Question

(7 points) 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. IJ Person A C DE F Test A 108 101 122 75 87 104 96 11889 94 Test B 105 102 122 76 87 102 100 1188697 1. Consider (Test A - Test B). Use a 0.05 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. Two Sample z B. Two Sample t C. Matched Pairs (b) The null hypothesis is Hdiff = 0. What is the alternate hypothesis? A. diff 7 0 B. Hdiff < 0 C. Hdiff> 0 (c) The test statistic is 0.3274 (d) The p-value is 0.6246 (e) Is there sufficient evidence to support the claim that people do better on the second test? A. No B.Yes 2. Construct a 95% confidence interval for the mean of the differences. Again, use (Test A - Test B) 1683

Answer 1

This a two sample dependent test since the sample variables are same.

For this we calculate D (difference in the individual test) and its parameters

A	B	D=A-B
108	105	3
101	102	-1
122	122	0
75	76	-1
87	87	0
104	102	2
96	100	-4
118	118	0
89	86	3
94	97	-3

$\bar{D}=-0.1\;\;S.D._{D}=2.331$$\mu_{d}=0$

(c) Test Statistics = $|\frac{\bar{D}-\mu_{d}}{S_{D}/\sqrt{n}}|$

Test Stat = $|\frac{-0.1-0}{2.331/\sqrt{10}}|$ = 0.13566

$Final \;Ans : Test \;Statistic = 0.13566$

(d) $p - value = P (t > Test\; stat)$ with n - 1 degrees of freedom

$p - value = P (t_{9} > 0.13566)$

Final Ans:  $p - value = 0.4475$      (t - score can be found using t-dist tables online or using excel 'tdist' function)

Since p-value > 0.05 we reject the null hypothesis and conclude that there no sufficient evidence to support the claim that test B are better than test A.

(e) 95% confidence interval for mean of the diff is

$(\bar{D}\;\bar{&plus;}\;t_{n-1,\alpha/2}*S_{d}/\sqrt{n})$

$(-0.1-2.2795*2.331/\sqrt{10},\;\;-0.1&plus;2.2795*2.331/\sqrt{10})$

Final Ans: (-1.7803, 1.5803)