Question

(3 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. Person A B C D E F G H I J Test A 94 104 86 110 78 92 102 97 115 126 Test B 97 102 87 108 80 91 102 96 119 129 1. Consider (Test A - Test B). Use a 0.01 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 OC. Two Samplet (b) The test statistic is (c) Is there sufficient evidence to support the claim that people do better on the second test? A. Yes B. No 2. Construct a 99% confidence interval for the mean of the differences. Again, use (Test A - Test B). u<

Answer 1

1) A matched-pairs test is used to test whether there is a significant mean difference between two sets of paired data.

Option B is correct

2)   The sample mean and sample standard deviation of the difference is

d = -0.7, sa 2.21

The value of the test static is

$t={-0.7\over 2.21/\sqrt{10}}\approx -1.002$

The test is left tailed test and the number of degrees of freedom is 10-1=9

So P-value is $\Pr( t <-1.002)=0.1713$

c) As P-value is 0.1713>0.01

So we fail to reject the null hypothesis

There is not enough evidecnce to support the claim.

Option 2nd is correct

d) The   confidence interval is given by

(1 - a) 100%

$\left( \bar{d}-{s_d\over \sqrt{n}}t_{\alpha/2}~,~ \bar{d}+{s_d\over \sqrt{n}}t_{\alpha/2} \right )$

Use the above values and the fact that for 99% confidence interval $t_{\alpha/2}= 3.25$

Hence the interval is