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Comparing 2 sample means of 2 different samples

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    sample 1: size -10, mean -124, variance -1681
    sample 2: size -5, mean -68, variance - 481

    Assuming equal variance, independent and normality holds

    a)Is there any evidence that that the mean in population 2 is is less than the mean in population 1? Use alpha = 0.01. What is the p-value?

    b)Test the 2 variances. Alpha = 0.05

    2. Relevant equations



    3. The attempt at a solution

    for a), would I use the following test statistic[tex]t = \frac{\overline{x} - c}{s/\sqrt{n}}[/tex]~[tex]t_{n-1}[/tex]. Just not sure how to incorporate the 2 different sample variances.
     
  2. jcsd
  3. Sep 25, 2009 #2

    statdad

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    Homework Helper

    For a), don't you want to test [tex] H_0 \colon \mu_1 - \mu_2 > 0 [/tex]? What test statistic is appropriate for that?
     
  4. Sep 25, 2009 #3
    For 2 independent Normal samples, with common unknown variance [tex]\sigma^2[/tex]

    [tex]\frac{(\overline{y}_1 - \overline{y}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{n^{-1}_1 -+n^{-1}_2}}[/tex] is a t distribution with (n1 + n2 - 2) degrees of freedom.

    Where S2p = [tex]\frac{(n_1 - 1)S^2_1 + (n_2 -1)S^2_2}{n_1 + n_2 - 2} = 1311.8[/tex]

    Not sure how to use the formula though since I don't know mu1 and mu2
     
    Last edited: Sep 25, 2009
  5. Sep 25, 2009 #4

    statdad

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    Your complete hypotheses would be (I typed [tex] H_0 [/tex] instead of [tex] H_a [/tex] and > rather than [tex] \le [/tex] at first, sorry)

    [tex]
    \begin{align*}
    H_0 \colon & \mu_1 - \mu_2 \le 0 \\
    H_a \colon & \mu_1 - \mu_2 > 0
    \end{align*}
    [/tex]

    You don't need to know the actual value of [tex] \mu_1 - \mu_2 [/tex] to use the test statistic: you use the boundary between the two possibilities, which is 0.
     
  6. Sep 25, 2009 #5
    [tex]H_0 \colon & \mu_1 - \mu_2 = 0[/tex]

    is that another acceptable way to write H0?
     
    Last edited: Sep 25, 2009
  7. Sep 28, 2009 #6

    Mark44

    Staff: Mentor

    No, you want the inequality, since that's what matches your problem statement.
    From the quote just above (which represents the alternate hypothesis), the null hypothesis is that

    [tex] \mu_1 - \mu_2 \leq 0[/tex]

    As you wrote it, the probability of the null hypothesis would be zero.
     
  8. Sep 29, 2009 #7
    For some reason I thought that it was asking if there was a difference between the 2 means. My mistake. Thanks.
     
  9. Sep 29, 2009 #8

    statdad

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    "the probability of the null hypothesis would be zero."

    Not sure what you mean here (that would be the only way to write the null in the case of the alternative being two-sided), but many authors would write [tex] H_0 \colon \mu_1 - \mu_2 [/tex] as [tex] H_0 \colon \mu_1 - \mu_2 = 0 [/tex]. It's an odd convention, but it does happen.
     
  10. Sep 29, 2009 #9

    Mark44

    Staff: Mentor

    My mistake. I was thinking in terms of variables rather than statistics, if that makes any sense. IOW, along these lines: Pr(z = k) = 0.
     
  11. Sep 29, 2009 #10
    So I computed the p-value using a t-test, and got the value of 0.0079. Since alpha = 0.01, I can say that the p-value is statistically significant at the 1% level, and I reject the null hypothesis and accept the alternate hypothesis?
     
  12. Sep 29, 2009 #11

    statdad

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    That's rather what I thought - but I have learned to be hesitant making assumptions about the posts of others.

    I haven't checked your calculations, but if the numbers are correct, so is your final statement. However, "accept the alternative hypothesis" is rarely the form a faculty member or researcher wants as the final answer: if you know that the alternative is to be accepted, you should clearly state what this means about the two population means.
     
  13. Sep 29, 2009 #12
    Does this sound better:

    At the 1% level, the data provides evidence against the null hypothesis in favour of the alternate hypothesis, which implies that the mean of population 2 is less than the mean if population 1.
     
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