Question

Two classes of warriors are arguing over who comes from the more powerful school of martial arts. The Wing Chun warrior claims that his school defeats more enemies, whereas the Jeet Kune Do warrior argues that his school defeats more enemies. Their friend proposes a solution: they each take a random sample of 39 days and compare the average number of enemies defeated on each day. He proposes they conduct a two-sample t-test with the hypotheses:

H₀: μ_{Wing Chun} - μ_{Jeet Kune Do} = 0
H_a: μ_{Wing Chun} - μ_{Jeet Kune Do} ≠ 0

After taking the samples and running through the correct procedures, the warriors produce a test statistic of 3.5282 and degrees of freedom 43.579. They return to their friend for some help:

(a) What is the p-value for their test? (3 decimal places)  

(b) At α = 0.05, what should the warriors conclude?

There is evidence that the two schools have different average numbers of enemies defeated each day.

We do not have enough evidence to reject the null hypothesis; there is not evidence of a difference in the average numbers of enemies defeated each day.  

   The school of Jeet Kune Do seems to have a higher average number of enemies defeated per day.

There is insufficient evidence to conclude that the school of Wing Chun has a lower number of enemies defeated per day, on average.

Answer 1

Since , the null and alternative hypothesis is ,

The test is two-tailed test.

(a)Since , df=degrees of freedom=43.57944

t-stat=3.5282

The p-value is ,

p-value= ; The Excel function is , =TDIST(3.5282,44,2)

(b) Decision : Here , p-value=0.001<0.05

Therefore , reject Ho.

Conclusion : There is sufficient evidence that the two schools have different average numbers of enemies defeated each day.