A football coach claims that players can increase their strength by taking a cer

Question

A football coach claims that players can increase their strength by taking a certain supplement. To test the theory, the coach randomly selects 9 athletes and gives them a strength test using a bench press. Thirty days later, after regular training using the supplement, they are tested again, results are listed below. Test the claim that the supplement is effective in increasing the athletes' strength Use alpha = 0.05. Be sure to include the following in your answer Null Hypothesis: Alternate Hypothesis: P-value

Explanation / Answer

XB =222.222     s2B = 967.944      nB=9

XA =226.667     s2A = 1233.5        nA=9

To test the equality of population variance:

H0 = The population variances are equal i.e., 2B = 2A

H1 = The population variances are not equal i.e., 2B 2A

F = 2A / 2B = 1233.5/967.944 = 1.2743 (The bigger variance in numerator)

F critical value =F0.05,(8,8) = 3.4381

P- Value = 0.7399

Since, Critical value = 3.4381 > F value =1.2743, and P-value =0.7399 > =0.05, we accept the null hypothesis and conclude that the population variances are equal.

To test the equality in mean:

H0 = There is no significant difference in strength due to supplement i.e., µB = µA

H1 = There is an increase in strength due to supplement i.e., µB < µA

Since the population variances are equal,

The t statistic is given by:

t = XB - XA / [Sp(1/nB + 1/nA)]

where Sp is the pooled standard deviation.

S2p = (nB -1) s2B + (nA-1) s2A /( nB + nA -2) = [(8*967.944)+(8*1233.5)]/16 = 1100.7222

Sp = 1100.7222 = 33.1771

t = 222.222 – 226.667 /[ 33.1771 0.222] = -0.2842

P-value = 0.39

Critical value = 1.7459

Since, tcal = -0.2842 > ttab = -1.7459 and the p-value= 0.39 > = 0.05, we accept the null hypothesis and hence conclude that the supplements do not increase the athletes’ strength.

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