Determine the minimum sample size required when you want to be 90% confident that the sample mean is within one unit of the population mean and o = 11.3. Assume the population is normally distributed. A 90% confidence level requires a sample size of: (Round up to the nearest whole number as needed.)

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**Title: Calculating Minimum Sample Size for a 90% Confidence Level**

**Introduction:**
When conducting statistical analysis, determining the appropriate sample size is crucial to ensure the reliability of the results. Below is an explanation of how to calculate the minimum sample size required to be 90% confident that the sample mean is within one unit of the population mean, given a standard deviation \( \sigma = 11.3 \). The assumption is that the population is normally distributed.

**Problem Statement:**
Determine the minimum sample size required for a 90% confidence level given:
- Desired margin of error (E) = 1 unit
- Population standard deviation (\( \sigma \)) = 11.3

**Calculation:**
1. **Z-Score for 90% Confidence Level:**
   - The Z-score associated with a 90% confidence level can be found in statistical Z-tables or using a standard normal distribution calculator. It is approximately 1.645.

2. **Formula for Sample Size (n):**
   \[
   n = \left( \frac{Z \cdot \sigma}{E} \right)^2
   \]
   - Where \( n \) is the sample size,
   - \( Z \) is the Z-score (1.645 for 90% confidence),
   - \( \sigma \) is the population standard deviation (11.3),
   - \( E \) is the margin of error (1 unit).

3. **Plug in the Values:**
   \[
   n = \left( \frac{1.645 \cdot 11.3}{1} \right)^2
   \]
   \[
   n = \left( 18.5885 \right)^2
   \]
   \[
   n \approx 345.64
   \]

4. **Rounding:**
   - Round up to the nearest whole number to ensure the sample size is sufficient.
   - Therefore, the minimum sample size required is 346.

**Conclusion:**
For a 90% confidence level with a margin of error of 1 unit, you should use a sample size of approximately 346 to ensure accuracy, assuming the population is normally distributed.
Transcribed Image Text:**Title: Calculating Minimum Sample Size for a 90% Confidence Level** **Introduction:** When conducting statistical analysis, determining the appropriate sample size is crucial to ensure the reliability of the results. Below is an explanation of how to calculate the minimum sample size required to be 90% confident that the sample mean is within one unit of the population mean, given a standard deviation \( \sigma = 11.3 \). The assumption is that the population is normally distributed. **Problem Statement:** Determine the minimum sample size required for a 90% confidence level given: - Desired margin of error (E) = 1 unit - Population standard deviation (\( \sigma \)) = 11.3 **Calculation:** 1. **Z-Score for 90% Confidence Level:** - The Z-score associated with a 90% confidence level can be found in statistical Z-tables or using a standard normal distribution calculator. It is approximately 1.645. 2. **Formula for Sample Size (n):** \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] - Where \( n \) is the sample size, - \( Z \) is the Z-score (1.645 for 90% confidence), - \( \sigma \) is the population standard deviation (11.3), - \( E \) is the margin of error (1 unit). 3. **Plug in the Values:** \[ n = \left( \frac{1.645 \cdot 11.3}{1} \right)^2 \] \[ n = \left( 18.5885 \right)^2 \] \[ n \approx 345.64 \] 4. **Rounding:** - Round up to the nearest whole number to ensure the sample size is sufficient. - Therefore, the minimum sample size required is 346. **Conclusion:** For a 90% confidence level with a margin of error of 1 unit, you should use a sample size of approximately 346 to ensure accuracy, assuming the population is normally distributed.
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