The population standard deviation is estimated to be $300. If a 98% confidence interval is used and an interval of +/-$75 is desired, how many cardholders should be sampled?
98% confidence interval = ± 2.33 Standard Error of the mean (SEm) = ± 75
SEm = SD/√(n-1)
(If n is large, do not need to subtract 1.)
I'll let you do the calculations.
To determine the sample size needed for a given confidence interval and desired level of precision, we can use the formula:
\(n = \left(\frac{Z \cdot \sigma}{E}\right)^2\)
Where:
- \(n\) is the sample size
- \(Z\) is the z-score corresponding to the desired confidence level
- \(\sigma\) is the estimated population standard deviation
- \(E\) is the desired margin of error
In this case, the desired confidence level is 98%, which corresponds to a z-score of approximately 2.33 (referencing a standard normal distribution table). The estimated population standard deviation is given as $300, and the desired margin of error is +/-$75.
Substituting these values into the formula:
\(n = \left(\frac{2.33 \cdot 300}{75}\right)^2\)
\(n = \left(\frac{699}{75}\right)^2\)
\(n \approx 8.792^2\)
\(n \approx 77.18\)
Since we cannot have a non-integer sample size, we need to round up to the nearest whole number. Hence, we would need to sample at least 78 cardholders to achieve a 98% confidence interval with a margin of error of +/-$75.