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antiseptic1488 [7]
2 years ago
6

Please solve the following question.

Mathematics
1 answer:
insens350 [35]2 years ago
7 0

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

<h3>How to determine the confidence interval?</h3>

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)

Read more about confidence interval at:

brainly.com/question/15712887

#SPJ1

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