Question 932828
For the 82 students in the highest quartile of the distribution, the mean score was x = 176.90. Assume a population standard deviation of σ = 7.85. These students were all classified as high on their need for closure. Assume that the 82 students represent a random sample of all students who are classified as high on their need for closure. Find a 95% confidence interval for the population mean score μ on the "need for closure scale" for all students with a high need for closure. (Round your answers to two decimal places.)
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x-bar = 176.9
ME = 1.96*(7.85/sqrt(85)) = 1.67
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95% CI:: 176.9-1.67 < u < 176.67+1.67
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cheers,
Stan H.
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