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(Urgent) Howto Compare variances(Please look at my calculations)

  1. Feb 12, 2008 #1
    Hi

    1. The problem statement, all variables and given/known data

    I have been on some statistics and I am not sure that I am on the right track

    Given two meassurement zones of radioactive vast:

    [tex]\left(\begin{array}{cc} zone 1 & zone 2 \\ 55 & 55 \\ 44 & 24 \\ 47 & 57 \\ 61 & 37 \\ 15 & 51 \\ 36 & 33 \\ 29 & 39 \\ 69 & 11 \\ 8 & 42 \\ 32 & 80 \\ \ & 21 \\ \ & 26 \end{array} \right) [/tex]

    Show that the variances can be assumed to the same in the two zones.

    2. The attempt at a solution

    I have looked up in my statischics book and found that the F-test looks to the best way to compare two variances.

    Here goes:

    [tex]\mu_{zone1} = \frac{\sum(X)}{N} = 39.6 [/tex] and [tex]\mu_{zone2} = \frac{\sum(X)}{N} = 39.6667 [/tex]

    Next the Variance for the two zones

    [tex]s^2_1 = \frac{\sum(X-\mu_1)^2}{N-1} = 367,56 [/tex] and

    [tex]s^2_{2} = \frac{\sum(X-\mu_2)^2}{N-1}= 507,04 [/tex]

    Then according to the F-test

    [tex]f = \frac{s^2_{2}}{s^2_1} \approx 1.38 [/tex]

    which is according to my Schaum's outlines of Statistics and Economics that gives arround 75 degrees of freedom, and of the f-test(the two zones lies in the same population) and therefore the variances can be assumed to be the same.

    I hope somebody will look at this and maybe help me to conclude if I have understood the problem correctly?

    Sincerely Yours
    Hummingbird
     
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2

    EnumaElish

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    Did you mean to write

    [tex]s^2_1 = \frac{\sum(X-\mu_1)^{squared}}{N-1},[/tex]

    and similarly for [itex]s_2^2[/itex]? You should check your calculations.

    You should compute one degrees of freedom for the numerator and another for the denominator. Also, in a 2-sided test, you should remember to divide the probability level ("alpha") by 2.
     
    Last edited: Feb 12, 2008
  4. Feb 12, 2008 #3
    Howdy and thank you for your answer,

    I have looked at the calculations and it was only in here that I had forgot the squares here.

    Anyway the f = 1.38 and I look up 1.38 in a f-distribution table I get a df(nummerator) = 40 and df(denominator) = 400 where it says(1.57) with bold writing.

    I devide 40 by 400?

    Sincerely Hummingbird
     
  5. Feb 12, 2008 #4

    EnumaElish

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    Here is a sample table: http://www.itl.nist.gov/div898/handbook/eda/section3/eda3673.htm#ONE-05-11-20

    Let's say your df for numerator (v1) was 20 and your df for denominator (v2) was 100. Then, the critical F0.05 value would be 1.676 such that (one-sided) probability to the right of the critical value is 0.05. Since your calculated F is less than 1.676, with these degrees of freedom (20 and 100), you would have been able to conclude that "the two variances are not different from one another at a 5% significance level in a one-sided test."
     
  6. Feb 12, 2008 #5
    Thank you EnumaElish,

    That leads to my second question, howto assume that the mean is the same in the two zones??

    Can that be concluded by that as show above that the difference in procent between the two means are less than a procent??

    Sincerely
    Hummingbird.
     
  7. Feb 12, 2008 #6

    EnumaElish

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  8. Feb 12, 2008 #7
    Hello again EnumaElish,

    Thank You for your answer,

    Good then using the t test which you kindly refered to with equal variance and un-equal sample sizes.

    I then get

    [tex]S_{(\mu_{1} - \mu_{2})} = \sqrt{\frac{(10-1) \cdot 367.56 + (12-1) \cdot 507.04 \cdot \frac{1}{10}+\frac{1}{12}}{10+12-2}} = 3.29035[/tex]

    Then [tex]\frac{39.6-39.6667}{3.29035} = -0.20271.[/tex]

    Does that look okay?

    Sincerely Yours
    Hummingbird.
     
