What is the Critical Value and 95% confidence interval

Question

Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below.

Person	A	B	C	D	E	F	G	H	I	J
Test A	96	108	117	88	100	83	96	75	110	130
Test B	95	110	113	91	103	82	99	75	113	132

1. Consider (Test A - Test B). Use a 0.050.05 significance level to test the claim that people do better on the second test than they do on the first. (Note: You may wish to use software.)

(a) What is the critical value?

(b) Construct a 95% confidence interval for the mean of the differences. Again, use (Test A - Test B).

Answer 1

Since the people are taking both tests and comparing the results, we'll use a paired t-test to answer this question. This means that we first find the differences (1, -2, 4, -3, -3, 1, -3, 0, -3, -2), mean (-10/10 = -1), and sample standard deviation (sum[(d-dbar)^2]/(10-1) ≈ 2.4037).

a) Since we have 9 degrees of freedom and we're constructing a 95% confidence interval, we simply pull that value from a t-table: t*=2.262

b) For a confidence interval, I find it easiest to first calculate the margin of error (MOE):

MOE = t*•s/√n

= 2.262•2.4037/√10

= 1.7194

Now add and subtract this from the mean:

CI = (-1-1.7194, -1+1.7194) = (-2.7194, 0.7194)

While this is not the same as running a hypothesis test, we would get the same result in either case: Because the confidence interval contains 0, we can be 95% confident that there is no significant difference between the two tests.