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# Answer to Question #2995 in Statistics and Probability for Andre Figueiredo

Question #2995
Suppose a sample of n = 25 items is drawn from a population described by a Normal distribution with standard deviation &sigma; = 15, and that the sample mean of the 25 values is 96.4. Derive a 95% confidence interval for &mu;, the (unknown) mean of this distribution, and use your confidence interval to test, at the 5% significance level, the hypothesis H0 : &mu; = 100 against the general (two-sided) alternative H1 : &mu; = 100.
1
2011-05-30T12:48:36-0400
95% confidence interval can be constructed in such way:
Upper limit: x + 1.96* σ/√n = 96.4 + 1.96* 15/√25 = 102.28
Lower limit: x - 1.96* σ/√n = 96.4 - 1.96* 15/√25 = 90.52
90.52 < μ =100 < 102.28 - thus at the significance level 5% the hypothesis H0 : μ = 100 is not rejected.

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