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Gennadij [26K]
3 years ago
5

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% confide

nce interval for the difference in proportion of men and women who wear seatbelts. g
Mathematics
1 answer:
Oduvanchick [21]3 years ago
8 0

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)

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