Answer to Question #151725 in Statistics and Probability for wan

Question #151725
Set up a 95% confidence interval estimate for the population mean, based on each of the
following sets of data, assuming that the population is normally distributed:
Set 1: 1, 1, 1, 1, 8, 8, 8, 8
Set 2: 1, 2, 3, 4, 5, 6, 7, 8
Explain why these data sets have different confidence intervals even though they have the
same mean and range.
1
Expert's answer
2020-12-18T11:12:15-0500

The critical value for "\\alpha=0.05" is "z_c=z_{1-\\alpha\/2}=1.96"

The corresponding confidence interval is computed as shown below:


"CI=(\\bar{x}-z_c\\times\\dfrac{\\sigma}{\\sqrt{n}}, \\bar{x}+z_c\\times\\dfrac{\\sigma}{\\sqrt{n}})"

The confidence interval is equal to two margins of errors and a margin of error is equal to about 2 standard errors (for 95% confidence). A standard error is the standard deviation divided by the square root of the sample size.

The width of the confidence interval increases as the standard deviation increases.


Set 1: 1, 1, 1, 1, 8, 8, 8, 8


"mean_1=\\bar{x}_1"

"=\\dfrac{1+1+1+1+8+8+8+8}{8}=4.5"


"Range_1=8-1=7"


"\\sigma_1^2=\\dfrac{1}{8}((1-4.5)^2+(1-4.5)^2+(1-4.5)^2+(1-4.5)^2"

"+(8-4.5)^2+(8-4.5)^2+(8-4.5)^2+(8-4.5)^2)"

"=12.25"

"\\sigma_1=\\sqrt{12.25}=3.5"

"CI_1=(\\bar{x}_1-z_c\\times\\dfrac{\\sigma_1}{\\sqrt{n_1}}, \\bar{x}_1+z_c\\times\\dfrac{\\sigma_1}{\\sqrt{n_1}})"

"=(4.5-1.96\\times\\dfrac{3.5}{\\sqrt{8}}, 4.5+1.96\\times\\dfrac{3.5}{\\sqrt{8}})"

"=(2.075,6.925)"

Therefore, based on the data provided, the 95%  confidence interval for the population mean is "2.075<\\mu<6.925," which indicates that we are 95% confident that the true population mean "\\mu"  is contained by the interval "(2.075,6.925)."


Set 2: 1, 2, 3, 4, 5, 6, 7, 8


"mean_2=\\bar{x}_2"

"=\\dfrac{1+2+3+4+5+6+7+8}{8}=4.5"


"Range_2=8-1=7"


"\\sigma_2^2=\\dfrac{1}{8}((1-4.5)^2+(2-4.5)^2+(3-4.5)^2+(4-4.5)^2"

"+(5-4.5)^2+(6-4.5)^2+(7-4.5)^2+(8-4.5)^2)"

"=5.25"

"\\sigma_2=\\sqrt{5.25}\\approx2.291288"




"CI_2=(\\bar{x}_2-z_c\\times\\dfrac{\\sigma_2}{\\sqrt{n_2}}, \\bar{x}_2+z_c\\times\\dfrac{\\sigma_2}{\\sqrt{n_2}})"

"=(4.5-1.96\\times\\dfrac{2.291288}{\\sqrt{8}}, 4.5+1.96\\times\\dfrac{2.291288}{\\sqrt{8}})"

"=(2.912,6.088)"

Therefore, based on the data provided, the 95%  confidence interval for the population mean is "2.912<\\mu<6.088," which indicates that we are 95% confident that the true population mean "\\mu"  is contained by the interval "(2.912,6.088)."



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