A LevelMathsEdexcelStatisticsFlashcardsData Presentation & InterpretationFurther Correlation & Regression

Further Correlation & Regression (Edexcel A Level Maths: Statistics): Flashcards

Exam code: 9MA0

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FrontPMCC & Non-linear Regression
Define the product moment correlation coefficient.
FrontPMCC & Non-linear Regression
What range of values can $r$ take, and what does each end of that range mean?
BackPMCC & Non-linear Regression
$r$ always lies between $- 1$ and 1:
$- 1 \leq r \leq 1$
$r = 1$ is perfect positive correlation, with every point exactly on a straight line of positive gradient
$r = - 1$ is perfect negative correlation, with every point exactly on a straight line of negative gradient
$r = 0$ is no correlation
The closer $r$ is to 1 or to $- 1$ , the stronger the correlation, so $r = 0.81$ is stronger than $r = 0.17$ , and $r = - 0.99$ is stronger than either.

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Cards in this collection (15)

Define the product moment correlation coefficient.
The product moment correlation coefficient, or PMCC, is a single number measuring the strength and direction of the linear correlation in a set of bivariate data.
It turns a judgement made by eye from a scatter diagram into a value that can be compared, tested and quoted.
The PMCC calculated from a sample is written $r$ .
What range of values can $r$ take, and what does each end of that range mean?
$r$ always lies between $- 1$ and 1:
$- 1 \leq r \leq 1$
- $r = 1$ is perfect positive correlation, with every point exactly on a straight line of positive gradient
- $r = - 1$ is perfect negative correlation, with every point exactly on a straight line of negative gradient
- $r = 0$ is no correlation
The closer $r$ is to 1 or to $- 1$ , the stronger the correlation, so $r = 0.81$ is stronger than $r = 0.17$ , and $r = - 0.99$ is stronger than either.
True or False?
Of two data sets that both lie exactly on a straight line, the one on the steeper line has the larger value of $r$ .
False.
Both have $r = 1$ exactly; the gradient makes no difference to $r$ at all.
$r$ measures how close to a straight line the points lie, not how steep that line is, so a gentle slope and a near-vertical one both score 1 as long as the points sit on them.
A negative gradient is the one thing that does change: any straight line sloping downwards gives $r = - 1$ .
A data set has $r$ close to 0. Does that mean the two variables are unrelated?
No: $r$ measures linear correlation only, so a value close to 0 means only that the points are not close to a straight line.
Two variables can be very strongly related and still give an $r$ near 0, if the relationship they follow is a curve.
That is exactly why non-linear models exist: a linear regression line is only worth fitting when $r$ is close to 1 or $- 1$ , and a curved pattern needs the data to be coded first.
A scatter diagram shows a clear pattern, but the points follow a curve rather than a straight line. What can be done?
The relationship may fit a model of the form $y = a x^{n}$ or $y = k b^{x}$ .
Taking logarithms of both sides turns either of them into the equation of a straight line, so the data can be coded and a linear regression line fitted to the coded values instead.
This is called changing the variables, and the regression line found from the coded data can then be turned back into a model for the original variables.
Which non-linear model needs $\log x$ plotted against $\log y$ , and which needs $x$ plotted against $\log y$ ?
$y = a x^{n}$ needs both logged, so plot $\log x$ against $\log y$ :
$\log y = \log a + n \log x$
$y = k b^{x}$ needs only $y$ logged, so plot $x$ against $\log y$ :
$\log y = \log k + x \log b$
The difference is where $x$ sits: in the first it is the base, so the power law brings $n$ down as a multiplier of $\log x$ ; in the second it is already the exponent.
The model $y = a x^{n}$ is coded with $X = \log x$ and $Y = \log y$ , giving the straight line $Y = n X + \log a$ . Complete what the line's gradient and intercept give you:
gradient $=______$
intercept $=______$
The completed statements are:
gradient $= n$
intercept $= \log a$
So the exponent of the original model is read straight off as the gradient, while the coefficient has to be recovered from the intercept by undoing the logarithm.
For the other model, $y = k b^{x}$ , both constants need undoing: the gradient is $\log b$ and the intercept is $\log k$ .
