Answer in Statistics and Probability for Blessing #248633
Given “n=400, p=0.25, q=1-p=1-0.25=0.75.”
“np=400(0.25)=100geq10, nq=400(0.75)=300geq10”
Then “hat{p}” has approximately a normal distribution with “mu_{hat{p}}=p=0.25” and “sigma_{hat{p}}=sqrt{dfrac{pq}{n}}=sqrt{dfrac{0.25(0.75)}{400}}=dfrac{sqrt{3}}{80}.”
Then
“P(hat{p}<0.18)=P(Z<dfrac{0.18-0.25}{sqrt{3}/80})”
“approx P(Z<-3.23316)approx0.000612”
The sample proportion is computed as follows, based on the sample size “N = 400” and the number of favorable cases “X = 152:”
“hat p = displaystyle frac{X}{N} = displaystyle frac{152}{400} = 0.38”
The critical value for “alpha = 0.1” is “z_c = z_{1-alpha/2} = 1.6449.”
The corresponding confidence interval is computed as shown below:
“CI(Proportion)=(hat{p}-z_csqrt{dfrac{hat{p}(1-hat{p})}{n}},nu200bnu200b”
“hat{p}+z_csqrt{dfrac{hat{p}(1-hat{p})}{n}})”
“=(0.38-1.6449sqrt{dfrac{0.38(1-0.38)}{400}},”
“0.38+1.6449sqrt{dfrac{0.38(1-0.38)}{400}})”
“=(0.34, 0.42)”
Therefore, based on the data provided, the “90%” confidence interval for the population proportion is “0.34 < p < 0.42,” which indicates that we are “90%” confident that the true population proportion “p” is contained by the interval “(0.34, 0.42).”
