use a sample n = 500, p^ = 0.90, and a 95% confidence level to construct a confidence interval estimate of the population proportion p: Answer 0.874 < p < 0.926 How do you get that answer?

Question

I want to know how to get the answer and explain step by step please!

Answer 1

This looks like a binomial distribution where mean=np and variance=np(1-p). In this question p=sample proportion p hat.

So, with mean=500×0.9=450,  variance=500×0.9×0.1=45, standard deviation=√45=6.708 approx.

The standard error is standard deviation/n=6.708/500=0.0134 approx (I think!).

Now we need a t test critical value for 95% confidence level. The large sample size means we can use 1.96 as the critical value, which we get from t tables or normal distribution (z) tables for a significance level of 0.05.

We apply this critical value to the standard error to get the margin of error for the required confidence interval=1.96×0.0134=0.026 approx.

The test statistic is p=0.90, and the 95% confidence level means that we can apply a range of values for p: 0.90±0.026.

Therefore 0.900-0.026<p<0.900+0.026, that is, 0.874<p<0.926.