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Level of significans and confidens interval of the mean and the variance

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data

    I am given the table

    [tex]\begin{array}{ccccc} samplesite & 1950 & 1975 & Change \\ 1 & 7.30 & 7.70 & 0.40\\ 2 & 6.14 & 6.72 & 0.58 \\ 3 & 6.47 & 6.32 & -0.15 \\4 & 5.87 & 6.38 & 0.51 \\ 5 & 6.06 & 6.34 & 0.28 \\ 6 & 4.71 & 5.78 & 1.07 \\ 7 & 5.45 & 5.59 & 0.14 \\ 8 & 6.17 & 6.33 & 0.16 \\9 & 5.83 & 5.50 & -0.33 \\ 10 & 5.55 & 5.55 & 0.00 \\ 11 & 5.50 & 6.08 & 0.58 \end{array}[/tex]

    I have to stachistics question where the above table is used.
    1. The problem statement, all variables and given/known data
    Question (1)

    Look at the change column: Is the change in polution levels from 1950 to 1975 significant?

    Using the significance test for unknown mean and unknown standard diviation.

    So I use the formula:

    [tex]t = \frac{\overline{x} - \mu_0}{\frac{s}{\sqrt{n}}}[/tex]

    where [tex]\overline{x} = \frac{\sum(x_i)}{N} = \frac{3.24}{11} = 0.2946[/tex]

    and where [tex]S^2 = \frac{1}{N-1} \sum(X -\overline{x})^2 = 0.154[/tex]

    where S = [tex]\sqrt{S^2} = 0.394228[/tex]

    Average change = average 1975 - average 1950. Under the null hypothesis, change = 0, then mu = 0

    Then I insert into the formula
    [tex]t = \frac{0.2946 - 0}{\frac{0.392428}{\sqrt{11}}} = 2.47813.[/tex]

    Looking that up in the t-table Level of significans is 95%. How does that sound?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    But what do I use for [tex]\mu_{0}[/tex] ??

    Question (2):

    Write the 95% confidence Interval for the mean and the variance for the change polution from 1950 og 1975.

    The 95% confidence interval for the mean is defined

    [tex]\overline{x} + t_{CL} \cdot {N} , \overline{x} - t_{CL} \cdot {N} [/tex]

    I know that [tex]\overline{x} = 0.2945[/tex] and if look at the t table for 95% t-level at degree of freedom df = N - 1 then I find a t-value called t = 2.98.

    Therefore (as I understand it) the 95% confidence level for the mean is
    [tex]0.2945 + 2.95 \cdot {11} , 0.2945 - 2.98 \cdot {11} [/tex] ???

    If the above is true how do I write the 95% Confidance Interval of the variance?

    Sincerely Yours
    Math Frank.
     
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 19, 2008 #2

    EnumaElish

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    Last edited: Feb 19, 2008
  4. Feb 19, 2008 #3
    Hi I have changed the answer to question 1) With Your corrections.

    Regarding question 2)

    Is my assumption regarding the 95% interval for the mean. Is that correct?

    If not I would very much appriacte if you could find it in your heart to correct 2) for me because, tomorrow I am going in for a bypass(believe it or not, I have had pain in my chest for some time now :( ), and I have to hand this in before I go to hospital. I know it much to ask, but it will only be this one time.

    Sincerely Yours
    Frank.
     
    Last edited: Feb 19, 2008
  5. Feb 20, 2008 #4

    EnumaElish

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    The 95% C.I. around the mean is [itex]\bar {x} \pm t_{0.025}(n-1)s/\sqrt{n}[/itex] where [itex]\bar {x}[/itex] and s are respectively the computed average and the computed standard deviation of the sample, n is the sample size and [itex]t_{0.025}(n-1)[/itex] is the critical t value with 0.05 total (two-sided) tail probability and n-1 degrees of freedom.

    The 95% C.I. around the variance is [L , U] = [ [itex](n-1)s^2\left/\chi^2_{0.025}(n-1)[/itex] , [itex](n-1)s^2\left/\chi^2_{0.975}(n-1)[/itex] ], where [itex]\chi^2_p(n-1)[/itex] is the critical chi-squared value with probability = p (to the left of the critical value) and degrees of freedom = n-1.

    Good luck with the bypass.
     
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