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Ten randomly selected people took an IQ test A, and next day they took a very si

ID: 3265206 • Letter: T

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.



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. Two Sample z
B. Matched Pairs
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).
<<

Person A B C D E F G H I J Test A 100 74 104 116 91 112 126 93 87 96 Test B 104 73 108 117 96 112 129 95 87 100

Explanation / Answer

(a) The test to be used is Paired t test, as the values of Test A and Test B are paired (same people involved).

The R code is:

testA <- c(100,74,104,116,91,112,126,93,87,96)
testB <- c(104,73,108,117,96,112,129,95,87,100)
t.test(testA,testB,paired = T, tail=left)

The output is:

Paired t-test

Paired t-test

data: testA and testB

t = -3.3166, df = 9, p-value = 0.008989

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-3.7005453 -0.6994547

sample estimates:

mean of the differences

-2.2

(b) The test statistic is: t = -3.3166

(c) The critical value for 0.05 and df=9 is: 1.833

(d) Since the absolute value of the test statistic is greater than the absolute critical value. the null hypothesis is rejected, and there is sufficient evidence to support the claim that the people do better on the second test. Yes

(e) The 95% confidence interval of the difference is: (-3.7005453, -0.6994547)

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