Answer:
Explanation:
1. We can use the formula for the minimum sample size required to estimate a population proportion:
n = (z^2 * p * (1 - p)) / (E^2)
where:
z is the z-score corresponding to the desired level of confidence, which is 1.645 for a 90% confidence level.
p is the estimated population proportion, which is 0.7 based on the study.
E is the maximum error of estimate, which is 0.05 (5 percentage points).
Substituting the known values, we get:
n = (1.645^2 * 0.7 * (1 - 0.7)) / (0.05^2) ≈ 228
Therefore, the minimum sample size required to estimate the population proportion with a 90% confidence level and a maximum error of 5 percentage points is approximately 228.
Therefore, the answer is (B) 228.
2. We can use the formula for the margin of error for a confidence interval for the population mean:
E = t*(s/sqrt(n))
where:
t is the t-score corresponding to the desired level of confidence and degrees of freedom. For a 90% confidence level with 24 degrees of freedom (n-1), the t-score is 1.711 (using a t-table or calculator).
s is the sample standard deviation, which is 12.7.
n is the sample size, which is 25.
Substituting the known values, we get:
E = 1.711*(12.7/sqrt(25)) ≈ 4.3
Therefore, the margin of error is approximately 4.3.
Therefore, the answer is (D) 4.3.