AFTER Practice, do football players have DIFFERENT mean CPK values compared to s

Question

AFTER Practice, do football players have DIFFERENT mean CPK values compared to soccer players? Test this claim by performing a hypothesis test, using a=0.10.

Football players n=25 After practice: Mean=225.6 Standard deviation= 132.6

Soccer Players n=15 After pratice: Mean=173.8 Standard deviation= 64.4

H0:u=u

H1: u not equal to u

P-value= .11>a=.10

Do no reject H0, Reject h1 We do not have evidence to support the claim.

I was wondering if this was the right answer I feel like it is not. If it is not I would like to know what answer and what thing to use for the TI89 calculator and what I did wrong.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 31.30

DF = 38
t = [ (x1 - x2) - d ] / SE

t = 1.65

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 38 degrees of freedom is more extreme than -1.65; that is, less than -1.65 or greater than 1.65.

Thus, the P-value =0.11

Interpret results. Since the P-value (0.11) is greater than the significance level (0.10), we chave to reject the null hypothesis.

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