Suppose we wish to analyze the distribution of the number of children per family

Question

Suppose we wish to analyze the distribution of the number of children per family in a population. Let µ denote the unknown true population mean number of children per family. Observations Xi have been collected on 400 families, with the following summary values. Here, X_bar is the sample mean. X1+ X2+...+X400 = 832, (X1 - X_bar)^2 + (X2 - X_bar)^2 +...+(X400 - X_bar)^2 =324. Compute a 95% confidence interval and a 99 % confidence interval for µ and compare these two estimates. Why do you think there is such a difference?

Explanation / Answer

var= sum(Xi-mean)^2/(n-1) = 324/399 = 0.81203

sd = sqrt(var) = 0.901127

mean = 832/400 = 2.08

95% CI

= mean +- z(a/2)*(sd/sqrt(n)) = mean +- z(0.05/2)*(sd/sqrt(n))

99% CI

= mean +- z(a/2)*(sd/sqrt(n)) = mean +- z(0.01/2)*(sd/sqrt(n))

My confidence interval corresponding to 99% confidence is wider than 95% confidence.Comparison:

as the level of confidence increases, the width of interval also increases. ME = Z(A/2)*(sd/sqrt(n))

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