SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

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### Estimating SAT Scores

**Scenario:**

SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You wish to estimate the average SAT score of first-year students at your college. To achieve this, you want the margin of error of your 95% confidence interval to be within 25 points. 

### Question:

How many students should you sample to attain this level of precision in your estimate? 

### Requirements:

Submit a whole number answer for the number of students needed.

[Input Box]

---

### Explanation:

To solve this, you can use the formula for the margin of error in a confidence interval, which is:

\[ E = Z \times \left(\frac{\sigma}{\sqrt{n}}\right) \]

Where:
- \(E\) is the margin of error
- \(Z\) is the Z-value corresponding to the desired confidence level
- \(\sigma\) is the standard deviation
- \(n\) is the sample size

For a 95% confidence level, the Z-value is approximately 1.96. 

Rearrange the formula to solve for \(n\):

\[ n = \left(\frac{Z \times \sigma}{E}\right)^2 \]

Substitute the given values into the formula:

- \(Z = 1.96\)
- \(\sigma = 300\)
- \(E = 25\)

\[ n = \left(\frac{1.96 \times 300}{25}\right)^2 \]

Calculate the sample size and round to the nearest whole number to get your answer.

### Note:
Providing an exact number ensures you're adequately sampling the population to achieve the desired confidence interval within the specified margin of error.
Transcribed Image Text:### Estimating SAT Scores **Scenario:** SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You wish to estimate the average SAT score of first-year students at your college. To achieve this, you want the margin of error of your 95% confidence interval to be within 25 points. ### Question: How many students should you sample to attain this level of precision in your estimate? ### Requirements: Submit a whole number answer for the number of students needed. [Input Box] --- ### Explanation: To solve this, you can use the formula for the margin of error in a confidence interval, which is: \[ E = Z \times \left(\frac{\sigma}{\sqrt{n}}\right) \] Where: - \(E\) is the margin of error - \(Z\) is the Z-value corresponding to the desired confidence level - \(\sigma\) is the standard deviation - \(n\) is the sample size For a 95% confidence level, the Z-value is approximately 1.96. Rearrange the formula to solve for \(n\): \[ n = \left(\frac{Z \times \sigma}{E}\right)^2 \] Substitute the given values into the formula: - \(Z = 1.96\) - \(\sigma = 300\) - \(E = 25\) \[ n = \left(\frac{1.96 \times 300}{25}\right)^2 \] Calculate the sample size and round to the nearest whole number to get your answer. ### Note: Providing an exact number ensures you're adequately sampling the population to achieve the desired confidence interval within the specified margin of error.
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