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Math Help - Confidence interval

  1. #1
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    Confidence interval

    Electricity usage in homes was studied over a period.
    For Year 1:
    Mean = 1130
    Std dev. = 196
    n = 16

    For Year 2:
    Mean = 1350
    Std dev. = 302
    n = 21

    Calculate a 95% confidence interval for the average usage year 2?

    I got the answer
    x-bar = 1350 +/- ME
    where ME = t(n-1), (alpha/2) x s / sq root of n

    Am I on the right track here?
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  2. #2
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    Quote Originally Posted by Pinto9 View Post
    Electricity usage in homes was studied over a period.
    For Year 1:
    Mean = 1130
    Std dev. = 196
    n = 16

    For Year 2:
    Mean = 1350
    Std dev. = 302
    n = 21

    Calculate a 95% confidence interval for the average usage year 2?

    I got the answer
    x-bar = 1350 +/- ME
    where ME = t(n-1), (alpha/2) x s / sq root of n

    Am I on the right track here?
    Since this is exactly the formula your textbook and classnotes would say to use, I don't see why you are usure.
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  3. #3
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    Well, thanks. I have another question though.
    The question:
    Test the hypothesis that the std dev has increased in year 2 compared to year 1?

    So H(0): var(x) < or = var(y)
    and H(a): var(x) > var(y) ?

    But as s(x) < s(y)
    (if x= year 1 and y= year 2)
    how do you state the rejection of H(0)?
    Is it reject H(0) if
    F(s(x)^2 / s(y)^2) < or > F n(x)-1, n(y)-1, alpha
    ?
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  4. #4
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    I would state your null hypothesis that there is "no difference between the variances" vs. "the variance in year 2 is greater than in year 1".

    Your decision rule should be to reject the null hypothesis using a one-sided test. You have a two-sided test listed there. Since your teacher asked specifically to test "increased in year 2", then this to me would indicate a one-sided test.
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