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Thread: Concluding 95% confidance

  1. #1
    Nov 2008

    Cool Concluding 95% confidance

    Im stuck on a homework question and help would be appreciated

    Milton and Webster, two brothers with very different tastes in televison shows, have been arguing for months over who should get to watch their father's giant LCD tv. Milton always complains that Webster gets too much time to watch the programs he wants to watch leaving Milton to use the smaller television. Webster counters though that Milton gets just as much time as he does because Milton watches everyday whereas Webster, though he admits to watching for hours at a time, only does so two or three times a week. Their father, a Vice President at a large corporation, and having taken Business Statistics several years ago, remembers that he can test to determine whether or not Milton's claim is accurate. Over the course of 21 days, he records the amount of time that each brother spends watching the big screen Tv. For Milton, the average time is 84.05 minutes with a variance of 1156. For Webster, the average time is 106.1 minutes with a variance 2435. Can he conclude with 95% confidence that Webster uses the TV more than Milton?
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  2. #2
    Jul 2008
    This requires a hypothesis test. You will have to assume (I assume) that the 2 random variables ( X and Y, the times that the Milton and Webster - respectively watch per day) are Normal random variables.

    Let \mu_M be the population mean of the time that Milton watches per day. (i.e. his true average time per day), and let \mu_W be the same for Webster.

    H_0:\ \mu_W - \mu_M \leq 0\ (\mu_W \leq \mu_M)

    H_1:\ \mu_W - \mu_M > 0\ (\mu_W > \mu_M)

    In theory you could do this using matched pairs (1 sample test based on the differences X-Y since there is a nice pairing (assuming that the father observed his kids on the same 21 days. However, the problem seems to be pushing you to do a 2-sample test.

    Can you finish the problem?? (Note that N=21)
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