Question

(1 point) Ten randomly selected people took an IQ test A, and the next day they...

(1 point) Ten randomly selected people took an IQ test A, and the 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 102 107 88 109 77 96 99 117 83 128
Test B 101 108 91 112 78 100 102 119 84 130



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.

(a) What test method should be used?

A. Two Sample z
B. Two Sample t
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. No
B. Yes


2. Construct a 95% confidence interval for the mean of the differences. Again, use (Test A - Test B).
<?<<μ<

Homework Answers

Answer #1

Step 1

C. Matched Pairs

To Test :-

H0 :- µd = 0

H1 :- µd < 0

Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(18.9 / 10-1) = 1.4491
d̅ = Σ di/n = -19 / 10 = -1.9
t = d̅ / ( S(d) / √(n) )
t = -1.9 / ( 1.4491 / √(10) )
t = -4.1461


Test Criteria :-
Reject null hypothesis if t < - t(α)
Critical value t(α) = t(0.05) = 1.83311
t < - t(α) = -4.1461 < -1.83311
Result :- Reject null hypothesis

B. Yes, we  support the claim that people do better on the second test.

Step 2

Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.05 /2) = 1.8331
-1.9 ± t(0.05/2) * 1.4491/√(10)
Lower Limit = -1.9 - t(0.05/2) 1.4491/√(10)
Lower Limit = -2.9365
Upper Limit = -1.9 + t(0.05/2}) 1.4491/√(10)
Upper Limit = -0.8635
95% Confidence interval is ( - 2.9365 , - 0.8635 )

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