Question

Ten randomly selected people took 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	80	120	113	81	97	112	90	103	106	95
Test B	80	123	109	80	96	111	88	108	105	96

Calculate (Test B - Test A) to find the differences. Use a 0.050.05 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 Dependent Samples t-test (Matched Pairs)
B. Two Independent Samples z-test
C. Two Independent Samples t-test

(b) The test statistic is

(c) The p-value is

Answer 1

a)

Since the observations were taken on the same persons, that is, the test A and test B has given on same peoples but at the different time period, therefore the given problem is of two dependent Samples t-test (Matched Pairs).

b)

Let be the true mean difference = mean of scores in test B - mean of scores in test A.

We have to test the claim that people do better on the second test than they do on the first. That is, we have to test that is greater than zero.

Therefore, the hypothesis is,

The null hypothesis, H₀ : True mean difference (?_d) is equal to zero. That is, the performance of the peoples in both the test is same.

The alternative hypothesis, H₁ : True mean difference (?_d) is greater than zero. That is, people do better on the second test than they do on the first.

Let. di = score of test B -score of test A on each pair.

then di = 0, 3, -4, -1, -1, -1, -2, 5, -1, 1

and standard deviation of di = SD_d= 2.5582.

Now, the test statistic is,

therefore the value of test statistic is -0.1236.

c)

The sampling distribution of test statistic is t- distribution with (n-1) degrees of freedom.

Therefore, the p-value is 0.5478.

Since the p-value > 0.05 (Level of significance), we fail to reject the null hypothesis and conclude that there is no evidence to claim that people do better on the second test than they do on the first at 5 % level of significance.