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 233233 cars owned by students had an average age of 6.626.62 years. A sample of 280280 cars owned by faculty had an average age of 7.947.94 years. Assume that the population standard deviation for cars owned by students is 2.132.13 years, while the population standard deviation for cars owned by faculty is 3.143.14 years. Determine the 90%90% 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.

Step 2 of 2:

Construct the 90%90% confidence interval. Round your answers to two decimal places.

Answer 1

	1	2
	6.62	7.94	mean
	2.13	3.14	std. dev.
	233	280	n
		-1.32000	difference (1 - 2)
		0.23385	standard error of difference
		0	hypothesized difference
		-5.64	z
		1.65E-08	p-value (two-tailed)
		-1.70464	confidence interval 90.% lower
		-0.93536	confidence interval 90.% upper
		0.38464	margin of error

Step 1 of 2:

Find the critical value that should be used in constructing the confidence interval.

If there are two blanks then ans is -1.645 and 1.645.

1.64485 or 1.645 [Please follow the round off instructions i have done round off upto three decimal places]

Step 2 of 2:

Construct the 90%90% confidence interval. Round your answers to two decimal places.

(-1.70,-0.94)