The regression line for coded data is $Y = - 3.5 X$ , where $X = \log h$ and $Y = \log t$ . How do you turn that into a model relating $t$ and $h$ ?
Substitute the codings back in, then use the laws of logarithms to remove them:
$\log t = - 3.5 \log h$
$\log t = \log (h^{- 3.5})$
$t = h^{- 3.5}$
The power law is what does the work, turning the multiplier $- 3.5$ back into an exponent.
Here the line has no intercept, so there is no constant to recover; an intercept of $\log a$ would give a factor of $a$ in front.
A value has been predicted from a regression line fitted to coded data. What has to be done to it, and what is the trap?
The coding has to be reversed, because the prediction comes out as a coded value rather than as a value of the original variable.
The trap is which logarithm was used: if the coding was $\log$ , undo it with $10^{value}$ ; if it was $\ln$ , undo it with $e^{value}$ .
So, for example, a coded prediction of $Y = 0.16015$ from $Y = \log t$ gives $t = 10^{0.16015} = 1.45$ to 3 significant figures; label your values as you go, since it is easy to lose track of which are coded.
Why is a hypothesis test needed for a correlation coefficient at all?
Because the coefficient you can calculate is not the one you want to know about.
The population correlation coefficient, written $ρ$ and pronounced "rho", would need data on every member of the population, which is almost never possible; what you have instead is $r$ , calculated from a sample.
The test asks whether an $r$ this far from zero is strong enough evidence that the whole population has correlation too, or whether a sample this size could easily have produced it by chance.
A test is carried out for positive correlation in a population. Complete the hypotheses:
$H_{0} : ρ =______$
$H_{1} : ρ______0$
The completed hypotheses are:
$H_{0} : ρ = 0$
$H_{1} : ρ > 0$
The null hypothesis is always $ρ = 0$ , meaning no correlation in the population. Only the alternative changes: $ρ > 0$ for positive correlation, $ρ < 0$ for negative, and $ρ \neq 0$ for a two-tailed test.
Both hypotheses are about $ρ$ , not $r$ : $r$ is a number you already know, so there is nothing to hypothesise about it.
True or False?
A question asking you to test whether there is correlation between two variables calls for a one-tailed test if the sample value of $r$ is positive.
False.
The wording of the question decides it, and nothing else: "test for correlation" means test for correlation in either direction, which is a two-tailed test, however positive the data happens to look.
A test is one-tailed only if you are asked to test specifically for positive correlation or for negative correlation.
Letting the data choose the direction after you have seen it would make the test much easier to pass than the stated significance level claims.
How do you find a critical value from the table of critical values for correlation coefficients?
Read off the value where the row for the sample size $n$ meets the column for the significance level.
Two adjustments are needed, because the table only gives one-tailed values and only positive ones:
- for a two-tailed test, halve the significance level first, then read the table as though the test were one-tailed
- for negative correlation, look up the matching positive value and change its sign
So a two-tailed test at the 5% level uses the 0.025 column, and gives two critical values, one positive and one negative.
You have the sample value $r$ and the critical value. How do you decide the outcome of the test?
Compare their sizes, ignoring signs: the result is significant, and $H_{0}$ is rejected, when
$|r| > |critical value|$
because $r$ then lies in the critical region.
If it does not, you do not reject $H_{0}$ : there is insufficient evidence of correlation in the population at that significance level.
Say "do not reject", not "accept": failing to find evidence against $H_{0}$ is not the same as showing it to be true, and mark schemes can treat "accept $H_{0}$ " as a contradictory statement.
How do the sample size and the significance level each change the critical value in a test for correlation?
Both make the test easier to pass as they increase, so the critical value gets smaller.
A larger sample is stronger evidence, so a weaker correlation will do: at the 5% level $n = 10$ needs 0.5494, while $n = 100$ needs far less.
A larger significance level sets a lower bar, so again a smaller $|r|$ will do; as a check, a critical value that rises with the sample size means you have read the table wrongly.

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