Answer to Question #204789 in Statistics and Probability for sam21ucd

Question #204789

The following observations were randomly sampled from a Normal distribution. Using this data, construct a 99% confidence interval for the mean and interpret the interval. 8.11 12.06 10.01 12.01 8.59 9.35 10.79 11.74 7.42 11.72 8.44 7.67 10.93 8.42 8.50


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Expert's answer
2021-06-11T02:37:33-0400
xˉ=i=1nxin=i=115xi15\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{\displaystyle\sum_{i=1}^{15}x_i}{15}

=115(8.11+12.06+10.01+12.01+8.59=\dfrac{1}{15}(8.11+12.06 +10.01 +12.01+ 8.59

+9.35+10.79+11.74+7.42+11.72+9.35+ 10.79 +11.74+ 7.42+ 11.72

+8.44+7.67+10.93+8.42+8.50=145.7615+ 8.44+ 7.67+ 10.93+ 8.42+ 8.50=\dfrac{145.76}{15}

9.717333\approx9.717333

s2=i=115(xixˉ)21512.843450s^2=\dfrac{\displaystyle\sum_{i=1}^{15}(x_i-\bar{x})^2}{15-1}\approx 2.843450

s=s21.686253s=\sqrt{s^2}\approx1.686253

The critical value for α=0.01\alpha=0.01 and df=n1=151=14df=n-1=15-1=14 degrees of freedom is tc=2.976842.t_c=2.976842.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉtc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}-t_c\times\dfrac{s}{\sqrt{n}})

=(9.7173332.976842×1.68625315,=(9.717333-2.976842\times\dfrac{1.686253}{\sqrt{15}},

9.717333+2.976842×1.68625315)9.717333+2.976842\times\dfrac{1.686253}{\sqrt{15}})

(8.421,11.013)\approx(8.421, 11.013)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 8.421<μ<11.013,8.421<\mu <11.013, which indicates that we are 99% confident that the true population mean μ\mu is contained by the interval (8.421,11.013).(8.421, 11.013).



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