Suppose that the results of a survey show that 46% of people consider summer as their favorite season. How many people should be surveyed in order to have a margin of error of 2%? Round up to the nearest whole number, if necessary. In order to have a margin of error of 2%, people should be surveyed.

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**Survey Sample Size Calculation**

**Problem Statement:**
Suppose that the results of a survey show that 46% of people consider summer as their favorite season. How many people should be surveyed in order to have a margin of error of 2%? Round up to the nearest whole number, if necessary.

**Calculation Instruction:**
To calculate the required sample size (n), you can use the formula for the margin of error in proportion surveys, which is given by:

\[ ME = Z \times \sqrt{\frac{p(1 - p)}{n}} \]

Where:
- \( ME \) is the margin of error,
- \( Z \) is the Z-score for the confidence level (for a 95% confidence level, Z is approximately 1.96),
- \( p \) is the estimated proportion (0.46 in this case),
- \( n \) is the sample size.

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

\[ n = \left( \frac{Z^2 \times p(1 - p)}{ME^2} \right) \]

Plug in the values:

\[ n = \left( \frac{1.96^2 \times 0.46 \times 0.54}{0.02^2} \right) \]

Calculate the sample size and round up to the nearest whole number if necessary.

**Answer:**

In order to have a margin of error of 2%, [input box] people should be surveyed.
Transcribed Image Text:**Survey Sample Size Calculation** **Problem Statement:** Suppose that the results of a survey show that 46% of people consider summer as their favorite season. How many people should be surveyed in order to have a margin of error of 2%? Round up to the nearest whole number, if necessary. **Calculation Instruction:** To calculate the required sample size (n), you can use the formula for the margin of error in proportion surveys, which is given by: \[ ME = Z \times \sqrt{\frac{p(1 - p)}{n}} \] Where: - \( ME \) is the margin of error, - \( Z \) is the Z-score for the confidence level (for a 95% confidence level, Z is approximately 1.96), - \( p \) is the estimated proportion (0.46 in this case), - \( n \) is the sample size. Rearranging the formula to solve for \( n \): \[ n = \left( \frac{Z^2 \times p(1 - p)}{ME^2} \right) \] Plug in the values: \[ n = \left( \frac{1.96^2 \times 0.46 \times 0.54}{0.02^2} \right) \] Calculate the sample size and round up to the nearest whole number if necessary. **Answer:** In order to have a margin of error of 2%, [input box] people should be surveyed.
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