Question

Please solve the following question.

Answer 1

The confidence interval for the difference p1 - p2 of the population proportion is (0.063, 0.329)

How to determine the confidence interval?

The given parameters are:

n₁ = 93; n₂ = 80

p₁ = 0.814;  p₂ = 0.618

The critical value at 95% confidence interval is

z = ±1.96

So, we have:

$CI = (p_1 - p_2) \pm z * \sqrt{\frac{p_1(1 - p_1}{n_1} + \frac{p_2(1 - p_2)}{n_2}}$

Substitute known values in the above equation

$CI = (0.814 - 0.618) \pm 1.96 * \sqrt{\frac{0.814 * (1 - 0.814)}{93} + \frac{0.618 * (1 - 0.618)}{80}}$

Evaluate

$CI = 0.196 \pm 1.96 * \sqrt{0.001628 + 0.00295095}$

This gives

CI = 0.196 ± 0.133

Expand

CI = (0.196 - 0.133, 0.196 + 0.133)

Evaluate

CI = (0.063, 0.329)

Hence, the confidence interval is (0.063, 0.329)

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