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	125	100	114	99	91	110	87	73	104
Test B	97	125	102	116	99	90	112	89	75	104

1. Consider (Test A - Test B). Use a 0.010.01 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. Two Sample z
B. Matched Pairs
C. Two Sample t

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

A. μdiff<0μdiff<0
B. μdiff>0μdiff>0
C. μdiff≠0μdiff≠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. No
B. Yes

2. Construct a 9999% confidence interval for the mean of the differences. Again, use (Test A - Test B).
<μ<<μ<

Answer 1

Q 1) a) Answer: Matched Pairs

Q b)

Null Hypothesis

Alternative Hypothesis

c) Difference d = Type A- Type B

The Calculation table is given below:

Person	A	B	C	D	E	F	G	H	I	J
Test A	95	125	100	114	99	91	110	87	73	104
Test B	97	125	102	116	99	90	112	89	75	104	Total
d= Type A-Type B	-2	0	-2	-2	0	1	-2	-2	-2	0	-11
d^2	4	0	4	4	0	1	4	4	4	0	25

Under H0, the test statistic is

Degrees of freedom = n-1= 10-1= 9

d) The P-Value is .0087

e) Since p value is less than significance level, Reject H0.

Hence,at 1% significance level, we have sufficient evidence to support the claim that people do better on the second test

Answer: Yes

Q2) 99% CI

= (-0.130, 2.330)

Answer: The 99% CI is