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?

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%

Step-by-step explanation:

The given parameters are;

Sample size of the women sample, n₁ = 200

Percentage of the women that prefer beach, p₁ = 45%, = 0.45

Sample size of the men sample n₂ = 300

Percentage of the men that prefer beach, p₂ = 52%, = 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;

= ±2.57 from the z score tables

Plugging in the vales, we have;

Which gives;

-0.1872 < < 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%.

answer: sorry

step-by-step explanation: can you tell me pls how to find the rational of two integers

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