A college admissions officer takes a simple random sample of 100 entering freshmen and computes their mean mathematics SAT score to be 441. Assume the population standard deviation is 115.

(a) Construct an 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 426 < u < 456.

If the confidence levels were 95% rather than 80%, would the margin of error be larger or smaller than the result in part (a)? Explain. 

The margin of error would be (larger or smaller), since (a decrease or increase) in the confidence level will (decrease or increase) the critical value.


To solve part (a) and answer the follow-up question, we need to construct the confidence interval and understand the relationship between confidence levels and the margin of error.

### Part (a) Constructing an 80% Confidence Interval

Given dataTo solve part (a) and answer the follow-up question, we need to construct the confidence interval and understand the relationship between confidence levels and the margin of error.

### Part (a) Constructing an 80% Confidence Interval

Given 
- Sample mean (\(\bar{x}\)) = 441
- Population standard deviation (\(\sigma\)) = 115
- Sample size (\(n\)) = 100
- Confidence level = 80%

First, we need to find the critical value for an 80% confidence interval. For an 80% confidence level, the critical value (z*) is approximately 1.28 (this value is derived from the standard normal distribution table).

The formula for the margin of error (ME) is:
\[ ME = z^* \left( \frac{\sigma}{\sqrt{n}} \right) \]

Plugging in the values:
\[ ME = 1.28 \left( \frac{115}{\sqrt{100}} \right) \]
\[ ME = 1.28 \left( \frac{115}{10} \right) \]
\[ ME = 1.28 \times 11.5 \]
\[ ME = 14.72 \]

Now, we construct the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 441 - 14.72 = 426.28 \]
\[ \text{Upper limit} = \bar{x} + ME = 441 + 14.72 = 455.72 \]

Rounding to the nearest whole number, the 80% confidence interval is:
\[ 426 < \mu < 456 \]

### Part (b) Margin of Error for a 95% Confidence Interval

If the confidence level were 95%, the margin of error would be different. A 95% confidence level has a critical value (z*) of approximately 1.96.

The new margin of error (ME) is:
\[ ME = 1.96 \left( \frac{115}{\sqrt{100}} \right) \]
\[ ME = 1.96 \left( \frac{115}{10} \right) \]
\[ ME = 1.96 \times 11.5 \]
\[ ME = 22.54 \]

Since the new margin of error (22.54) is larger than the original margin of error (14.72), we can conclude:

The margin of error would be **larger** since an **increase** in the confidence level will **increase** the critical value.

Would you like more details or have any questions?

Here are 5 questions you can ask:
1. How do you find the critical value for different confidence levels?
2. Can you explain how to interpret confidence intervals?
3. What is the relationship between sample size and margin of error?
4. How does the standard deviation affect the confidence interval?
5. Can you demonstrate constructing a 99% confidence interval for this problem?

**Tip:** Confidence intervals provide a range of values within which the true population parameter is likely to fall, helping to estimate the precision of your sample statistics.

Learn how to construct confidence intervals and calculate margin of error for sample mean calculations using the normal distribution and critical values. Explore how confidence levels affect the precision of statistical estimates.

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