SAT scores: A college admissions officer takes a simple random sample of 100 entering freshmen and computes their mean mathematics SAT score to be 452. Assume the population standard deviation is \( \sigma=115 \). Part: \( 0 / 4 \) Part 1 of 4 (a) Construct a \( 95 \% \) confidence interval for the mean mathematics SAT score for the entering freshman class. Round the answer to the nearest whole number. A \( 95 \% \) confidence interval for the mean mathematics SAT score is \( 429<\mu<475 \). Part: \( 1 / 4 \) Part 2 of 4 (b) If the sample size were 110 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 \( \square \) smaller , since \( \square \) an increase in the sample size will \( \square \) decrease the standard error. Part: \( 2 / 4 \) Part 3 of 4 (c) If the confidence level were \( 80 \% \) rather than \( 95 \% \), would the margin of error be larger or smaller than the result in part (a)? Explain. The margin of error would be \( \square \) (Choose one) \( \bar{V} \), since (Choose one) in the confidence level will \( \square \) (Choose one) the critical value \( z_{\alpha / 2} \). larger smaller \( \square \) \( \square \)

Let's go through each part of the problem step by step. ### Part 1 of 4 (a) To construct a \( 95\% \) confidence interval for the mean mathematics SAT score, we can use the formula for the confidence interval: \[ \text{CI} = \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \] Where: - \(\bar{x} = 452\) (sample mean) - \(\sigma = 115\) (population standard deviation) - \(n = 100\) (sample size) - \(z_{\alpha/2}\) is the critical value for a \(95\%\) confidence level. For a \(95\%\) confidence level, \(z_{\alpha/2} \approx 1.96\). Now, we can calculate the standard error (SE): \[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{115}{\sqrt{100}} = \frac{115}{10} = 11.5 \] Next, we calculate the margin of error (ME): \[ \text{ME} = z_{\alpha/2} \cdot \text{SE} = 1.96 \cdot 11.5 \] Now, let's calculate the margin of error. ### Part 2 of 4 (b) If the sample size were \(110\) rather than \(100\), we can analyze the effect on the margin of error. The new standard error would be: \[ \text{SE}_{new} = \frac{\sigma}{\sqrt{n_{new}}} = \frac{115}{\sqrt{110}} \] Since the sample size increases, the standard error will decrease, leading to a smaller margin of error. Thus, 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 level were \(80\%\) rather than \(95\%\), we need to consider the critical value \(z_{\alpha/2}\) for \(80\%\). The critical value for \(80\%\) is approximately \(1.28\). Since the critical value decreases when the confidence level decreases, the margin of error would be **smaller**, since a decrease in the confidence level will **decrease** the critical value \(z_{\alpha/2}\). Now, let's calculate the margin of error for part (a) and finalize the answers. #### Calculating the Margin of Error for Part (a) \[ \text{ME} = 1.96 \cdot 11.5 \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(1.96\times 11.5\) - step1: Multiply the numbers: \(22.54\) The margin of error (ME) calculated is approximately \( 22.54 \). Now, we can construct the confidence interval: \[ \text{CI} = \bar{x} \pm \text{ME} = 452 \pm 22.54 \] Calculating the lower and upper bounds: - Lower bound: \( 452 - 22.54 \approx 429.46 \) - Upper bound: \( 452 + 22.54 \approx 474.54 \) Rounding to the nearest whole number, we have: \[ 429 < \mu < 475 \] ### Final Answers: 1. **Part (a)**: The \( 95\% \) confidence interval for the mean mathematics SAT score is \( 429 < \mu < 475 \). 2. **Part (b)**: The margin of error would be **smaller**, since an increase in the sample size will **decrease** the standard error. 3. **Part (c)**: The margin of error would be **smaller**, since a decrease in the confidence level will **decrease** the critical value \( z_{\alpha/2} \).