Cora wants to determine a 80 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.02? Assume we have no prior estimate of the proportion and want a conservative choice for the sample size.
[Round to the smallest integer that works.] n =
To determine the sample size required to obtain a margin of error less than 0.02 for a 80% confidence interval, we need to use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
- n is the required sample size
- Z is the z-score corresponding to the desired confidence level (80% in this case)
- p is the estimated proportion (since we have no prior estimate, we can use 0.5 as a conservative choice)
- E is the desired margin of error (0.02 in this case)
First, we need to find the z-score corresponding to the desired confidence level of 80%. The z-score can be found using a standard normal distribution table or a statistical calculator. For an 80% confidence level, the z-score is approximately 1.282.
Next, substitute the values into the formula:
n = (1.282^2 * 0.5 * (1-0.5)) / 0.02^2
Simplifying the expression:
n = (1.642724 * 0.5 * 0.5) / 0.0004
n = 0.410681 / 0.0004
n = 1026.7025
Since we need to round to the smallest integer that works, we round up to the nearest whole number:
n = 1027
Therefore, Cora would need a sample size of at least 1027 to obtain a margin of error less than 0.02 for the 80% confidence interval.