A random sample of 42 college graduates who worked during their program revealed that a student spent an average 5.5 years on the job before being promoted. The sample standard deviation was 1.1 years. Using the 0.99 degree of confidence, what is the confidence interval for the population mean?
a. 5.04 and 5.96b. 5.06 and 5.94c. 2.67 and 8.33d. 4.40 and 6.60

So on this case the 99% confidence interval would be given by (5.04;5.96)

And the best option would be:

a. 5.04 and 5.96

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".

represent the sample mean for the sample  

population mean (variable of interest)

s=1.1 represent the sample standard deviation

n=42 represent the sample size  

Solution to the problem

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

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.99 or 99%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,41)".And we see that

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

So on this case the 99% confidence interval would be given by (5.04;5.96)

And the best option would be:

a. 5.04 and 5.96

correct answer:

b.(19,-2)

cant open file sorry!

step-by-step explanation: