Question

sat scores: a college admissions officer takes a simple random sample of 100 entering freshmen and computes their mean mathematics sat score to be 470. assume the population standard deviation is σ = 116.
(a) construct a 80% confidence interval for the mean mathematics sat score for the entering freshman class. round the answer to the nearest whole number.
a 80% confidence interval for the mean mathematics sat score is < μ <.
(b) if the sample size were 125 rather than 100, would the margin of error be larger or smaller than the result in part (a)? explain.
the margin of error would be since in the sample size will the standard error.

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E = z\times\frac{\sigma}{\sqrt{n}}\), where \(z\) is the z - score (for 80% confidence level, \(z = 1.28\)), \(\sigma\) is the population standard deviation (\(\sigma=116\)), and \(n\) is the sample size.

Step2: Analyze the effect of sample size on margin of error

We have \(E = z\times\frac{\sigma}{\sqrt{n}}\). When \(n\) increases, \(\sqrt{n}\) increases. Since \(E\) is inversely proportional to \(\sqrt{n}\) (because \(E=\frac{z\sigma}{\sqrt{n}}\), with \(z\) and \(\sigma\) constant in this comparison), when \(n_1 = 100\) and \(n_2=125\) (\(n_2>n_1\)), \(\sqrt{n_2}>\sqrt{n_1}\).

Let \(E_1=z\times\frac{\sigma}{\sqrt{n_1}}\) and \(E_2 = z\times\frac{\sigma}{\sqrt{n_2}}\). Then \(\frac{E_2}{E_1}=\frac{\sqrt{n_1}}{\sqrt{n_2}}\) (since \(z\) and \(\sigma\) cancel out). Substituting \(n_1 = 100\) and \(n_2 = 125\), \(\frac{E_2}{E_1}=\sqrt{\frac{100}{125}}=\sqrt{\frac{4}{5}}\approx0.894<1\), so \(E_2

Answer:

The margin of error would be smaller. Because the margin of error formula \(E = z\times\frac{\sigma}{\sqrt{n}}\) (where \(z\) and \(\sigma\) are constant in this context) shows that margin of error is inversely proportional to the square - root of the sample size. As the sample size \(n\) increases from \(100\) to \(125\), \(\sqrt{n}\) increases, and thus the margin of error \(E\) decreases.