Answer to Question #238433 in Statistics and Probability for baber

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) $s=9.302.$