Find the minimum sample size to estimate the population proportion for margin of

Question

Find the minimum sample size to estimate the population proportion for margin of error 0.008; 99% confidence level; p and q unknown. A sample of 125 SAT scores has a sample mean of 1522.The standard deviation is 333. Construct a 90% confidence interval of the mean SAT scores. For problems 10 and 11, use the confidence level and sample data to find a confidence interval for estimating the population mean. Round answer to the same number of decimal places as the sample mean. Test scores: n= 76; x(bar) = 46.1; standard deviation = 5.7; 98 % confidence.

Explanation / Answer

Answer to question 8:

.

Given : margin of error (E) = 0.008

Confidence level (c) = 99% or 0.99

The Z critical value for c=0.99 can be obtained from the Z table

Zc for c=0.99 is 2.575

p and q are unknown. When p and q are unknown they are taken as 0.5

That is p = q = 0.5

.

To find: Minimum sample size (n)

Formula to be used :

n = p * q * (Zc/E)^2

.

On plugging in all the values, we get

n = 0.5 * 0.5 * (2.575/0.008)^2

n = 25900.8789062

Since n = sample size and sample size can never be in decimals so we convert it to the next integeral value

Hence the minimum sample size = 25,901 [Answer]

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