8. Sampling Distributions & Confidence Intervals: Proportion
Confidence Intervals for Population Proportion
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A previous study found that your school consists of White/Caucasian students. You want the confidence interval for the proportion of White/Caucasian students to be no more than away from the true proportion. How many students must you include in a sample to create this confidence interval?
A
520
B
519
C
542
D
260
Verified step by step guidance1
Step 1: Understand the problem requirements. We need to find the sample size required to achieve a 98% confidence interval with a margin of error of 0.05 for the proportion of White/Caucasian students, given that the proportion is 0.60.
Step 2: Recall the formula for the margin of error in a proportion confidence interval: \( ME = z \times \sqrt{\frac{p(1-p)}{n}} \), where \( ME \) is the margin of error, \( z \) is the z-score corresponding to the confidence level, \( p \) is the estimated proportion, and \( n \) is the sample size.
Step 3: Determine the z-score for a 98% confidence level. The z-score for 98% confidence is approximately 2.33. This value can be found using a standard normal distribution table or calculator.
Step 4: Set up the equation using the margin of error formula: \( 0.05 = 2.33 \times \sqrt{\frac{0.60 \times (1 - 0.60)}{n}} \). Solve this equation for \( n \) to find the required sample size.
Step 5: Rearrange the equation to solve for \( n \): \( n = \frac{(2.33)^2 \times 0.60 \times 0.40}{(0.05)^2} \). Calculate \( n \) using this formula to determine the sample size needed.
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