Question

To examine performance on two aptitude tests, 4 students were enlisted to take both tests.

Their scores were recorded as follows:

Student 1 2 3 4

Test1 Score 72 87 64 67

Test2 Score 74 74 79 74

Calculate a 99% confidence interval for the difference between the test scores.

Answer 1

Test 1	Test 2	Difference
72	74	-2
87	74	13
64	79	-15
67	74	-7

∑d =    -11  

∑d² =    447  

n =    4  

Mean , x̅d = Ʃd/n =    -11/4 =    -2.7500

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] =    √[(447-(-11)²/4)/(4-1)] =    11.7863

99% Confidence interval for the difference between the test scores:  

At α = 0.01 and df = n-1 = 3, two tailed critical value, t-crit = T.INV.2T(0.01, 3) =    5.841

Lower Bound = x̅d - t-crit*sd/√n = -2.75 - 5.841 * 11.7863/√4 =    -37.1713

Upper Bound = x̅d + t-crit*sd/√n = -2.75 + 5.841 * 11.7863/√4 =    31.6713

-37.1713 < µd < 31.6713