Answer to Question #346753 in Statistics and Probability for kath

Question #346753

Compute the population proportion interval estimate given n, p, and the confidence level


a. Confidence Level= 99%, p=0.4, n=40

b. Confidence Level= 90%, p=0.15, n=55


1
Expert's answer
2022-06-02T13:54:38-0400

a. The critical value for α=0.01\alpha = 0.01 is zc=z1α/2=2.5758.z_c = z_{1-\alpha/2} = 2.5758.

The corresponding confidence interval is computed as shown below:



CI(Proportion)=(p^zc×p^(1p^)n,CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},p^+zc×p^(1p^)n)\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})=(0.42.5758×0.4(10.4)40,=(0.4-2.5758\times \sqrt{\dfrac{0.4(1-0.4)}{40}},0.4+2.5758×0.4(10.4)40)0.4+2.5758\times \sqrt{\dfrac{0.4(1-0.4)}{40}})=(0.20048,0.59952)=(0.20048, 0.59952)

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.20048<p<0.59952,0.20048 < p < 0.59952, which indicates that we are 99% confident that the true population proportion pp is contained by the interval (0.20048,0.59952).(0.20048, 0.59952).


b. The critical value for α=0.10\alpha = 0.10 is zc=z1α/2=1.6449.z_c = z_{1-\alpha/2} =1.6449.

The corresponding confidence interval is computed as shown below:



CI(Proportion)=(p^zc×p^(1p^)n,CI(Proportion)=(\hat{p}-z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},p^+zc×p^(1p^)n)\hat{p}+z_c\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})=(0.151.6449×0.15(10.15)55,=(0.15-1.6449\times \sqrt{\dfrac{0.15(1-0.15)}{55}},0.15+1.6449×0.15(10.15)55)0.15+1.6449\times \sqrt{\dfrac{0.15(1-0.15)}{55}})=(0.06975,0.23025)=(0.06975, 0.23025)

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.06975<p<0.23025,0.06975 < p < 0.23025, which indicates that we are 90% confident that the true population proportion pp is contained by the interval (0.06975,0.23025).(0.06975, 0.23025).

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