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# stats7 - 64(20.74 107.26 If 64 is the PE then it falls...

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The population mean is 88.96 and the standard deviation is 69.79. If the point estimate equals 88.96 then it falls within the confidence interval of 45.7 and 132.22. This would be 95% of confidence level for the population value. Sample Means: Confidence Intervals: 114.8 (71.54, 158.06) If 114.8 is the point estimate then it falls within the confidence interval of 71.54, 158.06. 84.8 (41.54, 128.06) If 84.8 is the PE then it falls within the CI of 41.54, 128.06. 116.8 (73.54, 160.06) If 116.8 is the PE then it falls within the CI of 73.54, 160.06. 78.9 (35.64, 122.16) If 78.9 is the PE then it falls within the CI of 35.64, 122.16. 112.7 (69.44, 155.96) If 112.7 is the PE then it falls within the CI of 69.44, 155.96. 77.1 (33.84, 120.36) If 77.1 is the PE then it falls within the CI of 38.84, 120.36. 102 (58.74, 145.26) If 102 is the PE then it falls within the CI of 58.74, 145.26. 119.2 (75.94, 162.46) If 119.2 is the PE then it falls within the CI of 75.94, 162.46.
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Unformatted text preview: 64 (20.74, 107.26) If 64 is the PE then it falls within the CI of 20.74, 107.26. 97 (53.74, 140.26) If 97 is the PE then it falls within the CI of 53.74, 140.26. All of these have a 95% confidence level for the sample values. If the standard deviation is known it is fairly easy to fit it into the formula and figure out the confidence interval. This also means less work for the person who is figuring out the confidence intervals. If the standard deviation is unknown then you would have to use the student’s t distribution. If the sample size is large enough you can substitute the sample SD in, but if the sample size is small then you have to use the student’s t distribution. The interval changes on student’s t distribution are that it has thicker tails and a lower peak. You would have to estimate two numbers for the problem instead of only one with known SD....
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