Chapter 10.3, Problem 6P

Basic Computation: Testing $p_1-p_2$ For one binomial experiment, 200 binomial trials produced 60 successes. For a second independent binomial experiment. 400 binomial trials produced 156 successes. At the 5% level of significance, test the claim that the probability of success for the second binomial experiment is greater than that for the first.

(a) Compute the pooled probability of success for the two experiments.

(b) Check Requirements What distribution does the sample test statistic follow? Explain.

(c) State the hypotheses.

(d) Compute ${\hat{p}}_1-{\hat{p}}_2$ and the corresponding sample test statistic.

(c) Find the P-value of the sample test statistic.

(f) Conclude the test.

(g) Interpret the results.

To determine

To find: The pooled probability of success for the two experiments.

Answer to Problem 6P

Solution: The pooled probability of success for the two experiments is $\overline{p}=0.36\text{and}\overline{q}=0.64$.

Explanation of Solution

Calculation: Using, $r_1=60,n_1=156,r_2=200,n_2=400$

Now, the pooled probability of success is calculated as follows:

$\begin{array}{l}\overline{p}=\frac{r_1+r_2}{n_1+n_2}\\ \overline{p}=\frac{60+156}{200+400}\\ \overline{p}=0.36\end{array}$

The pooled probability of success is 0.36.

Now the value of the $\overline{q}$ is

$\begin{array}{l}\overline{q}=1-\overline{p}\\ \overline{q}=1-0.36\\ \overline{q}=0.64\end{array}$

The pooled probability of failure is 0.64.

To determine

Whether we should use normal distribution or not.

Answer to Problem 6P

Solution: The sample distribution follows standard normal distribution.

Explanation of Solution

The number of binomial trials is large enough that each of the products $n_1\overline{p},n_1\overline{q},n_2\overline{p},n_2\overline{q}$ is greater than 5. Thus, it is appropriate to use standard normal distribution.

To determine

The null and alternate hypothesis.

Answer to Problem 6P

Solution: The hypotheses are $H_0:{\overline{p}}_1-{\overline{p}}_2=0\text{and}{H}_1:{\overline{p}}_1-{\overline{p}}_2<0$

Explanation of Solution

Since, we want to conduct a test of the claim that the probability of success for the second binomial experiment is greater than that for the first. Therefore the null hypothesis is $H_0:{\overline{p}}_1-{\overline{p}}_2=0$ v/s alternative hypothesis is $H_1:{\overline{p}}_1-{\overline{p}}_2<0$.

To determine

To find: The difference of sample proportion ${\overline{p}}_1-{\overline{p}}_2$ and the corresponding sample test statistic.

Answer to Problem 6P

Solution: The difference of sample proportion ${\overline{p}}_1-{\overline{p}}_2$ is - 0.09 and the corresponding sample test statistic is z = - 2.2.

Explanation of Solution

Calculation:

The difference of ${\overline{p}}_1-{\overline{p}}_2$ is calculated as follows:

$\begin{array}{l}{\widehat{p}}_1-{\widehat{p}}_2=\frac{r_1}{n_1}-\frac{r_2}{n_2}\\ {\widehat{p}}_1-{\widehat{p}}_2=\frac{60}{200}-\frac{156}{400}\\ {\widehat{p}}_1-{\widehat{p}}_2=-0.09\end{array}$

Thus, the difference of sample proportion ${\overline{p}}_1-{\overline{p}}_2$ is - 0.09.

The sample test statistic is calculated as follows:

$\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}}}\\ z=\frac{-0.09}{\sqrt{\frac{0.36*0.64}{200}+\frac{0.36*0.64}{400}}}\\ z=-2.2\end{array}$

Thus, the sample test statistic is - 2.2

To determine

To find: The P-value of the sample test statistic.

Answer to Problem 6P

Solution: The P-value of test statistic is approximately 0.0134.

Explanation of Solution

Calculation:

The P-value is calculated as follows:

$\begin{array}{l}P-\text{value}=Areatotheleftof-2.2\\ P-\text{value}=P(z<-2.2)\end{array}$

Using table 3 from Appendix, we get

$P-\text{value}=0.0134$

Thus the P-value of the sample test statistic is approximately 0.0134.

To determine

Whether we should reject or fail to reject the null hypothesis for a 5% level of significance.

Answer to Problem 6P

Solution: The $P-\text{value < }\alpha$, hence we have to reject the null hypothesis $H_0$.

Explanation of Solution

The P-value is less than the level of significance ($\alpha$) of 0.05. Therefore we have enough evidence to reject the null hypothesis $H_0$ i.e. the P-value is statistically significant.

To determine

Interpretation for the result.

Answer to Problem 6P

Solution: We have sufficient evidence in the favor of the claim that the probability of success for the second binomial is greater than that for the first.

Explanation of Solution

The P-value is less than the level of significance ($\alpha$) of 0.05. Therefore we have enough evidence to reject the null hypothesis $H_0$ i.e. the P-value is statistically significant. We have sufficient evidence in the favor of the claim that the probability of success for the second binomial is greater than that for the first.