Suppose that scores on the mathematics part of the National Assessment of Educational Progress (NAEP) test for eighth-grade students follow a Normal distribution with standard deviation $\sigma=40$. You want to estimate the mean score within \pm 1 with $90 \%$ confidence. How large an SRS of scores must you choose? Give your answer rounded up to the nearest whole number.
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Answer
So, to estimate the mean score within \(\pm 1\) with \(90 \%\) confidence, you must choose a sample size of \(\boxed{4329}\).
Step 1 ：Suppose that scores on the mathematics part of the National Assessment of Educational Progress (NAEP) test for eighth-grade students follow a Normal distribution with standard deviation \(\sigma=40\). You want to estimate the mean score within \(\pm 1\) with \(90 \%\) confidence.
Step 2 ：First, we need to find the z-score for a \(90\%\) confidence level. A \(90\%\) confidence level corresponds to \(95\%\) in the standard normal distribution. Using a standard normal distribution table or a calculator, we find that the z-score is approximately \(1.6448536269514722\).
Step 3 ：Next, we calculate the sample size using the formula \(n = \lceil (Z \cdot \sigma / E)^2 \rceil\), where \(Z\) is the z-score, \(\sigma\) is the standard deviation, and \(E\) is the margin of error.
Step 4 ：Substituting the given values into the formula, we get \(n = \lceil (1.6448536269514722 \cdot 40 / 1)^2 \rceil = 4329\).
Step 5 ：So, to estimate the mean score within \(\pm 1\) with \(90 \%\) confidence, you must choose a sample size of \(\boxed{4329}\).
