Question

Ten randomly selected people took an IQ test A, and the 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	113	122	89	96	80	94	110	85	105	102
Test B	120	123	87	99	80	95	110	88	105	102

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 the software.)

(a) What test method should be used?

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

(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

1)

Matched pairs

b)

Test statistic,
t = (dbar - 0)/(s(d)/sqrt(n))
t = (-1.3 - 0)/(2.4967/sqrt(10))
t = -1.647

c)

Critical value of t is -1.833.

d)

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

e)

sample mean, xbar = -1.3
sample standard deviation, s = 2.4967
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.25

ME = tc * s/sqrt(n)
ME = 3.25 * 2.4967/sqrt(10)
ME = 2.566

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (-1.3 - 3.25 * 2.4967/sqrt(10) , -1.3 + 3.25 * 2.4967/sqrt(10))
CI = (-3.87 , 1.27)

-3.87< mu < 1.27