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n=10 before: mean=255.63, SD=115.50 after: mean=284.75, SD=132.63 difference: mean=29.13, SD=21 conduct hypothesis test with a significance...

n=10

before: mean=255.63, SD=115.50

after: mean=284.75, SD=132.63

difference: mean=29.13, SD=21

conduct hypothesis test with a significance level of 0.05 to determine theater the mean levels differ significantly before and after. report test statistic, p-value, DF (show work)

Math   Statistics-And-Probability   
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Homework Answers

Answer #1

H0: Null Hypothesis: = 0

HA: Alternative Hypothesis: 0

= 29.13

sd = 21

SE = sd/

= 21/ = 6.6407

Test statistic is:

t = /SE

= 29.13/6.6407 = 4.3866

= 0.05

ndf = 10 - 1 = 9

From Table, critical values = t = 2.2622

Since calculated value of t = 4.3866 is greater than critical value of t = 2.2622, the difference is significant. Reject null hypothesis.

Conclusion:
The mean levels differ significantly before and after.

t score = 4.3866

ndf = 9

By Technology, p-value = 0.0018

Since p - value is less than , Reject null hypothesis, as already calculated.

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