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	81	105	91	116	102	74	95	122	99	112
Test B	84	105	89	122	103	77	98	122	102	116

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. Round calculated answers to three decimal places.

(a) What test method should be used?

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

(b) The null hypothesis is μdiff=0μdiff=0. What is the alternate hypothesis?

A. μdiff≠0
B. μdiff>0
C. μdiff<0

(c) The test statistic is

(d) The p-value is

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

A. Yes
B. No

2. Construct a 9595% confidence interval for the mean of the differences. Again, use (Test A - Test B).
____ <μ< ____

Answer 1

1)a) Option - A) Matched pairs

b) Option - C) < 0

c) = ((-3) + 0 + 2 + (-6) + (-1) + (-3) + (-3) + 0 + (-3) + (-4))/10 = -2.1

s_d = sqrt((((-3 + 2.1)^2 + (0 + 2.1)^2 + (2 + 2.1)^2 + (-6 + 2.1)^2 + (-1 + 2.1)^2 + (-3 + 2.1)^2 + (-3 + 2.1)^2 + (0 + 2.1)^2 + (-3 + 2.1)^2 + (-4 + 2.1)^2)/9) = 2.331

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

                                 = (-2.1 - 0)/(2.331/)

                                = -2.85

d) P-value = P(T < -2.85)

                  = 0.0095

e) Since the P-value is less than the significance level(0.0095 < 0.05), so we should reject H₀.

Yes, there is sufficient evidence to support the claim that people do better on the second test.

2) At 95% confidence interval the critical value is t^* = 2.262

The 95% confidence interval is

+/- t^* * s_d/

= -2.1 +/- 2.262 * 2.331/

= -2.1 +/- 1.667

= -3.767, -0.433