Answer to Question #192111 in Statistics and Probability for sagun poudel

Question

A researcher would like to test whether there is any significant difference between the proportion of

safety consciousness of men and women while driving a car. In a sample of 300 men, 130 said that

they used seat belts. In a sample of 300 women, 90 said that they used seat belts. Test the claim that

there is no significant difference between the proportion of safety consciousness of men and women

while driving a car at 5% level of significance. (Given that Z0.025 = 1.96)

Answer 1

Given,

$n_1=300\\X_1=130\\X_2=90\\n_2=300$

Let $P_1$ be the population of male use seat belts and $P_2$ be the proportion of female use seat belts.

$\hat P_1=\dfrac{X_1}{n_1}=\dfrac{130}{300}=0.4333\\\ \\\hat P_2=\dfrac{X_2}{n_2}=\dfrac{90}{300}=0.3$

Now,

$P=\dfrac{n_1\hat P_1+n_2\hat P_2}{n_1+n_2}\\P=\dfrac{(300\times0.4333)+(300\times 0.3)}{600}\\P=0.3667$

We want to test

Null Hypothesis $H_0:P_1=P_2$

Alternate Hypothesis $H_1:P_1\neq P_2$

$\therefore Test\ Statistics=\dfrac{\hat P_1-\hat P_2}{\sqrt{P(1-P)(\frac{1}{n_1}+\frac{1}{n_2})}}$

$\Rightarrow\dfrac{0.4333-0.3}{\sqrt{0.3667\times0.6333\times(\frac{1}{300}+\frac{1}{300})}}\\\ \\\Rightarrow\dfrac{0.1333}{0.0395}=3.3877$

$Z_{cal}=3.3877$

If $Z_{cal}>Z_{\alpha/2}$ we reject $H_0$

Here, $Z_{cal}>1.96$ we may reject $H_0$

Hence, There is significant difference in proportion of safety of men and women.