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 93 83 113 102 108 73 106 92 100 128 Test B 93 84 109 105 111 71 109 89 101 129 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. (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. Yes B. No 2. Construct a 95% confidence interval for the mean of the differences. Again, use (Test A - Test B). <μ

Answer 1

a) matched pairs

b)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
93	93	0.000	0.090
83	84	-1.000	0.490
113	109	4.000	18.490
102	105	-3.000	7.290
108	111	-3.000	7.290
73	71	2.000	5.290
106	109	-3.000	7.290
92	89	3.000	10.890
100	101	-1.000	0.490
128	129	-1.000	0.490

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	998	1001	-3.000	58.100

mean of difference ,    D̅ =ΣDi / n =   -0.300                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.5408                  
                          
std error , SE = Sd / √n =    2.5408   / √   10   =   0.8035      
                          
t-statistic = (D̅ - µd)/SE = (   -0.3   -   0   ) /    0.8035   =   -0.37

c)

Ho :   µd=   0
Ha :   µd <   0

Degree of freedom, DF=   n - 1 =    9  
t-critical value , t* =        -1.833   [excel function: =t.inv(α,df) ]

d)

test stat >-1.833, do not reject Ho

NO

2)

sample size ,    n =    10          
Degree of freedom, DF=   n - 1 =    9   and α =    0.05  
t-critical value =    t α/2,df =    2.2622   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.5408          
                  
std error , SE = Sd / √n =    2.5408   / √   10   =   0.8035
margin of error, E = t*SE =    2.2622   *   0.8035   =   1.8176
                  
mean of difference ,    D̅ =   -0.300          
confidence interval is                   
Interval Lower Limit= D̅ - E =   -0.300   -   1.8176   =   -2.118
Interval Upper Limit= D̅ + E =   -0.300   +   1.8176   =   1.518
                  
so, confidence interval is (   -2.1176   < µd <   1.5176   )