Chapter 8.5, Problem 29P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 29P

The level of significance is 0.05.

The null hypothesis is $H_0:p_1=p_2$.

The alternative hypothesis is $H_1:p_1\ne p_2$.

Explanation of Solution

Calculation:

Let $p_1$ denotes the population proportion of women who favor spending more federal tax dollars on the arts, $p_2$ denotes the population proportion of women who favor spending more federal tax dollars on the arts.

From the given information the value of $\alpha$ is 0.05, and there is difference between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts.

Hence, the level of significance is 0.05.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:p_1=p_2$

Alternative hypothesis:

$H_1:p_1\ne p_2$

(b)

To determine

Identify the sampling distribution to be used.

Mention the assumption to test.

Find the sample test statistic.

Find the z value.

Answer to Problem 29P

The sampling distribution to be used is normal distribution.

The sample test statistic is –1.13.

The z value is $\pm 1.96$.

Explanation of Solution

Calculation:

Conditions:

Consider binomial experiment 1 with $n_1$ number of trails, $r_1$ number of success, probability of success for each trail $p_1$ and probability of failure $q_1=1-p_1$. And binomial experiment 2 with $n_2$ number of trails, $r_2$ number of success, probability of success for each trail $p_2$ and probability of failure $q_2=1-p_2$.

Also, ${\widehat{p}}_1=\frac{r_1}{n_1}$ is the sample proportion of first sample where $r_1$ is the successes out of $n_1$ trials for the first population and ${\widehat{p}}_2=\frac{r_2}{n_2}$ is the sample proportion of second sample where $r_2$ is the successes out of $n_2$ trials for the second population.

For sufficiently larger number of trails the following four conditions must be satisfied to use sample test statistic ${\widehat{p}}_1-{\widehat{p}}_2$ with z value are,

• $n_1\overline{p}>5$
• $n_1\overline{q}>5$
• $n_2\overline{p}>5$
• $n_2\overline{q}>5$

Where, $\overline{p}=\frac{r_1+r_2}{n_1+n_2}$ is the pooled estimate of proportion, with total number of successes $\left(r_1+r_2\right)$ and total number of trials $\left(n_1+n_2\right)$.

$\overline{q}=1-\overline{p}$ is the pooled estimate of proportion of failure.

The two sample z statistic for proportion is,

$z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_1}}}$

In the formula,

${\widehat{p}}_1=\frac{r_1}{n_1}$ is the sample proportion of first sample where $r_1$ is the successes out of $n_1$ trials for the first population.

${\widehat{p}}_2=\frac{r_2}{n_2}$ is the sample proportion of first sample where $r_2$ is the successes out of $n_2$ trials for the first population.

$\overline{p}=\frac{r_1+r_2}{n_1+n_2}$ is the pooled estimate of proportion, with total number of successes $\left(r_1+r_2\right)$ and total number of trials $\left(n_1+n_2\right)$.

$\overline{q}=1-\overline{p}$ is the pooled estimate of proportion of failure.

The pooled probability of success for the two experiments is,

$\begin{array}{c}\overline{p}=\frac{59+56}{220+175}\\ =\frac{115}{395}\\ =0.2911\end{array}$

The pooled probability of success for the two experiments is 0.2911.

Checking conditions:

$\begin{array}{c}{n}_1\overline{p}=220\left(0.2911\right)\\ =64.042\\ >5\end{array}$

$\begin{array}{c}{n}_2\overline{p}=175\left(0.2911\right)\\ =50.9425\\ >5\end{array}$

$\begin{array}{c}{n}_1\overline{q}=n_1\left(1-\overline{p}\right)\\ =220\left(1-0.2911\right)\\ =220\left(0.7089\right)\\ =155.958\\ >5\end{array}$

$\begin{array}{c}{n}_2\overline{q}=n_2\left(1-\overline{p}\right)\\ =175\left(1-0.2911\right)\\ =175\left(0.7089\right)\\ =124.0575\\ >5\end{array}$

It can be observed that two of the conditions $n_1\overline{p}>5$, $n_2\overline{p}>5$, $n_1\overline{q}>5$, $n_1\overline{q}>5$ are satisfied. It is appropriate to use normal distribution.

Hence, distribution of the sample test statistic normal distribution.

The value of ${\widehat{p}}_1-{\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_1-{\widehat{p}}_2=\frac{59}{220}-\frac{56}{175}\\ =0.268-0.32\\ =-0.052\end{array}$

The value of ${\widehat{p}}_1-{\widehat{p}}_2$ is –0.052.

Z-statistic:

Substitute ${\widehat{p}}_1$ as 0.268, ${\widehat{p}}_2$ as 0.32, $\overline{p}$ as 0.2911, $\overline{q}$ as 0.7089, $n_1$ as 220, and $n_2$ as 175 in the test statistic formula

$\begin{array}{c}z=\frac{0.268-0.32}{\sqrt{\frac{0.2911\times 0.7089}{220}+\frac{0.2911\times 0.7089}{175}}}\\ =\frac{-0.052}{0.0460}\\ =-1.13\end{array}$

Hence, the sample test statistic is –1.13.

Critical value:

Use the Appendix II: Tables, Table 5 (c): Hypothesis Testing, Critical Values $z_0$.

• In the row of level of significance locate the level of significance $\alpha =0.05$.
• Locate the corresponding critical value $z_0$ for a two-tailed test for $\alpha =0.05$.
• The intersecting value of row and columns is $\pm 1.96$.

Hence, the z value is $\pm 1.96$.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 29P

The P-value is 0.2584.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tails, for the region of the curve to shade.
• Enter the X value as –1.13.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.1292 which is one sided value.

The P-value for the two-tailed value is,

$\begin{array}{c}P\text{-value}=2\times 0.1292\\ =0.2584\end{array}$

Hence, the P-value of the sample test statistic is 0.2584.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.05 or not.

Answer to Problem 29P

The null hypothesis is not rejected.

The data is not statistically significant at level 0.05.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.2584.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.2584 and the level of significance is 0.05.

The P-value is greater than the level of significance.

That is, $0.2584\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is not rejected.

Hence, the data is not statistically significant at level 0.05.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is not rejected. This shows that, there is no sufficient evidence that there is difference between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts at level of significance 0.05.