  9. Feb 12, 2008 #8

    EnumaElish

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    Possibly there is an error, see the parentheses I inserted around 1/10 + 1/12. You should check your calculation:

    [tex]S_{(\mu_{1} - \mu_{2})} = \sqrt{\frac{(10-1) \cdot 367.56 + (12-1) \cdot 507.04 \cdot \left(\frac{1}{10}+\frac{1}{12}\right)}{10+12-2}} = 3.29035[/tex]

    Once you check your calculation, you can compare your t-value against a t-table.
     
  10. Feb 12, 2008 #9
    [tex]S_{(\mu_{1} - \mu_{2})} = \sqrt{\frac{(10-1) \cdot 367.56 + (12-1) \cdot 507.04 }{10+12-2} \cdot (\frac{1}{10} + \frac{1}{12})} = 9,02498[/tex]

    where [tex]t = \frac{39.6667-39.6}{9,02498} = 0,007391.[/tex]

    I have been Looking for a t-value which resembles this one, but couldn't find it. But its very close to the t-value t = 0.005. Which is 99,5 %. That must mean that are in the same zone or population?

    How does it look now??

    Sincerely
    Hummingbird.
     
    Last edited: Feb 12, 2008
  11. Feb 12, 2008 #10

    EnumaElish

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    No. You should check the your calculated t statistic against the critical t-value from a t-table. For example, the critical t-value in a ONE-sided test at 5% significance with degs. of frdm. = n1 + n2 - 2 = 20 (see http://projectile.is.cs.cmu.edu/research/public/talks/t-test.htm#equalvariance) is 1.724718 (from http://www.statsoft.com/textbook/stathome.html?sttable.html&1, under p = 0.05 column).

    This means that if the calculated t statistic < 1.7247..., you cannot reject [itex]\mu_1 = \mu_2[/itex] at 5% in a ONE-sided test (that is, against the alternative hypothesis [itex]\mu_1 > \mu_2[/itex].

    Two-sided t-test

    If your alternative hypothesis is [itex]\mu_1 \ne \mu_2[/itex] then you need to run a two-sided t-test. At a 5% significnace level, that would mean [itex]\alpha/2[/itex] = 0.025 on each side (adding up to a total "tail" probability of [itex]\alpha[/itex] = 0.05). If you'd like to use the above-referenced t-table, you now need to look under the p = 0.025 column. For degs of frdm = 20, the critical t is 2.08596, which is the value to compare the calculated t statistic against. If calc. t stat < 2.08596, then one cannot reject the null hypothesis [itex]\mu_1 = \mu_2[/itex] against [itex]\mu_1 \ne \mu_2[/itex].
     
    Last edited: Feb 12, 2008
  12. Feb 12, 2008 #11
    I will do that and then return, but just one question. Is my t-value correct??

    Sincerely Hummingbird.
     
  13. Feb 12, 2008 #12

    EnumaElish

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    Assuming you correctly calculated the means and the s, yes.
     
  14. Feb 12, 2008 #13
    Since the two means aren't not completely a like i use the two sided test. Don't I ?
     
  15. Feb 12, 2008 #14

    EnumaElish

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    Very rarely the two calculated means are equal. The question (of 1-sided vs. 2-sided test) is theoretical not empirical.

    For example, suppose you'd like to prove "the French aren't shorter than the Germans." Then you should run a one-sided test. However, if your hypothesis is "the French and the Germans are equally tall" then you should run a 2-sided test.
     
  16. Feb 12, 2008 #15
    Hi Enuma,

    Looking at the t-table here http://mips.stanford.edu/public/classes/stats_data_analysis/principles/t_table.html

    Then 20 degres of freedom and t approx 0.01 that gives 2.528. Excuse me for asking, but does this sound reasonable?

    Best Regards
    Hummingbird.
     
  17. Feb 12, 2008 #16
  18. Feb 13, 2008 #17

    EnumaElish

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    2.528 is right.

    The estimated t statistic is what you computed as [itex](\mu_1-\mu_2)/s_{\bar{x_1}-\bar{x_2}}[/itex] in a previous post.

    The 95% C.I. around the common 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 entire sample, [itex]n = n_1 + n_2[/itex] 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 common 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.
     
    Last edited: Feb 13, 2008
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