\right)\\ =42\\ >10\end{array}$

  Substitute n as 75 and ${\widehat{q}}_0$ as $0.44\left(=1-{\widehat{p}}_0\right)$,

  $\begin{array}{c}n{\widehat{q}}_0=75\left(0.44\right)\\ =33\\ >10\end{array}$

Therefore, all the conditions are satisfied.

c.

To determine

Find the value of test statistic.

Answer to Problem 11E

The value of test statistic is –0.71.

Explanation of Solution

Calculation:

Denote p as the true population proportion.

The given test hypotheses are:

Null hypothesis:

 $H_0:p=0.6$.

That is, the true proportion of the individuals under the category of interest is 0.6.

Alternate hypothesis:

$H_1:p\ne 0.6$.

That is, the true proportion of the individuals under the category of interest is different from 0.6.

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 find the test statistic using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Proportion.
• In Number of trials, enter the Sample size as 75 and in Number of Events, Enter Number of observations as 42.
• Enter Hypothesized proportion as 0.6.
• Select Options.
• Choose $\text{Proportion}\ne \text{hypothesized proprotion}$ in Alternative hypothesis.
• Click OK in all dialogue boxes.

The output using Minitab is given below:

From the MINITAB output, the test statistic, that is, the z-value is –0.71.

Thus, the value of test statistic z is –0.71.

d.

To determine

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

Answer to Problem 11E

The null hypothesis $H_0$ is not rejected at $\alpha =0.05$ level.

Explanation of Solution

From previous part (c), it has been found that the value of test statistic z is –0.71.

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

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

Therefore, the critical region is $z\le -1.96\text{or}z\ge 1.96$.

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 –0.71.

Here, the test statistic value of –0.71 lies within the critical region of $\pm 1.96$.

That is, $-z_{\frac{\alpha }{2}}\left(-1.96\right)<-0.71<z_{\frac{\alpha }{2}}\left(1.96\right)$.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence that the true proportion of the individuals under the category of interest is 0.6.