# Solved-A recent sample of 200 Canadians, 100 men and 100 women, found that 121 were in favour of legalizing recreational marijuana usage

#### ByDr. Raju Chaudhari

Oct 7, 2020

A recent sample of 200 Canadians, 100 men and 100 women, found that 121 were in favour of legalizing recreational marijuana usage. Further, 75 men and 46 women supported legalization.

a. Treating these as random samples and letting $p_1$ be the proportion of men who support legalization and $p_2$ be the proportion of women who support legalization, calculate the margin of error for a 99 percent confidence interval for $p_1-p_2$.
b. Construct a 99 percent confidence interval for $p_1-p_2$.
c. Explain whether one can infer that men and women have a differing opinion on legalization.

#### Solution

Let $X_1$ men supported legalization out of $n_1$ men and $X_2$ women supported legalization out of $n_2$ women.

. Men Women
Sample size $n_1=100$ $n_2=100$
Support legalization $X_1=75$ $X_2=46$

We wish to determine $99$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.
Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{75}{100}=0.75$ and
the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{46}{100}=0.46$.

a. The margin of error for the difference $(p_1-p_2)$ is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ & = 2.58 \sqrt{\frac{0.75*(1-0.75)}{100}+\frac{0.46*(1-0.46)}{100}}\\ &= 0.1703. \end{aligned}

b. $100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned}
where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.005} = 2.58$.

$99$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.75-0.46) - 0.1703 & \leq (p_1-p_2) \leq (0.75-0.46) + 0.1703\\ 0.1197 & \leq (p_1-p_2) \leq 0.4603 \end{aligned}

Thus, $99$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(0.1197,0.4603)$.

c. Because the $99$% confidence interval does not include zero one can infer that men and women have a differing opinion on legalization.