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Statistics 16 Online
OpenStudy (anonymous):

find the linear transformation model. x y ------------------ 0 8 1 39 2 195 3 960 4 4,738 5 23, 375 a) log y^hat= 0.6935 *log x + 0.9013 b) log y^hat= 0.9013x + 0.6935 c) log y^hat= 0.6935x + 0.9013 d) log y^hat= 0.9013 *log x + 0.6935

OpenStudy (anonymous):

@aggie12341

OpenStudy (anonymous):

@agent0smith

OpenStudy (anonymous):

@BTaylor @dan815 can you please help me ?

OpenStudy (anonymous):

hey skinny

OpenStudy (anonymous):

let me see if I can help you

OpenStudy (anonymous):

skinny, do you have to choose one option?

OpenStudy (anonymous):

yes I do then I have to use that option to predict what y will be if x is 12

OpenStudy (anonymous):

well, the equation would have to fit x=0, so a) and d) are discarted

OpenStudy (anonymous):

because log(0) doesn't exists, right?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

ok, with x=0 you have the following calculations for y, with b) and c): b) y=10^(0.9013*0 + 0.6935) = 10^0.6935 = 5 (aprox) c) y=10^(0.6935*0 + 0.9035) = 10^0.9035 = 8 (aprox)

OpenStudy (anonymous):

so, it has to be c)

OpenStudy (anonymous):

let's check with one more value for x=3: c) y=10^(0.6935*3 + 0.9035) = 10^(2.0805+0.9035) = 10^2.984 = 963 (aprox) so, it fits ok

OpenStudy (anonymous):

another way is to make a table for x and log(y):

OpenStudy (anonymous):

how did you get .9035? it is .9013

OpenStudy (anonymous):

x log(y) 0 0.9030 1 1.5910 2 2.2900 3 2.9822 4 3.6756 5 4.3688

OpenStudy (anonymous):

you're right!!! sorry!!! but the result is very similar, it fits ok

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so, from that table you could calculate the best values for a and b, to form the equation: log y = a*x + b

OpenStudy (anonymous):

so if you plug in x = 12 in the equation I should get 1,672,245,362 right?

OpenStudy (anonymous):

yes, I have the same numbre

OpenStudy (anonymous):

*number

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

that's your prediction you're welcome!!!

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