@jim_thompson5910
Do you know stats?
@jim_thompson5910
I know some stats, but not all. It will depend on the problem.
i have the question and answer but i do not understand how we got the answer
it's 9 but why?
one sec while I think
according to this page http://onlinestatbook.com/2/chi_square/contingency.html we simply add up the values in column 1 to get 8+6=14 then we add the values in row 1 to get 8+17 = 25 Afterwards, we multiply 14*25 = 350, then divide by 39 to get 350/39 8.97435897435898 which rounds to 9
This is from the part that says E = (Ti * Tj)/T where E = expected value in cell i,j (row i, column j) Ti = sum of all the values in row i Tj = sum of all the values in column j
THANK YOU!
you're welcome
also one more question
what's that?
it has to do with degrees of freedom
i was thinking it was 3 since there are 4 categories so you must do 4-1=3?
from this page http://www.ling.upenn.edu/~clight/chisquared.htm it says... "The degrees of freedom for a Chi-square grid are equal to the number of rows minus one times the number of columns minus one: that is, (R-1)*(C-1). In our simple 2x2 grid, the degrees of independence are therefore (2-1)*(2-1), or 1!"
so it basically says df = (R-1)*(C-1) df = degrees of freedom R = # of rows C = # of columns
but i thought it had to do with the # of categories rather than columns and rows?
okay so im guessing it would be like this (3-1)*(3-1)= 4 Df=4
I think that's if you have like 2 columns of data maybe (and each row can be interchanged)
or would i have to foil?
it's a 2x2 table
so R = 2, C = 2
df = (R-1)*(C-1) df = (2-1)*(2-1) df = 1*1 df = 1
i think you forogt about the column including complained,no,yes
so it would be 3x3
but that's not a data column, it's just a label
so you can know what each row means
ohh!
haha so yeah (2-1)x(2-1) = DF=1
you're amazing hopefully you have time for one more?
sure I can do one more
but, when i give the question let me do it on my own and you correct me. lol
alright
Z=XBAR-Mean/standard deviation SQuareroot n
is this the right equation?
no, first you must use this formula \[\Large \chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i}\] to compute the chi-square test statistic
the formula is given on that same page posted above http://www.ling.upenn.edu/~clight/chisquared.htm
so you'll need the expected frequencies for each cell
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