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The National Assessment of Educational Progress periodically administers tests on different subjects to high school students. In 2000, the grade 12 students in the sample averaged 301 on the mathematics test; the SD was 30. Can you say what the likely size of the chance error in the 301 is if a simple random sample of 1000 students was tested?

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Let $X_1, X_2,\dots,X_1000$ be the achievement results in the sample, and let $$Y=\frac{X_1+X_2+\cdots+X_{1000}}{1000}.$$ Then the expectation of $Y$ is $301$. The variance of $Y$ is $\frac{1}{1000}$ times the variance of $X$. So the standard deviation of $Y$ is $\dfrac{30}{\sqrt{1000}}$, approximately $0.9847$.

Moreover, because $Y$ is an average of identically distributed essentially independent random variable, we can assume that $Y$ has a nearly normal distribution. (We are presumably not sampling without replacement, but the student population is probably so large in comparison with the sample size that the variance is about the same as for sampling with replacement.)

It seems that you are being asked about how well the sample mean approximates the population mean $301$. More precisely, or imprecisely, you are asked for the "likely" size of the chance error.

Depends what one means by likely. For example, with probability about $0.9$, a normal is within $1.645$ standard deviation units of the true mean. In our case, $1.645$ standard deviation units is about $1.56$, so with probability about $90\%$ the sample mean will be within distance $1.56$ of the population mean of $301$.

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