Question

Construct the discrete series. Count unbiased estimates of the general mean and general variance. Find the confidence interval for expectation with a confidence level of 0.05 5, 11, 13, 9, 11, 5, 7, 7, 5, 9, 13, 13, 11, 9, 5, 9, 11, 9, 5, 9, 11, 9, 9, 5.

Answer 1

(i)

$n=24$

$$\bar{x}=\dfrac{1}{24}(5+11+13+9+11+5+7+7$$

$$+5+9+13+13+11+9+5+9+11+9$$

$$+5+9+11+9+9+5)=8.75$$

Unbiased estimates of the general mean is $8.75.$

$$Var(X)=s^2=\dfrac{1}{24-1}((5-8.75)^2+(11-8.75)^2$$

$$+(13-8.75)^2+(9-8.75)^2+(11-8.75)^2$$

$$+(5-8.75)^2+(7-8.75)^2+(7-8.75)^2$$

$$+(5-8.75)^2+(9-8.75)^2+(13-8.75)^2$$

$$+(13-8.75)^2+(11-8.75)^2+(9-8.75)^2$$

$$+(5-8.75)^2+(9-8.75)^2+(11-8.75)^2$$

$$+(9-8.75)^2+(5-8.75)^2+(9-8.75)^2$$

$$+(11-8.75)^2+(95-8.75)^2+(9-8.75)^2$$

$$+(5-8.75)^2)=7.413043$$

Unbiased estimates of the general variance is $7.413043.$

$$s=\sqrt{s^2}=2.7227$$

(ii)The critical value for $\alpha = 0.05$ and $df = n-1 = 23$  degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.068658.$

The corresponding confidence interval is computed as shown below:

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

$$=(8.75-2.068658\times\dfrac{2.7227}{\sqrt{24}},$$

$$8.75+2.068658\times\dfrac{2.7227}{\sqrt{24}})$$

$$=(7.6003, 9.8997)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $7.6003 < \mu < 9.8997,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(7.6003,9.8997).$