Chapter 9.2, Problem 41E

a.

To determine

The $P-$ value and its meaning are to be estimated.

Answer to Problem 41E

  $P-$ value is less than the significance level that means reject hypothesis.

Explanation of Solution

Given:

  $\begin{array}{l}z=2.19\\ n=200\\ \alpha =0.5\\ H_0:p=0.5\\ H_a:p>0.5\end{array}$

Calculation:

Now the probability value of the $z$ score $z=2.19$ is:

  $\begin{array}{l}P\left(x<z\right)=0.98574\\ P\left(x>z\right)=1-0.98574=0.014262\end{array}$

If the $P-$ value is less than or equal to the significance level then the null hypothesis is rejected.

If the $P-$ value is greater than the significance level that means fail to reject null hypothesis.

From the answer:

  $P=0.014262$

So, $P$ value is $0.014262$ which less than the given significance level is of $0.5$ . So, it leads to reject the null hypothesis $H_0:p=0.5$ .

Conclusion:

Hence, the null hypothesis $H_0$ is rejected.

b.

To determine

The conclusions need to interpret if the significance level changes as $\alpha =0.01$ and $\alpha =0.05$ .

Answer to Problem 41E

If the significance level changes as:

• $\alpha =0.01$: fail to reject the null hypothesis.
• $\alpha =0.05$: reject the null hypothesis

Explanation of Solution

Given:

  $\begin{array}{l}\text{}\alpha =0.01\\ \alpha =0.05\end{array}$

The higher significance level raises the levels of probability to reject the null hypothesis.

Know that:

  $P=0.014262$

So, $P$ value is $0.014262$ which less than the given significance level is of $0.5$ . So, it leads to reject the null hypothesis $H_0:p=0.5$ .

If the significance level is $\alpha =0.01$ :

  $P$ -value is $0.014262$ is greater than the given significance level is of $\alpha =0.01$ . So, it leads to fail to reject the null hypothesis $H_0:p=0.5$ .

If the significance level is $\alpha =0.05$ :

  $P$ -value is $0.014262$ is less than the given significance level is of $\alpha =0.05$ . So, it leads to reject the null hypothesis $H_0:p=0.5$ .

Conclusion:

Hence, the $P$ -value changes due to the significance level changes.

c.

To determine

The $\widehat{p}$ value is to be estimated.

Answer to Problem 41E

The $\widehat{p}$ value is $0.57774$ .

Explanation of Solution

Given:

  $\begin{array}{l}n=200\\ p=0.5\\ z=2.19\end{array}$

Concept Used:

The test statistic is given by the following expression:

  $Z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Where,

Sample proportion: $\widehat{p}$

Sample size: $n$

Calculation:

The test statistic is calculated by using the following expression:

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Put the values from given data in the above eq.

  $\begin{array}{l}z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}\\ 2.19=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{200}}}\\ 2.19=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(0.5\right)}{200}}}\\ 2.19=\frac{\widehat{p}-0.5}{0.03535}\\ \widehat{p}=\left(2.19\times 0.03535\right)+0.5\\ \widehat{p}=0.57774\end{array}$

Conclusion:

Hence, $\widehat{p}=0.57774$