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A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home falls between which of the following intervals? 7.3 and 9.7 7.7 and 9.3 8.4 and 8.6 6.5 and 10.5 8.1 and 8.9

Question

A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home falls between which of the following intervals? 7.3 and 9.7 7.7 and 9.3 8.4 and 8.6 6.5 and 10.5 8.1 and 8.9

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Solution

To answer this question, we need to calculate the 95% confidence interval for the mean. The formula for a confidence interval is:

CI = X ± Z * (σ/√n)

where:

  • X is the sample mean
  • Z is the Z-score (for a 95% confidence interval, the Z-score is approximately 1.96)
  • σ is the population standard deviation
  • n is the sample size

Substituting the given values into the formula:

CI = 8.5 ± 1.96 * (3.6/√81)

This simplifies to:

CI = 8.5 ± 0.784

So, the 95% confidence interval is (8.5 - 0.784, 8.5 + 0.784) = (7.716, 9.284)

Therefore, the mean number of weekly hours that U.S. adults use computers at home falls between 7.7 and 9.3 hours with 95% confidence.

This problem has been solved

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