Answer in Statistics and Probability for sam21ucd #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
“=dfrac{1}{15}(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=dfrac{145.76}{15}”
“approx9.717333”
“s^2=dfrac{displaystylesum_{i=1}^{15}(x_i-bar{x})^2}{15-1}approx t2.843450”
“s=sqrt{s^2}approx1.686253”
The critical value for “alpha=0.01” and “df=n-1=15-1=14” degrees of freedom is “t_c=2.976842.”
The corresponding confidence interval is computed as shown below:
“CI=(bar{x}-t_ctimesdfrac{s}{sqrt{n}}, bar{x}-t_ctimesdfrac{s}{sqrt{n}})”
“=(9.717333-2.976842timesdfrac{1.686253}{sqrt{15}},”
“9.717333+2.976842timesdfrac{1.686253}{sqrt{15}})”
“approx(8.421, 11.013)”
Therefore, based on the data provided, the 99% confidence interval for the population mean is “8.421<mu <11.013,” which indicates that we are 99% confident that the true population mean “mu” is contained by the interval “(8.421, 11.013).”
