vcaa 2007

[nooshnoosh95]

Question 9 on 2007 VCAA exam 1
i cant get the answer on my calculator
it keeps giving me log y=0.52x+2.02
help???


A student uses the following data to construct the scatterplot
x   1  2    3   4   5     6    7     8     9
y 12 25 33 58  98 168 345 397 869
To linearise the scatterplot, she applies a log y transformation; that is, a log transformation is applied to the
y-axis scale.
She then Þ ts a least squares regression line to the transformed data.
With x as the independent variable, the equation of this least squares regression line is closest to
A. log y = –217 + 88.0 x
B. log y = –3.8 + 4.4 x
C. log y = 3.1 + 0.008 x
D. log y = 0.88 + 0.23 x
E. log y = 1.58 + 0.002 x

thanks

[StumbleBum]

Which calculator are you using?

[FlorianK]

Using T-Nspire
First of all solve for y.
Let's say I use log y = 0.88 + 0.23 x
This would be y=10^(0.88 + 0.23x) - Don't know how to do it with further math knowledge

Then you open a graphing page, type in the equation.
Now press ctrl+t a table of values will pop up and then you can look if they match the values of of the table.

The equation from above will be the answer

[StumbleBum]

Using T-Nspire
First of all solve for y.
Let's say I use log y = 0.88 + 0.23 x
This would be y=10^(0.88 + 0.23x) - Don't know how to do it with further math knowledge

Then you open a graphing page, type in the equation.
Now press ctrl+t a table of values will pop up and then you can look if they match the values of of the table.

The equation from above will be the answer

I wouldn't think this would be a very feasible method in an exam as you would have to potentially check every one of the five possibilities.

[Daenerys Targaryen]

I got D.
I've inputed the x values and y values into my spreadsheet.
In another column ive done a log 10 (y)
and then applied a linear regression

[StumbleBum]

I got D.
I've inputed the x values and y values into my spreadsheet.
In another column ive done a log 10 (y)
and then applied a linear regression

Yeh that's what I did, and the answer I got too. However after trying numerous combinations of log(y), y, x and even log(x) i still can't manage to come up with the regression line noosh got. Perhaps you put the values in incorrectly noosh?

[Felicity Wishes]

Using T-Nspire
First of all solve for y.
Let's say I use log y = 0.88 + 0.23 x
This would be y=10^(0.88 + 0.23x) - Don't know how to do it with further math knowledge

Then you open a graphing page, type in the equation.
Now press ctrl+t a table of values will pop up and then you can look if they match the values of of the table.

The equation from above will be the answer

Yeah I did it the long way as well and got D. Most likely a wrong number OP or have you tried more than once?

[nooshnoosh95]

ohh man its because i put in log to the base e not 10

thanks guys!

[Stick]

So glad I don't have to worry about this with the Casio Classpad (although I have other things to worry about in Methods and Specialist...).

[StumbleBum]

So glad I don't have to worry about this with the Casio Classpad (although I have other things to worry about in Methods and Specialist...).

What do you mean? Can't the TI's do a log transformation of another list?

[Stick]

No, that's not really what I was referring to. When I want to do a log transformation, I can just type in "log(list1)", for example, and it will do it correctly for me.

[StumbleBum]

No, that's not really what I was referring to. When I want to do a log transformation, I can just type in "log(list1)", for example, and it will do it correctly for me.

Yeh I know, I also use the ClassPad. However what I was saying is, why would the TI be disadvantaged on a question such as this?

[Stick]

ohh man its because i put in log to the base e not 10

thanks guys!

She had to define hers, whereas we don't need to.

[StumbleBum]

She had to define hers, whereas we don't need to.

Ohhh, now I'm with it haha.

[FlorianK]

The way wasn't "the long way" - I don't have to put in the values into any spreadsheet.
Secondly there is no mistake or calculation error you could make using my method.
Lastly this method costs still less than 1 minute