Question

b) mean and standard deviation? (b) What is the confidence interval for population mean at 95 per cent confidence interval The following data have been collected for a sample from a normal population: 5,11, 8, 11, 18, 13, 33, 27, 22 (a) What is the point estimate of population? 

Answer 1

(a)

$$mean=\bar{x}=\dfrac{1}{9}(5+11+8+11+18+13+33$$

$$+27+22)=\dfrac{148}{9}\approx16.444$$

$$s^2=\dfrac{1}{9-1}((5-\dfrac{148}{9})^2+(11-\dfrac{148}{9})^2$$

$$+(8-\dfrac{148}{9})^2+(11-\dfrac{148}{9})^2$$

$$+(18-\dfrac{148}{9})^2+(13-\dfrac{148}{9})^2$$

$$+(33-\dfrac{148}{9})^2+(27-\dfrac{148}{9})^2$$

$$+(22-\dfrac{148}{9})^2)=86.528$$

$$standard\ deviation=s=\sqrt{s^2}=9.302$$

(b) The critical value for $\alpha=0.05$ and $df=n-1=9-1=8$ degrees of freedom is $t_c=z_{1-\alpha/2, n-1}=2.306002.$

The corresponding confidence interval is computed as shown below:

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

$$=(16.444-2.306\times \dfrac{9.302}{\sqrt{9}}, 16.444+2.306\times \dfrac{9.302}{\sqrt{9}})$$

$$=(9.294, 23.594)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $9.294<\mu<23.594,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(9.294, 23.594).$

The point estimate of population mean is the sample mean $\bar{x}=16.444.$

The sample standard deviation (s) is a point estimate of the population standard deviation (σ) $s=9.302.$