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:
[tex]CI = (p_1 - p_2) \pm z * \sqrt{\frac{p_1(1 - p_1}{n_1} + \frac{p_2(1 - p_2)}{n_2}}[/tex]
Substitute known values in the above equation
[tex]CI = (0.814 - 0.618) \pm 1.96 * \sqrt{\frac{0.814 * (1 - 0.814)}{93} + \frac{0.618 * (1 - 0.618)}{80}}[/tex]
Evaluate
[tex]CI = 0.196 \pm 1.96 * \sqrt{0.001628 + 0.00295095}[/tex]
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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