Answer to Question #178241 in Statistics and Probability for Abeer

Question #178241

Assume the population is normally distributed with: 

Xbar = 96.44           S =10.2        n=15                 t=1.76

Construct a​ 90% confidence interval estimate for the population​ mean, μ


1
Expert's answer
2021-04-15T07:40:52-0400

"c = 0.90 \\\\\n\nn = 15 \\\\\n\n\\bar{x} = 96.44 \\\\\n\nt_{\u03b1\/2}= 1.76"

The margin of error is:

"E = t_{\u03b1\/2} \\times \\frac{S}{\\sqrt{n}} = 1.76 \\times \\frac{10.2}{\\sqrt{15}} = 4.63"

The confidence interval:

"\\bar{x} -E < \u03bc < \\bar{x} +E \\\\\n\n96.44 -4.63 < \u03bc < 96.44 +4.63 \\\\\n\n91.81 < \u03bc < 101.07"


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