Question

In a random sample of n=100 observations we found the proportion of success to be 48% .The 95% confidence interval for the population proportion is
1. (0.449 ; 0.511)
2. (0.5 ; 0.95)
3. (0.480 ; 0.501)
4. (0.944 ; 0.151)
5. (0.48 ; 0.031)

Answer 1

n=100

$\hat{p}=0.48$

$\alpha=0.05$

$n\hat{p}=48,n(1-\hat{p})=52$ both are greater than 5. Thus, normal distribution can be used.

$CI=\hat{p}\pm{Z_{\frac{\alpha}{2}}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$Z_{0.025}=1.96$

$=0.48\pm{1.96}\sqrt{\frac{0.48(0.52)}{100}}$

={0.382;0.578}

No choice is correct. However, option one is the only choice with reasonable values since the lower limit should be less than 0.48 and the upper limit should be greater than 0.48.