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In a sample of males, 23 write with their left hand, and 217 do not. In another sample of females, 65 write with their left hand, and 455 do not.

It is claimed that the proportion of left-handed males is less than the proportion of right-handed females.

Null hypothesis:The rate of left-handedness among males is the same as that among females.

$H_0:p_1=p_2$

Alternate hypothesis: The rate of left-handedness among males is less than that among females.

$H_1:p_1<p_2$

The sample size$\left(n_1\right)$ is computed below:

$$\begin{gathered} n_1=23+217 \\ =240 \end{gathered}$$

The sample size$\left(n_2\right)$ is computed below:

$$\begin{gathered} n_2=65+455 \\ =520 \end{gathered}$$

Assume that, $x_1$and $x_2$are the number of males and females who write with their left hands respectively.

Let${\hat{p}}_1$ be the estimated proportion of left-handed males:

Thus,

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{23}{240} \\ =0.0958 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =0.9042 \end{gathered}$$

Let ${\hat{p}}_2$be the estimated proportion of left-handed females:

Thus,

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{65}{520} \\ =0.125 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =0.875 \end{gathered}$$

The value of the pooled sample proportion is equal to:

$$\begin{gathered} \stackrel{炉}{p}=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{23+65}{240+520} \\ =0.1158 \end{gathered}$$

Hence,

$$\begin{gathered} \stackrel{炉}{q}=1-\stackrel{炉}{p} \\ =1-0.1158 \\ =0.8842 \end{gathered}$$

The test statistic is equal to:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_1}+\frac{\stackrel{炉}{p}\stackrel{炉}{q}}{n_2}}} \\ =\frac{\left(0.0958-0.125\right)-0}{\sqrt{\frac{\left(0.1158\right)\left(0.8842\right)}{240}+\frac{\left(0.1158\right)\left(0.8842\right)}{520}}} \\ =-1.168 \end{gathered}$$

Referring to the standard normal distribution table, the critical value of z corresponding to $伪=0.01$for a left-tailed test is equal to -2.33.

The p-value for the z-score equal to -1.168 is equal to 0.01214.

a.

Since the test statistic value is greater than the critical value and the p-value is greater than 0.01, the null hypothesis is failed to reject.

There is insufficient evidence to support the claimthat the rate of left-handedness among males is less than that among females.

If the level of significance for a one-tailed test is equal to 0.01, then the corresponding confidence level to construct the confidence interval is equal to 98%.

The expression of the confidence interval is written below:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E$

The value of $z_{\frac{伪}{2}}$when is equal to 2.33.

E is the margin of error and has the following formula:

$$\begin{gathered} E=z_{\frac{伪}{2}}\sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =2.33脳\sqrt{\frac{\left(0.0958\right)\left(0.9042\right)}{240}+\frac{\left(0.125\right)\left(0.875\right)}{520}} \\ =0.0557 \end{gathered}$$

b.

Substituting the required values, the following confidence interval is obtained:

$$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E \\ (0.0958-0.125)-0.0557<p_1-p_2<(0.0958-0.125)+0.0557 \\ -0.0849<p_1-p_2<0.0265 \end{gathered}$$

Thus, the 98% confidence interval is equal to (-0.0848, 0.0264).

This confidence interval contains zero. This means there is a possibility that the two proportions will be equal.

Therefore, there is insufficient evidence to support the claimthat the rate of left-handedness among males is less than that among females.

c.

The sample proportion of left-handed males is equal to 9.58%.

The sample proportion of left-handed females is equal to 12.5%.

Thus, the proportion of left-handed males does not appear to be significantly less than the proportion of left-handed females.