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	95	89	124	105	99	114	108	73	103	84
Test B	99	91	123	110	99	119	105	72	106	86

1. Consider (Test A - Test B). Use a 0.01 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 test method should be used?

A. Matched Pairs
B. Two Sample z
C. Two Sample t

(b) The test statistic is:

(c) The critical value is:

(d) Is there sufficient evidence to support the claim that people do better on the second test?

A. Yes
B. No


2. Construct a 99% confidence interval for the mean of the differences. Again, use (Test A - Test B).
____ <μ< ____

Answer 1

a) Option - A) Matched pairs

H₀: = 0

H₁: < 0

b) = (-4 + (-2) + 1 + (-5) + 0 + (-5) + 3 + 1 + (-3) + (-2))/10 = -1.6

s_d = sqrt(((-4 + 1.6)^2 + (-2 + 1.6)^2 + (1 + 1.6)^2 + (-5 + 1.6)^2 + (0 + 1.6)^2 + (-5 + 1.6)^2 + (3 + 1.6)^2 + (1 + 1.6)^2 + (-3 + 1.6)^2 + (-2 + 1.6)^2)/9) = 2.7568

The test statistic t = ( - D)/(s_d/)

= (-1.6 - 0)/(2.7568/)

= -1.84

c) At alpha = 0.01, the critical value is t_{0.01, 9} = -2.821

d) Since the test statistic value is not less than the critical value (-1.84 > -2.821), so we should not reject the null hypothesis.

No, there is not sufficient evidence to support the claim that people do better on the second test.

2) At 99% confidence interval the critical value is t_{0.005, 9} = 3.250

The 99% confidence interval is

+/- t_{0.005, 9} * s_d/

= -1.6 +/- 3.25 * 2.7568/

= -1.6 +/- 2.83

= -4.53, 1.23

-4.53 < < 1.23