Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

Do you know stats?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

I know some stats, but not all. It will depend on the problem.

OpenStudy (anonymous):

i have the question and answer but i do not understand how we got the answer

OpenStudy (anonymous):

it's 9 but why?

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

one sec while I think

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

THANK YOU!

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

also one more question

jimthompson5910 (jim_thompson5910):

what's that?

OpenStudy (anonymous):

it has to do with degrees of freedom

OpenStudy (anonymous):

OpenStudy (anonymous):

i was thinking it was 3 since there are 4 categories so you must do 4-1=3?

jimthompson5910 (jim_thompson5910):

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!"

jimthompson5910 (jim_thompson5910):

so it basically says df = (R-1)*(C-1) df = degrees of freedom R = # of rows C = # of columns

OpenStudy (anonymous):

but i thought it had to do with the # of categories rather than columns and rows?

OpenStudy (anonymous):

okay so im guessing it would be like this (3-1)*(3-1)= 4 Df=4

jimthompson5910 (jim_thompson5910):

I think that's if you have like 2 columns of data maybe (and each row can be interchanged)

OpenStudy (anonymous):

or would i have to foil?

jimthompson5910 (jim_thompson5910):

it's a 2x2 table

jimthompson5910 (jim_thompson5910):

so R = 2, C = 2

jimthompson5910 (jim_thompson5910):

df = (R-1)*(C-1) df = (2-1)*(2-1) df = 1*1 df = 1

OpenStudy (anonymous):

i think you forogt about the column including complained,no,yes

OpenStudy (anonymous):

so it would be 3x3

jimthompson5910 (jim_thompson5910):

but that's not a data column, it's just a label

jimthompson5910 (jim_thompson5910):

so you can know what each row means

OpenStudy (anonymous):

ohh!

OpenStudy (anonymous):

haha so yeah (2-1)x(2-1) = DF=1

OpenStudy (anonymous):

you're amazing hopefully you have time for one more?

jimthompson5910 (jim_thompson5910):

sure I can do one more

OpenStudy (anonymous):

but, when i give the question let me do it on my own and you correct me. lol

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

OpenStudy (anonymous):

Z=XBAR-Mean/standard deviation SQuareroot n

OpenStudy (anonymous):

is this the right equation?

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

the formula is given on that same page posted above http://www.ling.upenn.edu/~clight/chisquared.htm

jimthompson5910 (jim_thompson5910):

so you'll need the expected frequencies for each cell

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!