What is the smallest sample size that guaranteesfthat the margin of error is less than 28 when constructing a $98 \%$ confidence interval for a population mean with the assumed population standard deviation of 551.05 ?
(Round UP to the nearest integer)
Answer
However, since we can't have a fraction of a sample, we need to round up to the nearest whole number. Therefore, the smallest sample size that guarantees a margin of error less than 28 is \(\boxed{2103}\).
Step 1 ：We are given that the margin of error (E) is 28, the population standard deviation (σ) is 551.05, and the Z-score (Z) for a 98% confidence interval is 2.33.
Step 2 ：We want to find the smallest sample size (n) that guarantees a margin of error less than 28. We can use the formula for the margin of error to solve for n: \[n = \left(\frac{Z \cdot \sigma}{E}\right)^2\]
Step 3 ：Substituting the given values into the formula, we get: \[n = \left(\frac{2.33 \cdot 551.05}{28}\right)^2\]
Step 4 ：Calculating the above expression, we find that n is approximately 2102.6.
Step 5 ：However, since we can't have a fraction of a sample, we need to round up to the nearest whole number. Therefore, the smallest sample size that guarantees a margin of error less than 28 is \(\boxed{2103}\).
