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(Urgent) Howto Compare variances(Please look at my calculations) 
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#1
Feb1208, 09:23 AM

P: 86

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 Ftest 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}{N1} = 367,56 [/tex] and [tex]s^2_{2} = \frac{\sum(X\mu_2)^2}{N1}= 507,04 [/tex] Then according to the Ftest [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 ftest(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 


#2
Feb1208, 09:54 AM

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Did you mean to write
[tex]s^2_1 = \frac{\sum(X\mu_1)^{squared}}{N1},[/tex] and similarly for [itex]s_2^2[/itex]? You should check your calculations. 


#3
Feb1208, 10:14 AM

P: 86

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 fdistribution 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 


#4
Feb1208, 10:27 AM

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(Urgent) Howto Compare variances(Please look at my calculations)
Here is a sample table: http://www.itl.nist.gov/div898/handb...m#ONE051120
Let's say your df for numerator (v1) was 20 and your df for denominator (v2) was 100. Then, the critical F_{0.05} value would be 1.676 such that (onesided) 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 onesided test." 


#5
Feb1208, 10:38 AM

P: 86

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. 


#6
Feb1208, 10:44 AM

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#7
Feb1208, 11:59 AM

P: 86

Thank You for your answer, Good then using the t test which you kindly refered to with equal variance and unequal sample sizes. I then get [tex]S_{(\mu_{1}  \mu_{2})} = \sqrt{\frac{(101) \cdot 367.56 + (121) \cdot 507.04 \cdot \frac{1}{10}+\frac{1}{12}}{10+122}} = 3.29035[/tex] Then [tex]\frac{39.639.6667}{3.29035} = 0.20271.[/tex] Does that look okay? Sincerely Yours Hummingbird. 


#8
Feb1208, 12:51 PM

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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{(101) \cdot 367.56 + (121) \cdot 507.04 \cdot \left(\frac{1}{10}+\frac{1}{12}\right)}{10+122}} = 3.29035[/tex] Once you check your calculation, you can compare your tvalue against a ttable. 


#9
Feb1208, 01:07 PM

P: 86

where [tex]t = \frac{39.666739.6}{9,02498} = 0,007391.[/tex] I have been Looking for a tvalue which resembles this one, but couldn't find it. But its very close to the tvalue 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. 


#10
Feb1208, 01:51 PM

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This means that if the calculated t statistic < 1.7247..., you cannot reject [itex]\mu_1 = \mu_2[/itex] at 5% in a ONEsided test (that is, against the alternative hypothesis [itex]\mu_1 > \mu_2[/itex]. Twosided ttest If your alternative hypothesis is [itex]\mu_1 \ne \mu_2[/itex] then you need to run a twosided ttest. 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 abovereferenced ttable, 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]. 


#11
Feb1208, 02:04 PM

P: 86

Sincerely Hummingbird. 


#12
Feb1208, 02:23 PM

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#13
Feb1208, 02:31 PM

P: 86

Since the two means aren't not completely a like i use the two sided test. Don't I ?



#14
Feb1208, 03:18 PM

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For example, suppose you'd like to prove "the French aren't shorter than the Germans." Then you should run a onesided test. However, if your hypothesis is "the French and the Germans are equally tall" then you should run a 2sided test. 


#15
Feb1208, 04:03 PM

P: 86

Looking at the ttable here http://mips.stanford.edu/public/clas...s/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. 


#16
Feb1208, 05:16 PM

P: 86

Next 95% confidence interval for the commen mean and variance. Is that this http://en.wikipedia.org/wiki/Confidence_interval BR Hummingbird.. 


#17
Feb1308, 10:03 AM

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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}(n1)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}(n1)[/itex] is the critical t value with 0.05 total (twosided) tail probability and n1 degrees of freedom. The 95% C.I. around the common variance is [L , U] = [ [itex](n1)s^2\left/\chi^2_{0.025}(n1)[/itex] , [itex](n1)s^2\left/\chi^2_{0.975}(n1)[/itex] ], where [itex]\chi^2_p(n1)[/itex] is the critical chisquared value with probability = p (to the left of the critical value) and degrees of freedom = n1. 


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