Question

Researchers are interested in determining whether more women than men prefer the beach to the mountains. In a random sample of 200 women, 45% prefer the beach, whereas in a random sample of 300 men, 52% prefer the beach. What is the 99% confidence interval estimate for the difference between the percentages of women and men who prefer the beach over the mountains?

Answer 1

The 99% confidence interval estimate for the difference between the percentage of women and men who prefer the beach over the mountains is -18.72% < (p₁ - p₂) < 4.72%

Explanation:

The given parameters are;

Sample size of the women sample, n₁ = 200

Percentage of the women that prefer beach, p₁ = 45%,
`\hat p_1` = 0.45

Sample size of the men sample n₂ = 300

Percentage of the men that prefer beach, p₂ = 52%,
`\hat p_2` = 0.52

Confidence level of the confidence interval = 99%

α = 1 - 0.99 = 0.01, therefore, α/2 = 0.005

The equation for the confidence interval for the difference between two proportions is given as follows;

`\left (\hat{p}_(1) - \hat{p}_(2) \right )\pm z_(\alpha /2)\sqrt{\frac{\hat{p}_(1)(1-\hat{p}_(1))}{n_(1)}+\frac{\hat{p}_(2)(1-\hat{p}_(2))}{n_(2)}}`

`z_(\alpha /2)` = ±2.57 from the z score tables

Plugging in the vales, we have;

`\left (0.45 - 0.52 \right )\pm 2.57\sqrt{(0.45(1-0.45))/(200)+(0.52(1-0.52))/(300)}`

Which gives;

-0.1872 <
`(\hat p_1 - \hat p_2)` < 0.0472

The 99% confidence interval estimate for the difference between the percentage of women and men who prefer the beach over the mountains = -18.72% < (p₁ - p₂) < 4.72%.