Chapter 8.5, Problem 14P

(a)

To determine

Find the pooled probability of success for the two experiments.

Answer to Problem 14P

The pooled probability of success for the two experiments is 0.360.

Explanation of Solution

Calculation:

The pooled best estimate for the population probabilities of success is,

$\overline{p}=\frac{r_1+r_2}{n_1+n_2}$

Where, $r_1$ is the successes out of $n_1$ trials for the first population, and $r_2$ is the successes out of $n_2$ trials for the second population.

Substitute $r_1$ as 60, $r_2$ as 156, $n_1$ as 200, and $n_2$ as 400 in the estimate formula

$\begin{array}{c}\overline{p}=\frac{60+156}{200+400}\\ =\frac{216}{600}\\ =0.360\end{array}$

Hence, the pooled probability of success for the two experiments is 0.360.

(b)

To determine

Identify the distribution of the sample test statistic.

Answer to Problem 14P

The distribution of the sample test statistic normal distribution.

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.

Checking conditions:

$\begin{array}{c}{n}_1\overline{p}=200\left(0.36\right)\\ =72\\ >5\end{array}$

$\begin{array}{c}{n}_2\overline{p}=400\left(0.36\right)\\ =144\\ >5\end{array}$

$\begin{array}{c}{n}_1\overline{q}=n_1\left(1-\overline{p}\right)\\ =200\left(1-0.36\right)\\ =200\left(0.64\right)\\ =128\\ >5\end{array}$

$\begin{array}{c}{n}_2\overline{q}=n_2\left(1-\overline{p}\right)\\ =400\left(1-0.36\right)\\ =400\left(0.64\right)\\ =256\\ >5\end{array}$

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, the distribution of the sample test statistic normal distribution.

(c)

To determine

State the hypotheses.

Answer to Problem 14P

The hypotheses is,

Null hypothesis: $H_0:p_1=p_2$.

Alternative hypothesis: $H_1:p_1<p_2$.

Explanation of Solution

Calculation:

Let $p_1$ denotes the probability of success of the first sample, $p_2$ denotes the probability of success of the first sample.

From the given information the value of $\alpha$ is 0.05, and the claim that the probability of success for the second binomial experiment is greater than that for the first.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:p_1=p_2$

Alternative hypothesis:

$H_1:p_1<p_2$

(d)

To determine

Find the value of ${\widehat{p}}_1-{\widehat{p}}_2$.

Find the sample distribution value of ${\widehat{p}}_1-{\widehat{p}}_2$.

Answer to Problem 14P

The value of ${\widehat{p}}_1-{\widehat{p}}_2$ is –0.09.

The sample distribution value of ${\widehat{p}}_1-{\widehat{p}}_2$ is –2.16.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{\widehat{p}}_1-{\widehat{p}}_2=\frac{60}{200}-\frac{156}{400}\\ =0.3-0.39\\ =-0.09\end{array}$

Hence, the value of ${\widehat{p}}_1-{\widehat{p}}_2$ is –0.09.

Z-statistic:

Substitute ${\widehat{p}}_1$ as 0.3, ${\widehat{p}}_2$ as 0.39, $\overline{p}$ as 0.36, $\overline{q}$ as 0.64, $n_1$ as 200, and $n_2$ as 400 in the test statistic formula

$\begin{array}{c}z=\frac{0.3-0.39}{\sqrt{\frac{0.36\times 0.64}{200}+\frac{0.36\times 0.64}{400}}}\\ =\frac{-0.09}{0.0416}\\ =-2.16\end{array}$

Hence, the sample distribution value of ${\widehat{p}}_1-{\widehat{p}}_2$ is –2.16.

(e)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 14P

The P-value of the sample test statistic is 0.1676.

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 Left Tail, for the region of the curve to shade.
• Enter the X value as –2.16.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0154.

Hence, the P-value of the sample test statistic is 0.0154.

(f)

To determine

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

Answer to Problem 14P

The null hypothesis is failed to reject.

Explanation of Solution

Calculation:

From part (e), the P-value is 0.0154.

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 0.0154 and the level of significance is 0.05.

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

That is, $0.0154\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.05.

(g)

To determine

Interpret the results.

Explanation of Solution

Calculation:

From part (f), the null hypothesis is rejected. This shows that, there is sufficient evidence to conclude the probability of success for the second binomial experiment is greater than that for the first at 0.05 level of significance.