Question

A survey of 35 individuals who passed the seven exams and obtained the rank of Fellow in the actuarial field finds the average salary to be $150,000. If the standard deviation for the population is $15000, construct a 95% confidence interval for all Fellows

Answer 1

Answer:

$150000-2.032\frac{15000}{\sqrt{35}}=144847.94$    

$150000+2.032\frac{15000}{\sqrt{35}}=155152.06$

So on this case the 95% confidence interval would be given by (144847.94;155152.06)    

Step-by-step explanation:

Previous concepts

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".

$\bar X=150000$ represent the sample mean

$\mu$ population mean (variable of interest)

s=15000 represent the sample standard deviation

n=35 represent the sample size  

Solution to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=35-1=34$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,34)".And we see that $t_{\alpha/2}=2.032$

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

$150000-2.032\frac{15000}{\sqrt{35}}=144847.94$    

$150000+2.032\frac{15000}{\sqrt{35}}=155152.06$

So on this case the 95% confidence interval would be given by (144847.94;155152.06)