Answer to Question #290461 in Statistics and Probability for yoshi

Question

A random sample of size 144 is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are x−−=58.2 and s = 2.6.

Answer 1

a. We need to construct the 80% confidence interval for the population mean $\mu.$

The following information is provided:

Sample Mean $\bar{X}=58.2$

Sample Standard Deviation $s=2.6$

Sample Size $n=144$

The critical value for $\alpha=0.2$ and $df=n-1=143$ degrees of freedom is $t_c=t_{1-\alpha/2, df}=1.2875.$  

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(58.2-1.2875\times\dfrac{2.6}{\sqrt{144}}, 58.2+1.2875\times\dfrac{2.6}{\sqrt{144}})$$ $$=(57.921, 58.479)$$

Therefore, based on the data provided, the 80% confidence interval for the population mean is $57.921<\mu<58.479),$  which indicates that we are 80% confident that the true population mean $\mu$  is contained by the interval $(57.921, 58.479).$

b. We need to construct the 90% confidence interval for the population mean $\mu.$

The following information is provided:

Sample Mean $\bar{X}=58.2$

Sample Standard Deviation $s=2.6$

Sample Size $n=144$

The critical value for $\alpha=0.1$ and $df=n-1=143$ degrees of freedom is $t_c=t_{1-\alpha/2, df}=1.655579.$  

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(58.2-1.655579\times\dfrac{2.6}{\sqrt{144}},$$ $$58.2+1.655579\times\dfrac{2.6}{\sqrt{144}})$$ $$=(57.8413, 58.5587)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $57.8413<\mu<58.5587),$  which indicates that we are 90% confident that the true population mean $\mu$  is contained by the interval $(57.8413, 58.5587).$

$1-\alpha,$ is called the confidence level. The width of a confidence interval is a measure of the accuracy with which we can pinpointthe estimate of a parameter. The narrower the confidence interval, the more accurately we can specify the estimate for a parameter

The width of the confidence interval increases as the confidence level increases. The wider the confidence interval is, the more confident we can be that the interval contains the unknown parameter.