Chapter 10.3, Problem 5P

(a)

To determine

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

Answer to Problem 5P

Solution: The pooled probability of success is $\overline{p}=0.66\text{and}\overline{q}=0.34$.

Explanation of Solution

Calculation: Using, $r_1=45,n_1=75,r_2=70,n_2=100$

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

$\begin{array}{l}\overline{p}=\frac{r_1+r_2}{n_1+n_2}\\ \overline{p}=\frac{45+70}{75+100}\\ \overline{p}=0.66\end{array}$

Thus, the probability of success is 0.66.

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

$\begin{array}{l}\overline{q}=1-\overline{p}\\ \overline{q}=1-0.66\\ \overline{q}=0.34\end{array}$

Thus, the probability of failure is 0.34.

(b)

To determine

Whether we should use normal distribution or not.

Answer to Problem 5P

Solution: The sample test statistic 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.

(c)

To determine

The null and alternate hypothesis.

Answer to Problem 5P

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

Explanation of Solution

Since, we want to conduct a test of the claim that the probability of success of the two binomial experiments are differ. Therefore, the null hypothesis is $H_0:{\overline{p}}_1-{\overline{p}}_2=0$ v/s alternative hypothesis is $H_1:{\overline{p}}_1-{\overline{p}}_2\ne 0$.

(d)

To determine

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

Answer to Problem 5P

Solution: The difference of sample proportion ${\overline{p}}_1-{\overline{p}}_2$ is - 0.1 and the corresponding sample test statistic is z = -1.42.

Explanation of Solution

Calculation:

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

$\begin{array}{l}{\widehat{p}}_1-{\widehat{p}}_2=\frac{r_1}{n_1}-\frac{r_2}{n_2}\\ {\widehat{p}}_1-{\widehat{p}}_2=\frac{45}{75}-\frac{70}{100}\\ {\widehat{p}}_1-{\widehat{p}}_2=-0.1\end{array}$

Thus, thedifference of sample proportion ${\overline{p}}_1-{\overline{p}}_2$ is - 0.1

The sample test statistic is calculated as follows:

$\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}}}\\ z=\frac{-0.1}{\sqrt{\frac{0.66*0.34}{75}+\frac{0.66*0.34}{100}}}\\ z=-1.42\end{array}$

Thus, the sample test statistic is -1.42.

(e)

To determine

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

Answer to Problem 5P

Solution: The P-value of test statistic is approximately 0.1556.

Explanation of Solution

Calculation:

The P-value is calculated as follows:

$\begin{array}{l}P-\text{value}=\text{Area}\text{to}\text{the}\text{left}\text{of}-1.42+\text{Area to the right of 1}\text{.42}\\ P-\text{value}=P(z<-1.42)+P(z>1.42)\\ P-\text{value}=2P(z<-1.42)\end{array}$

Using table 3 from Appendix, we get

$\begin{array}{l}P-\text{value}=2(0.0778)\\ P-\text{value}=0.1556\end{array}$

Thus the P-value of the sample test statistic is approximately 0.1556.

(f)

To determine

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

Answer to Problem 5P

Solution: The $P-\text{value > }\alpha$, hence we have failed to reject $H_0$.

Explanation of Solution

The P-value is greater than the level of significance ($\alpha$) of 0.05. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$ i.e. the P-value is not statistically significant.

(g)

To determine

Interpretation for the result.

Answer to Problem 5P

Solution: We have insufficient evidence in the favor of the claim the probability of success for the two binomial experiments are different from each other.

Explanation of Solution

The P-value is greater than the level of significance ($\alpha$) of 0.05. Therefore, we don’t have enough evidence to reject the null hypothesis $H_0$ i.e. the P-value is not statistically significant. We have insufficient evidence in the favor of the claim the probability of success for the two binomial experiments are different from each other.