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In a survey of 1053 people, 726 people said that they voted in the recent presidential election. Voting records show that 66% of eligible voters actually did vote.

a) ( frac{726}/{1053} )

b) ( frac{66}/{100} )

c) ( frac{66}/{1053} )

d) ( frac{726}/{66} )

User Shark
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1 Answer

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Final answer:

To estimate the true proportion of college students who voted with a 95 percent confidence and a margin of error of 5 percent, a sample size of approximately 385 students is needed, using a conservative approach with p set at 0.5.

Step-by-step explanation:

The student is interested in conducting a survey to estimate the true proportion of college students on their campus who voted in the 2012 presidential election with a 95 percent confidence level and a margin of error no greater than 5 percent. To calculate the sample size needed for the survey, one can use the formula for sample size estimation in a proportion:

n = (Z^2 * p * (1 - p)) / E^2

Where:

  • n is the sample size.
  • Z is the z-value corresponding to the desired confidence level (1.96 for 95% confidence).
  • p is the estimated proportion of the population that exhibits the characteristic (in this case, the historical voter turnout which can be assumed to be 0.55 to 0.65).
  • E is the margin of error (0.05 for 5%).

However, without the exact value of p from past data, a conservative approach often taken is to use p = 0.5, as this maximizes the product p*(1-p), and thus the sample size. Therefore, the formula would be:

n = (1.96)^2 * 0.5 * (1 - 0.5) / (0.05)^2

After doing the math, we find that:

n ≈ 385

Therefore, approximately 385 students need to be interviewed to estimate the political awareness with a 95 percent confidence and a margin of error of no more than 5 percent.

User Fishtoprecords
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