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Math Help - Find 90% confidence interval

  1. #1
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    Find 90% confidence interval

    Random sample of math scores assumed to be N(\mu_{x},\sigma^{2}): X1=644, X2=493, X3=532, X4=462, X5=565
    Random sample of verbal scores assumed to be N(\mu_{y},\sigma^{2}): Y1=623, Y2=472, Y3=492, Y4=661, Y5=540, Y6=502, Y7=549, Y8=518
    How do I find a 90% confidence level for \mu_{x}-\mu_{y}?
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  2. #2
    Newbie drichie's Avatar
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    hint:

    If  X \sim N(\mu_X, \sigma^2_X) and  Y \sim N(\mu_Y, \sigma^2_Y) are independent normal random variables, then:
    V = X - Y \sim N(\mu_X - \mu_Y, \sigma^2_X + \sigma^2_Y).
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  3. #3
    MHF Contributor matheagle's Avatar
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    This one is finally well worded.
    You want to pool the sample variances because we are assuming that the two population variances are equal.

    It will be a t with n_1+n_2-2 degrees of freedom.

    \bar X_{n_1}-\bar X_{n_2}\pm t_{n_1+n_2-2, .05}s_p\sqrt{{1\over n_1}+{1\over n_2}}

    see.. Unequal sample sizes, equal variance (Ignore uequal sample size comment)
    at http://en.wikipedia.org/wiki/Student's_t-test
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  4. #4
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    Thank you! That site was extremely helpful. I am now working on plugging all the correct values in to get a "normal" sounding answer. Thank you for helping me.
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by drichie View Post
    hint:

    If  X \sim N(\mu_X, \sigma^2_X) and  Y \sim N(\mu_Y, \sigma^2_Y) are independent normal random variables, then:
    V = X - Y \sim N(\mu_X - \mu_Y, \sigma^2_X + \sigma^2_Y).

    NOTE that the two populations variances are equal \sigma^2_X= \sigma^2_Y.
    In this case you are supposed to pool the sample variances.
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  6. #6
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    \mu_{x}=539.2

    \mu_{y}=544.625<br />

    \sigma^{2}_{x}=3958.96 (seems high)

    \sigma^{2}_{y}=3786.984372

    t=\frac{\mu_{x}-\mu_{y}}{S_{xy}*\sqrt{1/5+1/8}}

    =\frac{539.2-544.625}{62.04450793*0.5700877125}

    =\frac{-5.425}{35.3708116}=-.1533750501

    This answer doesn't seem quite right.

    Then how do I do the confidence interval using the 90% CI, 1.645?

    I was going to just do  -5.425+/-1.645(\frac{\sqrt{3958.96+3786.984372}}{\sqrt{13}}  )=-5.425+/-40.15424024
    but neither of my answers seemed to make any sense.
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  7. #7
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    Thank you for all the help on this problem. I do not understand why the population variances are equal, or how you figure that. I have been going through different books and thought I could just use the equation that uses just the sample means and variances with the t value from the table. I'm sure I'm wrong, I just can't figure out how to tell that the pop variances are equal. Thanks
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  8. #8
    MHF Contributor matheagle's Avatar
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    Who ever gave you this problem stated that the variance of X and the variance of Y was \sigma^{2},
    otherwise they would have said

    X\sim N(\mu_{X},\sigma_X^{2}) and Y\sim N(\mu_{Y},\sigma_Y^{2}).

    IN this case you need to pool the variances.
    And it's a t percentile and not a z, as I stated a week ago.



    Quote Originally Posted by madgab View Post
    Random sample of math scores assumed to be N(\mu_{x},\sigma^{2}): X1=644, X2=493, X3=532, X4=462, X5=565
    Random sample of verbal scores assumed to be N(\mu_{y},\sigma^{2}): Y1=623, Y2=472, Y3=492, Y4=661, Y5=540, Y6=502, Y7=549, Y8=518
    How do I find a 90% confidence level for \mu_{x}-\mu_{y}?
    Last edited by matheagle; June 16th 2009 at 08:19 PM.
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  9. #9
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    I got it! Thank you! I'm sorry for bothering you again. It all makes sense now.
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