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 71 119 104 100 108 127 83 97 94 92 Test B 72 119 110 105 106 130 85 100 92 91 1. Consider (Test A - Test B). Use a 0.05 0.05 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 test method should be used? A. Two Sample t B. Two Sample z 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 95 95 % confidence interval for the mean of the differences. Again, use (Test A - Test B). <μ< <μ

please show your work and what function to use on the calculator/thank you!

Answer 1

Answer:

Given Data

Person	A	B	C	D	E	F	G	H	I	J
Test A	71	119	104	100	108	127	83	97	94	92
Test B	72	119	110	105	106	130	85	100	92	91

a)

option B

b)

S.No	Test A	Test B
1	71	72	1	1
2	119	119	0	0
3	104	110	6	36
4	100	105	5	25
5	108	106	-2	4
6	127	130	3	9
7	83	85	2	4
8	97	100	3	9
9	94	92	-2	4
10	92	91	-1	1

df = n - 1

= 10 - 1

= 9

Mean = 1.5

Standard deviation =

Here

Here let d = test B -test A

a) Here we are using the same set of persons for twst A and test B .

So we need to use the matched pains.

b) From the given data n = 10

  

= 1.5

= 2.7988

Test statistic

= 1.6949

c) df = 9

It is right tailed

Critical value = +1.833

d) here test statistic = 1.6949 < 1.833

So , do not reject  

No ,there is sufficient evidence to support the claim that people do better on the 2nd test.

2) At 95% , = 2.262

The 95% CI for d is

1.5 -2.00187 , 1.5+2.00187

-0.50187 , 3.50187

From the above 95%confidence interval for population mean = ( -0.50187 , 3.50187)

****Please like it..

Thank you...