Joey Jabroni and Chris Muss are running for class president at CRHS. You conduct a sample survey of the senior class on who plans on voting for Joey instead of Chris. You will tole... Joey Jabroni and Chris Muss are running for class president at CRHS. You conduct a sample survey of the senior class on who plans on voting for Joey instead of Chris. You will tolerate a 4% margin of error with a 95% confidence level. Assume each candidate is equally likely to be favored in the senior class. How large a sample is needed?
Understand the Problem
The question is asking for the necessary sample size for a survey regarding the voting preferences of a senior class, given a specific margin of error and confidence level. The primary focus is on determining how large a sample size is required to ensure accuracy in the results.
Answer
The required sample size is \( n = 601 \).
Answer for screen readers
The required sample size is ( n = 601 ).
Steps to Solve
- Identify the formula for sample size To calculate the sample size needed for estimating proportions, use the formula:
$$ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) $$
where:
- ( n ) = required sample size
- ( Z ) = Z-score corresponding to the confidence level
- ( p ) = estimated proportion of the population favoring one candidate
- ( E ) = margin of error
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Determine the Z-score for a 95% confidence level For a 95% confidence level, the Z-score is approximately 1.96.
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Estimate the proportion ( p ) Since each candidate is equally likely to be favored, set ( p = 0.5 ).
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Plug in margin of error Convert the margin of error from percentage to decimal:
$$ E = 0.04 $$
- Substitute values into formula Rearranging the sample size formula, we get:
$$ n = \frac{(1.96^2) \cdot 0.5 \cdot (1-0.5)}{(0.04)^2} $$
Now substitute the values:
$$ n = \frac{(1.96^2) \cdot 0.5 \cdot 0.5}{0.0016} $$
- Calculate ( n ) Now perform the calculations step-by-step:
- Calculate ( 1.96^2 ):
$$ 1.96^2 \approx 3.8416 $$
- Next, calculate:
$$ n = \frac{3.8416 \cdot 0.25}{0.0016} $$
This simplifies to:
$$ n = \frac{0.9604}{0.0016} $$
Finally, dividing gives:
$$ n \approx 600.25 $$
Since we can't survey a fraction of a person, round up to the nearest whole number.
The required sample size is ( n = 601 ).
More Information
To achieve a 95% confidence level with a margin of error of 4%, a sample size of at least 601 students is needed. This ensures that the results accurately reflect the voting preferences of the senior class.
Tips
- Not rounding up the sample size correctly: Always round up to ensure the sample is sufficient.
- Using an incorrect Z-score: Ensure you're using the Z-score that corresponds to the desired confidence level.
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