The head of a computer science department is interested in estimating the proportion of students
entering the department who will choose the new computer engineering option. A preliminary sample
indicates that the proportion will be around 0.25. Therefore, what size sample should the department
head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion?
.......And what would be your answer to this question?
http://www.surveysystem.com/sscalc.htm#one
To the left, is a menu that will teach you.
To determine the sample size required for estimating a proportion with a desired level of confidence, you can use the formula for sample size calculation:
n = (Z^2 * p * q) / E^2
Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired level of confidence (in this case, 95% confidence level)
- p is the estimated proportion from the preliminary sample (0.25)
- q is the complement of the estimated proportion (1 - p)
- E is the desired margin of error (0.10)
First, we need to find the Z-score for a 95% confidence level. The Z-score can be obtained from a standard normal distribution table or by using a statistical software. For a 95% confidence level, the Z-score is approximately 1.96.
Next, we can substitute the values into the formula:
n = (1.96^2 * 0.25 * (1 - 0.25)) / 0.1^2
n = 3.8416 * 0.25 * 0.75 / 0.01
n = 0.7209 / 0.01
n ≈ 72.09
Since the sample size should be a whole number, we round up to the nearest integer:
n ≈ 73
Therefore, the department head should take a sample size of 73 in order to be 95% confident that the estimate is within 0.10 of the true proportion.