Answer to Question #269153 in Statistics and Probability for kamal

Question

The mean weight (in kg) for 8 adults males are given as follows: 70 72 65 80 75 76 68 78 Construct a 90% confidence interval estimate for the mean weight for all adult males.  

Answer 1

$$\bar{x}=\dfrac{70+ 72 +65+ 80+ 75 +76+ 68+78}{8}=73$$

$$s^2=\dfrac{1}{8-1}((70-73)^2+(72-73)^2+(65-73)^2$$

$$+(80-73)^2+(75-73)^2+(76-73)^2$$

$$+(68-73)^2+(78-73)^2)=\dfrac{186}{7}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{186}{7}}\approx5.154748$$

The critical value for $\alpha = 0.1$ and $df = n-1 = 7$ degrees of freedom is$t_c = z_{1-\alpha/2; n-1} = 1.894577.$

The corresponding confidence interval is computed as shown below:

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

$$=(73-1.894577\times\dfrac{5.154748}{\sqrt{8}},$$

$$73+1.894577\times\dfrac{5.154748}{\sqrt{8}})$$

$$=(69.547, 76.453)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $69.547 < \mu < 76.453,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(69.547, 76.453).$