Answer with explanation:
Given : Sample size : n=425
The proportion of the population in Category A : [tex]0.27[/tex]
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
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 :-
[tex]E=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=(1.96)\sqrt{\dfrac{0.27(1-0.27)}{425}}\\\\=0.0422089859404\approx0.0422[/tex]
Hence, the margin of error = 0.0422
Now, the confidence interval for population proportion is given by :-
[tex]p\pm E \\\\=0.27\pm0.0422\\\\=(0.27-0.0422,\ 0.27+0.0422)\\\\=(0.2278,\ 0.3122)[/tex]
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 = [tex](0.2278,\ 0.3122)[/tex]
