A survey many years ago found that 58% of U.S, teenagers had suffered a sunburn in the last twelve months. Researchers think that the percentage has declined due to increased awareness of the harm caused by ultraviolet rays. They plan to construct and interpret a 90% confidence interval to estimate the current percentage of all U.S. teenagers who have suffered a sunburn in the last twelve months. Which of the following should be used to determine the sample size required for the margin of error to be at most 2.5 percentage points? (1.6452)(0.52) (A) n ≥ 0.0252 (1.6452) (0.58)(0.42) (B) n ≥ 0.052 (1.962)(0.52) (C) n ≥ 0.0252 (1.963) (0.58)(0.42) (D) n≥ 0.0252 (1.963) (0.58)(0.42) (E) n ≥ 0.052
Question
A survey many years ago found that 58% of U.S, teenagers had suffered a sunburn in the last twelve months. Researchers think that the percentage has declined due to increased awareness of the harm caused by ultraviolet rays. They plan to construct and interpret a 90% confidence interval to estimate the current percentage of all U.S. teenagers who have suffered a sunburn in the last twelve months. Which of the following should be used to determine the sample size required for the margin of error to be at most 2.5 percentage points? (1.6452)(0.52) (A) n ≥ 0.0252 (1.6452) (0.58)(0.42) (B) n ≥ 0.052 (1.962)(0.52) (C) n ≥ 0.0252 (1.963) (0.58)(0.42) (D) n≥ 0.0252 (1.963) (0.58)(0.42) (E) n ≥ 0.052
Solution
The formula for determining the sample size for a proportion in a population is given by:
n = (Z^2 * p * (1-p)) / E^2
where:
- n is the sample size
- Z is the z-value (which depends on the desired confidence level, for a 90% confidence level, Z = 1.645)
- p is the estimated proportion of the population (in this case, 0.58)
- E is the desired margin of error (in this case, 0.025 or 2.5%)
Substituting the given values into the formula, we get:
n = (1.645^2 * 0.58 * (1-0.58)) / 0.025^2
This is equivalent to option (A) in your question:
n ≥ (1.645^2 * 0.58 * 0.42) / 0.025^2
So, the correct answer is (A).
Similar Questions
Researchers surveyed 1,150 high school students in CA to estimate the proportion of all high school students who vape. 368 of the surveyed students said they do vape. What is the margin of error of the 95% confidence interval?
A group of marketing students at a large university wants to determine the proportion of first year students who use certain types of social media. The students want their estimate to be within 0.03 of the true proportion with a 95% level of confidence. Two years ago, a similar study determined the proportion to be 0.796. How large of a sample is required?
A survey was given to a random sample of the residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. The survey reported a confidence interval that between 43.5% and 48.5% of the residents supports the plan. What is the margin of error on the survey?
Question 22You’re going to draw a random sample of professional football players because you want to know what percentage have completed high school. You want to have a margin of error of up to 0.03 at a confidence level of 90%. How big should your sample be?1 pointMinimum 456At least 30Minimum 748Minimum 1068
A food store owner wants to determine the type of products customers prefer to buy. The owner surveyed 40 customers on a day when the store had 335 customers, and found that 40% of those surveyed preferred to buy organic products over non-organic products. Assuming a 95% confidence level, which statement holds true? A. As the sample size is too small, the margin of error is 0.127. B. As the sample size is too small, the margin of error cannot be trusted. C. As the sample size is appropriately large, the margin of error is 0.127. D. As the sample size is appropriately large, the margin of error is 0.152.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.