Questions: SAT scores: A college admissions officer takes a simple random sample of 100 entering freshmen and computes their mean mathematics SAT score to be 469. Assume the population standard deviation is σ=119. Part 1 of 4 (a) Construct a 99.5% confidence interval for the mean mathematics SAT score for the entering freshman class. Round the answer to the nearest whole number. A 99.5% confidence interval for the mean mathematics SAT score is 436.0 μ<502.0 . Part 2 of 4 (b) If the sample size were 120 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 smaller, since an increase in the sample size will decrease the standard error. Part 3 of 4 (c) If the confidence levels were 99.8% rather than 99.5%, would the margin of error be larger or smaller than the result in part (a)? Explain. The margin of error would be (Choose one) larger since (Choose one) in the confidence level will (Choose one) the (Choose one)

Solution Steps

To construct a \(99.5\%\) confidence interval for the mean mathematics SAT score, we use the formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 469\) (sample mean)
• \(\sigma = 119\) (population standard deviation)
• \(n = 100\) (sample size)
• \(z\) is the critical value for \(99.5\%\) confidence level, which is approximately \(2.807\).

Calculating the margin of error:

\[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 2.807 \cdot \frac{119}{\sqrt{100}} = 33.4037 \]

Thus, the confidence interval is:

\[ (469 - 33.4037, 469 + 33.4037) = (436.0, 502.0) \]

For a sample size of \(100\), the margin of error has already been calculated as:

\[ \text{Margin of Error} = 33.4037 \]

For a sample size of \(120\), we recalculate the margin of error using the same critical value \(z\):

\[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 2.807 \cdot \frac{119}{\sqrt{120}} = 30.4933 \]

For a \(99.8\%\) confidence level, the critical value \(z\) is approximately \(3.0902\). The margin of error is calculated as:

\[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 3.0902 \cdot \frac{119}{\sqrt{100}} = 36.7738 \]

Final Answer