(2 pts) Ten randomly selected people took an IQ test A, and next day they took a

Question

(2 pts) 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 a software program, such as Excel, to compute the differences, sample mean difference, and sample standard deviation of differences.)

B. Paired data t should be used

(a) The test statistic is

(b) The critical value is

(c) 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).
<d<

Person 1 2 3 4 5 6 7 8 9 10 Test A 76 123 93 83 103 100 104 89 114 108 Test B 77 126 97 85 106 101 106 90 115 106

Explanation / Answer

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 a software program, such as Excel, to compute the differences, sample mean difference, and sample standard deviation of differences.)

B. Paired data t should be used

Here the difference table is

Average of difference dbar =  1.6

Standard deviation sd = 1.6465

(a) The test statistic is

t = dbar / (Sd / sqrt(n)) = 1.6 / (1.6465/ sqrt(10) ) = 1.6/0.5207 = 3.073

(b) The critical value is

Here dF = 10 - 1 = 9

alpha = 0.05

tcritical = t9,0.05 = 1.8331

(c) Is there sufficient evidence to support the claim that people do better on the second test?

Here t > tcritical so we shall reject the null hypothesis and can conclude that there is significant difference in mean score in IQ score.
Yes option is correct.

Question 2

95%  confidence interval for the mean of the differences. = dbar - tcritical (sd/sqrt(n) < d <  dbar + tcritical (sd/sqrt(n)

=> 1.6 - 1.8331 * 0.5207 <  d < 1.6 +1.8331 * 0.5207

95% confidence interval = (0.6455, 2.5545)

Person Test A Test B Difference 1 76 77 1 2 123 126 3 3 93 97 4 4 83 85 2 5 103 106 3 6 100 101 1 7 104 106 2 8 89 90 1 9 114 115 1 10 108 106 -2

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