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Math Statistics Question - null/alternative Test Statistic, P-Value/Critical, Claim?


In a study of proctored and nonproctored exams in am online class, researchers obtained the following data.Group One (Proctored): n = 30, x bar = 73.4, s = 11.92 Group Two (Unproctored): n = 32, x bar = 86.6, s= 15.4 Test the claim that unproctored students did better than the proctored students at the 0.05 significance level.(Do not assume the population standard deviations are equal.) State the null and alternative hypothesis.


Find the test statistic.


Find the P-value or critical value.


Make a conclusion about this claim.
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1 Answer

Dear Nick,
You might want to check out one of the websites that "lays it all out" for you in an orderly fashion, such as
By the way, typically YOU must select the P-value you want to use (here it is specified for you). Note that you are never "proving" a claim of equivalence or non-equivalence, simply rejecting the null hypothesis (typically, that there is no difference) at some level of significance.
Now, take your example one step further: would a finding of non-equivalence imply a particular cause-effect relationship (such as, cheating!)? That's where the real scientists step up to the bat -- correlation ≠ cause-effect; unraveling a mechanism is not always easy, nor will you always convince some people who have their own different theories!
Cheers, -- S.