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	99	86	112	90	106	78	120	95	118	101
Test B	102	86	109	92	104	79	121	97	115	106

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. No
B. Yes


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

Answer 1

Let be mean test score for Test A

and be mean test score for Test B

Here we have to test

a) Here we have to use two sample t test since two samples are independent.

But first we have to check population variances are equal or not.For that we have to use F test

Test statistic:

P value from excel using command:

=FDIST(1.138,9,9)         (Degrees of freedom for numerator = n1-1=10-1= 9 and degrees of freedom

=0.4252   for denominator =n2-1=10-1 = 9)                       

P value = 2*0.4252 = 0.8505

Here P value > alpha = 0.01

So we do not reject H0

That means population variances are equal

So we can use here pooled two sample t test

b) Test statistic:

where

             

            

             = 13.3535              (Round to 4 decimal)

                            = 5.9719                    (Round to 4 decimal)

t = -0.100       (Round to 3 decimal)

Test statistic = -0.100

c) Critical value:

degrees of freedom = n1+n2 - 2 = 10 + 10 -2 = 18

alpha = level of significance = 0.01

Critical value = 2.552                 (From statistical table of t values)

Here |t| = 0.1 < critical value = 2.552

So we do not reject H0

d) No.There is not sufficient evidence to support the claim that people do better on the second test

2) 99% confidence interval for is

Where tc is t critical for degrees of freedom = n1+n2-2 = 10+10-2 = 18 and confidence level = c= 0.99

tc = 2.878        (From statistical table of t values)

99% confidence interval for mean of differences is (-17.7871 , 16.5871)