Question

Fifty people in the civilian labor force are randomly selected and the sample average age iscomputed to be 36.4.(a) Find a 90% confidence interval for the mean age, ?, of all people in the civilian laborforce. Assume that the population standard deviation for the ages of civilian labor force is12.1 years. Interpret the confidence interval.(b) It is being claimed that the mean age of the population of civilian labor force is 40. Whatdo you conclude based on the confidence interval?

Answer 1

Answer:

a) The 90% confidence interval would be given by (33.594;39.206)  

b) Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

$\bar X=36.4$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=12.1$ represent the population standard deviation  

n=50 represent the sample size  

90% confidence interval  

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$  

Now we have everything in order to replace into formula (1):  

$36.4-1.64\frac{12.1}{\sqrt{50}}=33.594$  

$36.4+1.64\frac{12.1}{\sqrt{50}}=39.206$  

So on this case the 90% confidence interval would be given by (33.594;39.206)  

Part b

Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.