Question

Suppose that administrators of a large school district wish to estimate the proportion of children in the district enrolling in kindergarten who attended preschool. They took a simple random sample of children in the district who are enrolling in kindergarten. Out of 60 children sampled, 39 had attended preschool. Construct a large-sample 95% z ‑confidence interval for p , the proportion of all children enrolled in kindergarten who attended preschool. Give the limits of the confidence interval as decimals, precise to at least three decimal places. Construct a plus four 95% z ‑

Answer 1

Answer:

95% z-confidence interval for the proportion of all children enrolled in kindergarten who attended preschool is between a lower limit of 0.528 and an upper limit of 0.772.

Step-by-step explanation:

Confidence interval = p + or - zsqrt[p(1-p) ÷ n]

p is sample proportion = 39/60 = 0.65

n is the number of children sampled = 60

Confidence level (C) = 95% = 0.95

Significance level = 1 - C = 1 - 0.95 = 0.05

Divide significance level by 2 to obtain critical value (z)

0.05/2 = 0.025 = 2.5%

z at 2.5% significance level = 1.96

zsqrt[p(1-p) ÷ n] = 1.96sqrt[0.65(1-0.65) ÷ 60] = 1.96sqrt[0.2275 ÷ 60] = 1.96sqrt(3.792×10^-3) = 1.96×0.062 = 0.122

Lower limit = p - 0.122 = 0.65 - 0.122 = 0.528

Upper limit = p + 0.122 = 0.65 + 0.122 = 0.772

95% confidence interval is between 0.528 and 0.772