Question

A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 104 cars owned by students had an average age of 5.42 years. A sample of 145 cars owned by faculty had an average age of 5.57 years. Assume that the population standard deviation for cars owned by students is 2.69 years, while the population standard deviation for cars owned by faculty is 2.46 years. Determine the 98% confidence interval for the difference between the true mean ages for cars owned by students and faculty. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval.

Answer 1

A sample of 104 cars owned by students had an average age of 5.42 years. A sample of 145 cars owned by faculty had an average age of 5.57 years. Assume that the population standard deviation for cars owned by students is 2.69 years, while the population standard deviation for cars owned by faculty is 2.46 years.

So,

When 1 denotes the student's car population and 2 refers to the faculty's car population. Note than,

, also,

and

now for 98% confidence interval, that is, 0.02 level of confidence,

here,

So,(1-)100% Confidence interval is given by:

putting, =0.02, we have , which is the critical value.

Using all the information, the 98% confidence interval for difference in mean is given by: [-0.93,0.63] (rounded up to 2 significant digits after decimal)