1. You are a teacher who is involved with a new, CD-ROM version of exercise phys

Question

1. You are a teacher who is involved with a new, CD-ROM version of exercise physiology laboratories, as opposed to the old method of actually testing everyone in the class using different fitness tests. You decide to give one of the two methods to two different classes in order to see if one method is better than the other. After you do this, you test all the students on their exercise physiology knowledge using a test with a possible score of 100. The data are:

CD-ROM: 88, 77, 69, 86, 69, 92, 99, 91, 85, 87

Old method: 76, 84, 79, 92, 82, 77, 65, 68, 73, 82, 84, 87

a. What t-test is used?

b. How many df are there?

c. What are the means?

d. What are the standard deviations?

e. What is the alpha level?

f. One or two-tailed test?

g. What is the t value?

h. What is the probability value?

i. What is the interpretation of the test?

Explanation / Answer

a. HEre we will use two sample t- test for equal variances.

b. dF = n1 + n2 -2 = 10+ 12 - 2 = 20

c. Means are

for CD - ROM x1 = 84.3

for old method , x2 = 79.0833

d. standard deviation of samples are

for CD - ROM, s1 = 9.81

for old method, s2 = 7.7864

e. alpha level is not given so we take it 0.05

f. Here we want to know if one method is better than the others so ofcourse it is a one - tailed test.

g. Here pooled standard deviation sp = sqrt [{(n1 -1)s12 + (n2 -1)s22 }/(n1 + n2 -2)]

sqrt [(11 * 9.812 + 9 * 7.78642 )/20] = 8.755

standard error of difference sed = sp * sqrt [1/n1 + 1/n2] = 8.755 * sqrt (1/10 + 1/12) = 3.7487

h. Test statistic

t = (x1 - x2)/ sed = (84.3 - 79.0833)/ 3.7487 = 1.392

(h) Here p - value = 0.0897

(i) Here we can interpret that the CD- ROM method is not better then the other old method.

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