A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least

Question

A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.40. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context. but for this problem, compute a 95% confidence interval (use z*=1.96), a 90% confidence interval (use z*=1.645) and a 99% confidence interval (use z*=2.576). Make sure you write conclusions/interpretations for all confidence intervals.

Explanation / Answer

95% confidence interval:

(0.52 - 1.96*0.024, 0.52 + 1.96*0.024)

(0.473, 0.567)

Conclusion: We are 95% confident that the true proportion of U.S. adult Twitter users who get at least some news on Twitter is between the above interval.

99% confidence interval:

(0.52 - 2.576*0.024, 0.52 + 2.576*0.024)

(0.458, 0.582)

Conclusion: We are 99% confident that the true proportion of U.S. adult Twitter users who get at least some news on Twitter is between the above interval.

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