Exponential models s1 question Watch

dont know it
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Essentially, a graph of log x against log t gives a product moment correlation coefficient of 0.9998.

https://activeteach-prod.resource.pe...sm2_ex1mix.pdf

^^ The table of values is in question 1 if you need to see it.

The question asks us to use this coefficient to explain why an equation of t=ax^n, where a and n are constants, is likely to be a good model for x and t.

I'm thinking it's because there's strong positive correlation and hence linear relationship which suggests exponential growth. Is that also correct or am I wrong? They give their own answer(part b on q1 on the link above)
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RDKGames
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(Original post by dont know it)
Essentially, a graph of log x against log t gives a product moment correlation coefficient of 0.9998.

https://activeteach-prod.resource.pe...sm2_ex1mix.pdf

^^ The table of values is in question 1 if you need to see it.

The question asks us to use this coefficient to explain why an equation of t=ax^n, where a and n are constants, is likely to be a good model for x and t.

I'm thinking it's because there's strong positive correlation and hence linear relationship which suggests exponential growth. Is that also correct or am I wrong? They give their own answer(part b on q1 on the link above)
You're on the right lines, but I would word it better.

r \approx 1 \implies \log t \approx n \log x + \log a

so

\log t = n \log x + \log a

is a decent model, from which you can rearrange to get their answer.
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dont know it
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(Original post by RDKGames)
You're on the right lines, but I would word it better.

r \approx 1 \implies \log t \approx n \log x + \log a

so

\log t = n \log x + \log a

is a decent model, from which you can rearrange to get their answer.
I'm a bit confused about the first line. Why is it an approximation sign after log t and not an equal.
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RDKGames
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(Original post by dont know it)
I'm a bit confused about the first line. Why is it an approximation sign after log t and not an equal.
Because the values of \log t against \log x are not exactly following a straight line, hence the straight line on the RHS is just an approximation.
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(Original post by RDKGames)
Because the values of \log t against \log x are not exactly following a straight line, hence the straight line on the RHS is just an approximation.
Nice, thanks.
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