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 90% confidence interval for p given that p Overscript ^ EndScripts equals 0.7 and n equals 110.

Answer 1

Answer: (0.63, 7.07)

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Given :$\hat{p} =0.7$ ;  $n=110$

Significance level : $1-0.90=0.1$

Critical value :  $z_{\alpha/2}=1.645$

Now, a 90% confidence interval for population proportion will be :-

$0.7\pm (1.645)\sqrt{\dfrac{0.7(1-0.7)}{110}}\\\\\approx0.7\pm0.07\\\\=(0.7-0.07,0.7+0.07)=(0.63,\ 7.07)$

Hence, a 90% confidence interval for population proportion = (0.63, 7.07)