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

To construct a 95% confidence interval for the average salary of all Fellows in the actuarial field, the formula Confidence Interval = Sample Mean ± (Z * Standard Deviation / √n) is used. Plugging in the given values, the 95% confidence interval for the average salary is approximately $145,737.51 to $154,262.49.

Step-by-step explanation:

To construct a 95% confidence interval for the average salary of all Fellows, we can use the formula:

Confidence Interval = Sample Mean ± (Z * Standard Deviation / √n)

Where:

Sample Mean = $150,000 (as given)

Z = Z-score corresponding to the desired confidence level (in this case, 95%; Z-score = 1.96 for a 95% confidence level)

Standard Deviation = $15,000 (as given)

n = Number of individuals surveyed = 35 (as given)

Plugging in the values:

Confidence Interval = $150,000 ± (1.96 * $15,000 / √35)

Simplifying further:

Confidence Interval = $150,000 ± $4,262.49

Therefore, the 95% confidence interval for the average salary of all Fellows is approximately $145,737.51 to $154,262.49.

Answer 2

`150000-2.032(15000)/(√(35))=144847.94`

`150000+2.032(15000)/(√(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)(s)/(√(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(15000)/(√(35))=144847.94`

`150000+2.032(15000)/(√(35))=155152.06`

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