Question

Help please !

Answer 1

You test calories for a food item. The brand name has a mean of 158.706 and a sample standard deviation =25.236, when seventeen are tested. The generic item has a mean of 122.471 and a sample standard deviation =24.583, when seventeen are tested. Which is a confidence interval of 95%?

Answer 2

i dont remember what deviation is.

Answer 3

I never gave any multiple choice answers but I can't give you a medal for just Google Searching it. The question isn't asking for the closest answer, I need an exact one. Sorry.

Answer 4

No need to be rude, friend.

Answer 5

Have a nice day. =)

Answer 6

a. 17.21 to 55.26
b. 17.79 to 54.67
c. 17.90 to 54.57
d. 18.83 to 53.64

Answer 7

im sorry but I cant do this. good luck though.

Answer 8

Do you have any idea what it could be? @amorfide

Answer 9

okay so

you want to work out the standard error of the means, to do this

\[error=\sqrt{\frac{ (\sigma_{1})^{2}+(\sigma_{2})^{2} }{ n}}\]

work out your degrees of freedom, which is (n -1)
n=17, where n is the sample size
but you have two sample sizes, 17 and 17 so we would just add them together

(n-1)+(n-1)=2n-2=34-2=32

so your degrees of freedom is 32

you want 95% confidence interval which is 1-0.95=0.05

im not sure if it is 2 tailed or 1 tailed test

you would use your t-table for 32 degrees of freedom, and if it is 2 tailed test, you will divide 0.05 by 2 to get 0.05/2=0.025, and look up 32df,0.025 on your table
if one tailed you will do 32df,0.05

I believe this is the method, it has been a while sorry I was checking some things to make sure to give the correct method