Question

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 99% confidence interval for the proportion of the population in Category A given that 27% of a sample of 425 are in Category A.

Answer 1

Answer with explanation:

Given : Sample size : n=425

The proportion of the population in Category A : $0.27$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Since in normal distribution, the best estimate for proportion for population p  is equals to the proportion for the sample.

Thus ,the best point estimate for p =0.27

Formula for margin of error :-

$E=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=(1.96)\sqrt{\dfrac{0.27(1-0.27)}{425}}\\\\=0.0422089859404\approx0.0422$

Hence, the margin of error = 0.0422

Now, the confidence interval for population proportion is given by :-

$p\pm E \\\\=0.27\pm0.0422\\\\=(0.27-0.0422,\ 0.27+0.0422)\\\\=(0.2278,\ 0.3122)$

Hence, the 99% confidence interval for the proportion of the population in Category A given that 27% of a sample of 425 are in Category A = $(0.2278,\ 0.3122)$