A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval (z-score = 2.58) for the number of minutes that an adult spends surfing the Internet per day?

asked by kevin

8 years ago

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2 answers

99% = mean ± Z SEm

SEm = SD/√n

answered by PsyDAG

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50min

answered by Anonymous

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Question

A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval (z-score = 2.58) for the number of minutes that an adult spends surfing the Internet per day?

2 answers

99% = mean ± Z SEm

SEm = SD/√n

50min

Ask a New Question or answer this question.

Anonymous · Accepted Answer

50min

PsyDAG · Answer

99% = mean ± Z SEm SEm = SD/√n