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Hummingbird25
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Hi
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
Homework Statement
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
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