Statistics/Probability questions... 1) How do I tell if the model accurately predicted the sales? 2) What is the "p" in this problem/what do I do with it? : "Assume p = 0.05" ___ I attached the full problem and a clearer version of what I think is called a 'Chi Squared Critical Values Chart.' ___ Could someone run me through how to do this problem?

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Statistics/Probability questions... 1) How do I tell if the model accurately predicted the sales? 2) What is the "p" in this problem/what do I do with it? : "Assume p = 0.05" ___ I attached the full problem and a clearer version of what I think is called a 'Chi Squared Critical Values Chart.' ___ Could someone run me through how to do this problem?

Statistics
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you shd start by finding the degree of freedom do u know how to do that?
No, I don't. I asked my proctor to help me, but he didn't know how either.

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ok there a formula for that (C-1)(R-1) where C represent the total # of column and R represent the total # of rows
(12-1)(35-1)?
no there 2 columns the expected and the observed ....and there 3 row.,vanilla, strawberry and chocolate
correct now take away and then multiply out the brackets to get the df
2
correct the next step is to find the chi square value and there a formula for that also
|dw:1435291295954:dw|
so we plug in the respective values in the formula and add them up note: we do them separately i will start u will finish it ok
Sounds good, thanks a bunch :)
|dw:1435291759806:dw|
6033 _____ ? 875
^(fraction)
yes but we usually take the decimal in statistics
which is 6.894857 now this is ur chi square value
what now? is this where that "p" comes in?
now the want the p value so we take the degree of freedom which is 2 and at p=0.05 which i dont see on this chi square tables??
im seeing 0.005 on the table not 0.05
and it there twice the table have a fault
u see it? im assuming that the first 0.005 suppose to be 0.05??
Maybe it is supposed to be the 5th column?
no actually there shdnt have the same p value on a chi square table
You're the expert; I'm in Algebra II. My proctor didn't know why it was included, and couldn't help me with it because he didn't know how to do stats either. There wasn't even a review or notes or anything, just questions. :/
im lookin at my chi square table and well the p value for 0.05 corresponds to the first 0.005 on your chi sq. table so lets used that 0k
Sure thing
so |dw:1435293790254:dw|
now we compare the 2 chi square value
oh wait, so that was right? Because 6.83 > 5.99, the correct answer is either A or D?
the answer shd be D because the p value was too low
for the chi square
Both A and D say x was too low
ok u see what the difference between the two values?
One says yes, and the other says no?
referring to whether or not the model predicted the milkshake sales accurately
but i don't know how to tell that from our data/if it should be yes or no
and there a saying that goes like this....if the p-value is low H0 must go and if the p-value is high then h0 can fly
ok the difference between the value is 0.84 can conclude D or A as the answer?
I understand how to get 0.84 as the difference, but not how to tell if it is Yes or No
https://www.youtube.com/watch?v=HwD7ekD5l0g
ok what we actually had to look at is if p>0.05 or p<0.05
I thought it said p=0.05 though?
yes but when we find the value we see if its greater ok lesser than 0.05
..
o no we were on the right track before i got mix up after watching that video that for critical value...so we do have to look at our 2 chi squared values to determine the answer
okay. what about them?
well the chi squared value is low and lets look at it this way the chi sq value doesnt equal our chi square value and once the chi squared values is too low then it says something about the test so i still think its D
ok maybe if we get a second opionon an this it would be nice cause i really dont know how to explain the no it was to low part @SolomonZelman
Thanks for all of your help so far @LynFran :)
ok
LynFran you're using the right formula \[\Large \chi^2 = \sum \frac{(O_i -E_i)^2}{E_i}\] \[\Large \chi^2 = \frac{(O_1 -E_1)^2}{E_1}+\frac{(O_2 -E_2)^2}{E_2}+\frac{(O_3 -E_3)^2}{E_3}\] \[\Large \chi^2 = \frac{(205-175)^2}{175}+\frac{(114-125)^2}{125}+\frac{(264-250)^2}{250}\] \[\Large \chi^2 = 6.894857\]
is there a typo in the table? I see 0.005 written twice along the top
I'm guessing that first 0.005 should be 0.05 ?
yes, there is
ok thought so
Degrees of Freedom df = k-1 = 3-1 = 2 look at the row that starts with 2 and look at the column that has 0.05 at the top
So we know test statistic = 6.89 critical value = 5.99
do you see how to finish up?
actually we did all of this is just how to compare and tell the correct answer part that a bit troublesome
Rule: if (test statistic) > (chi-square critical value), then reject the null
the goodness of fit test is most always right tailed. I've never seen any cases that are otherwise, but I don't know for sure since there may be some rare cases
@jim_thompson5910 okay, reject the null, but we can't determine whether it would be option A or D. (Yes, or No)
from the options at the bottom of this attachment
Null: the observed matches with the expected Alternate: the observed does not match with the expected we rejected the null, so the observed does not match with the expected The question was "did the model predict the sales correctly?" ie "does the expected and observed values match up?" so the the answer is "no". The reason why the null was rejected was because the chi-square test statistic was too high
Ohhhh! so it's actually B I was looking at it backwards
yeah it's B
wasnt the chic squared values 5.99? so how come that higher than 6.89??
Thank you both! I'll write your testimonials when I'm all done. (later tonight/early morning)
that 5.99 is the critical value
it's the cutoff point
I greatly appreciate all the help :)
anything higher than that cutoff point means you reject H0
o ok i see
|dw:1435292080595:dw|
|dw:1435292122002:dw|
yes ur right

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