Answer to Question #204789 in Statistics and Probability for sam21ucd

Question

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

Answer 1

$$\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{\displaystyle\sum_{i=1}^{15}x_i}{15}$$

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

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

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

$$\approx9.717333$$

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

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

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:

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

$$=(9.717333-2.976842\times\dfrac{1.686253}{\sqrt{15}},$$

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

$$\approx(8.421, 11.013)$$

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 $(8.421, 11.013).$