Hi there!
1. Margin of error: 0.005; Confidence level: 99%; `p and `q unknown
The z value is 2.575 (from a z table).
Plug that into the formula:
n = [Za/2]^2 0.25/E^2
n = 2.575^2 * 0.25 / 0.005^2
n = 66306.25
round that up to:
n = 66307
2. Margin of error: two percentage points; confidence level: 99%; from prior study, `p is estimated by the decimal equivalent of 14%.
n = [Za/2]^2 p(1-p)/E^2
n = 2.575^2 * 0.14*(1-0.14) / 0.02^2
n = 1995.818
n = 1996
3. A simple random sample of 125 SAT scores has a mean of 1522. Assume that SAT scores have a standard deviation of 333. a) Construct a 95% confidence interval estimate of the mean SAT score. The z value is 1.96.
Use the formula:
mean - z*sd/sqrt(N) to mean + z*sd/sqrt(N)
1522 - 1.96*333/sqrt(125) to 1522 + 1.96*333/sqrt(125)
Using a calculator:
1463.62 to 1580.38
b) Construct a 99% confidence interval estimate of the mean SAT score. The z value is 2.575.
1522 - 2.575*333/sqrt(125) to 1522 + 2.575*333/sqrt(125)
1445.31 to 1598.69
c) Which of the preceding confidence intervals is wider? Why?
The 99% confidence interval is larger, because by making the interval larger, we gain 4% more confidence that the true population mean is within the interval.
Let me know if you have any questions,
Scott
MIT Graduate
College degree in math... proficient in all levels -- from algebra to calculus