Question

In a sample of 200 men, 130 said they wore seatbelts. In a sample of 300 women, 140 said they wore seatbelts. Find a 95% confidence interval for the difference in proportion of men and women who wear seatbelts. g

Answer 1

Answer: (0.093,0.267)

Step-by-step explanation:

The confidence interval for the difference in proportion is given by :-

$(p_1-p_2)\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

Given : $n_1=200$;  $n_2=300$

Proportion of men said they wore seatbelts =$p_1=\dfrac{130}{200}=0.65$

Proportion of women said they wore seatbelts =$p_2=\dfrac{140}{300}\approx0.47$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=z_{0.025}=1.96$

Now, the 95% confidence interval for the difference in proportion of men and women who wear seatbelts will be :-

$(0.65-0.47)\pm (1.96)\sqrt{\dfrac{0.65(1-0.65)}{200}+\dfrac{0.47(1-0.47)}{300}}\\\\\approx0.18\pm0.087\\\\=(0.093,0.267)$

Hence, the 95% confidence interval for the difference in proportion of men and women who wear seatbelts = (0.093,0.267)