Question

to estimate the mean score \\( \mu \\) of those who took the medical college admission test on your campus, you will obtain the scores of an srs of students. from published information you know that the scores are approximately normal with standard deviation about 6.4. you want your sample mean \\( \bar{x} \\) to estimate \\( \mu \\) with an error of no more than 1 point in either direction.
how large an srs do you need in order to reduce the standard deviation of \\( \bar{x} \\) to the value you found?
give your answer rounded up to the nearest whole number.
srs size:

Explanation:

Step1: Recall the formula for the standard deviation of the sample mean (standard error)

The standard deviation of the sample mean \(\bar{x}\) (also called the standard error) is given by \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. We want \(\sigma_{\bar{x}}\leq1\) (since we want the error to be no more than 1 point, and the standard error is related to the error in estimation). We know that \(\sigma = 6.4\).

Step2: Set up the inequality and solve for \(n\)

We set up the inequality \(\frac{\sigma}{\sqrt{n}}\leq1\). Substituting \(\sigma = 6.4\), we get \(\frac{6.4}{\sqrt{n}}\leq1\). To solve for \(n\), we can first multiply both sides by \(\sqrt{n}\) (assuming \(n>0\), which it is since it's a sample size) to get \(6.4\leq\sqrt{n}\). Then we square both sides of the inequality: \(n\geq(6.4)^2\). Calculating \((6.4)^2 = 40.96\). Since the sample size \(n\) must be an integer, and we need to round up to the nearest whole number (because a sample size of 40 would give \(\frac{6.4}{\sqrt{40}}\approx\frac{6.4}{6.3246}\approx1.012\), which is more than 1, so we need to round up to 41? Wait, no, wait: Wait, actually, the formula is that the margin of error (for a 95% confidence interval, but here we are just talking about the standard error, but maybe the problem is using the standard error as the measure of deviation. Wait, the problem says "reduce the standard deviation of \(\bar{x}\) to the value you found?" Wait, no, the problem says "reduce the standard deviation of \(\bar{x}\) to the value you found?" Wait, no, re - reading: "How large an SRS do you need in order to reduce the standard deviation of \(\bar{x}\) to the value you found?" Wait, no, the original problem: "You want your sample mean \(\bar{x}\) to estimate \(\mu\) with an error of no more than 1 point in either direction." So the standard error \(\sigma_{\bar{x}}\) should be at most 1. So we have \(\frac{\sigma}{\sqrt{n}}\leq1\), so \(n\geq\frac{\sigma^{2}}{1^{2}}\). So \(n\geq\frac{(6.4)^{2}}{1}=40.96\). Since we can't have a fraction of a sample, we round up to the next whole number, so \(n = 41\)? Wait, no, wait: Wait, if \(n = 40\), \(\sqrt{40}\approx6.3246\), \(\frac{6.4}{6.3246}\approx1.012\), which is more than 1. If \(n = 41\), \(\sqrt{41}\approx6.4031\), \(\frac{6.4}{6.4031}\approx0.9995\), which is less than 1. So we need to round up 40.96 to 41? Wait, no, wait, the problem says "rounded up to the nearest whole number". So \(n=\lceil40.96
ceil = 41\)? Wait, no, wait, maybe I made a mistake. Wait, the formula for the standard deviation of the sample mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). We want \(\sigma_{\bar{x}}\leq1\), so \(n\geq\frac{\sigma^{2}}{1^{2}}\). So \(\sigma = 6.4\), so \(n\geq\frac{6.4^{2}}{1}=40.96\). Since the sample size must be an integer, and we need to have \(\sigma_{\bar{x}}\leq1\), we round up 40.96 to 41. Wait, but let's check: If \(n = 41\), \(\sigma_{\bar{x}}=\frac{6.4}{\sqrt{41}}\approx\frac{6.4}{6.4031}\approx0.9995\), which is less than 1. If \(n = 40\), \(\sigma_{\bar{x}}=\frac{6.4}{\sqrt{40}}\approx\frac{6.4}{6.3246}\approx1.012\), which is more than 1. So we need to round up 40.96 to 41.

Wait, but maybe the problem is simpler: maybe it's just asking for \(n\) such that \(\frac{\sigma}{\sqrt{n}}\leq1\), so \(n\geq\sigma^{2}\). Since \(\sigma = 6.4\), \(n\geq6.4^{2}=40.96\), so we round up to 41.

Answer:

41