Question

The California Health Survey involved 51,048 randomly selected adults in California. The author selected 100 of those subjects and obtained these results. Among the 53 females, the mean height is 63.7 inches and the standard deviation is 4.5 inches. Among the 47 males, the mean is 69.2 inches and the standard deviation is 3.1 inches.

The author selected subjects from the California Health Survey and 53 were female. Construct a 95% confidence interval for the percentage of adults Californians that are female. (10 points)

1. Confidence interval (5 points)

2. Conclusion. (5 points)

Answer 1

Solution-A:

confidence interval for proportion of women,p^=x/n=53/100=0.53

z crit for 95%=1.96

95% confidence interval for proportion f adults Californians that are female

=p^-z *sqrt(p^*(1-p^)/n,p^+z *sqrt(p^*(1-p^)/n

=0.53-1.96*sqrt(0.53*(1-0.53)/100),0.53+1.96*sqrt(0.53*(1-0.53)/100)

=0.4321766, 0.6278234

=0.4321766*100, 0.6278234*100

=43.22%,62.78%

conclusion:

we are 95% confident that the true percentage   of adults Californians that are female lies in between

43.22%, and 62.78%