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Chapter 8: Problem 41

America's young people are heavy Internet users; \(87 \%\) of Americans ages 12 to 17 are Internet users (The Cincinnati Enquirer, February 7,2006 ). MySpace was voted the most popular website by \(9 \%\) in a sample survey of Internet users in this age group. Suppose 1400 youths participated in the survey. What is the margin of error, and what is the interval estimate of the population proportion for which MySpace is the most popular website? Use a \(95 \%\) confidence level.

Short Answer

Expert verified
The margin of error is approximately 1.41%, and the interval estimate of the population proportion for which MySpace is the most popular website among Internet users aged 12 to 17 in America is (7.59%, 10.41%) with a 95% confidence level.

Step by step solution

01

Identify Given Values

Sample size (n) = 1400 Sample proportion (p) = 0.09 (as 9% of the survey participants voted MySpace as the most popular website) Confidence level = 95%
02

Find the Z-score

For a 95% confidence level, the Z-score (Z) is 1.96. This value can be found using a Z-table or a calculator that provides critical values of Z for different confidence levels.
03

Calculate Margin of Error

Now, we can plug the values into the Margin of Error formula: ME = Z * sqrt(p * (1 - p) / n) ME = 1.96 * sqrt(0.09 * (1 - 0.09) / 1400) ME ≈ 0.0141 The margin of error is approximately 0.0141 or 1.41%.
04

Calculate the Interval Estimate

Now we will calculate the interval estimate using the sample proportion and the margin of error: CI = p ± ME CI = 0.09 ± 0.0141 CI ≈ (0.0759, 0.1041) Thus, the interval estimate of the population proportion for which MySpace is the most popular website among Internet users aged 12 to 17 in America is (0.0759, 0.1041), or (7.59%, 10.41%) with a 95% confidence level.

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Most popular questions from this chapter

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