Question #94929
the mean of a random sample of 25 observations from a normal population whose standard deviation is 50, with a sample mean of 200. Estimate the population mean with 95% confidence.
b. repeat part a changing the population standard deviation to 25
c. repeat part a changing the population standard to 10
d. describe what happens to the confidence interval estimate when the standard deviation decreased.
1
Expert's answer
2019-09-23T09:33:09-0400

a.

95%CI=(xˉ1.96σn,  xˉ+1.96σn)==(2001.965025,  200+1.965025)=(180.4,  219.6).95\%CI=(\bar{x}-1.96\frac{\sigma}{\sqrt{n}},\;\bar{x}+1.96\frac{\sigma}{\sqrt{n}})=\\=(200-1.96\frac{50}{\sqrt{25}},\;200+1.96\frac{50}{\sqrt{25}})=(180.4,\;219.6).


b.

95%CI=(xˉ1.96σn,  xˉ+1.96σn)==(2001.962525,  200+1.962525)=(190.2,  209.8).95\%CI=(\bar{x}-1.96\frac{\sigma}{\sqrt{n}},\;\bar{x}+1.96\frac{\sigma}{\sqrt{n}})=\\=(200-1.96\frac{25}{\sqrt{25}},\;200+1.96\frac{25}{\sqrt{25}})=(190.2,\;209.8).


c.

95%CI=(xˉ1.96σn,  xˉ+1.96σn)==(2001.961025,  200+1.961025)=(196.08,  203.92).95\%CI=(\bar{x}-1.96\frac{\sigma}{\sqrt{n}},\;\bar{x}+1.96\frac{\sigma}{\sqrt{n}})=\\=(200-1.96\frac{10}{\sqrt{25}},\;200+1.96\frac{10}{\sqrt{25}})=(196.08,\;203.92).


d. When the standard deviation decreases, the confidence interval becomes narrower.



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