Regression, correlation and hypothesis testing 1B
1 a r = 0.9 is a good approximation, since the points lie roughly, but not exactly, on a straight line.
Remember that the value of r tells you how ‘close’ the data is to having a perfect positive or
negative linear relationship.
b Clearly r is negative, and the data is not as close to being linear as in part a. r = −0.7 is therefore a
good approximation.
c The data seems to have some negative correlation, but is rather ‘random’. Because so many points
would lie far away from a line of best fit, r = −0.3 is a good approximation.
2 a The product moment correlation coefficient gives the type (positive or negative) and strength of
linear correlation between v and m.
b By inputting the (ordered) data into your calculator, r = 0.870 (to 3 s.f.).
3 a r = −0.854 (to 3 s.f.)
b There is a negative correlation. The relatively older young people took less time to reach the
required level.
4 a The completed table should read:
Time, t 1 2 4 5 7
Atoms, n 231 41 17 7 2
log n 2.36 1.61 1.23 0.845 0.301
b r = −0.980 (to 3 s.f.)
c There is an almost perfect negative correlation with the data in the form log n against t, which
suggests an exponential decay curve. (This uses knowledge from the previous section.)
d
= y 2.487 − 0.320 x
⇒ log n =
2.487 − 0.320t
102.487 −0.320t = 102.487 ×10−0.320t
⇒n=
102.487 × (10−0.320 )
t
⇒n=
Therefore a = 102.487 = 307 (3 s.f.) =
and b 10
= −0.320
0.479 (3 s.f.).
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5 a
Width, w 3 4 6 8 11
Mass, m 23 40 80 147 265
log w 0.4771 0.6021 0.7782 0.9031 1.041
log m 1.362 1.602 1.903 2.167 2.423
b r = 0.9996
c A graph of log w against log m is close to a straight line as the value of r is close to 1, therefore
m = kwn is a good model for this data.
d
= y 0.464 + 1.88 x
⇒ log m = 0.464 + 1.88 log w
10(0.464+1.88log w)
⇒m=
⇒m = 100.464 × w1.88
Therefore k = 100.464 = 2.91 (3 s.f.) and n = 1.88 (3 s.f.).
6 a r = −0.833 (3 s.f.)
b −0.833 is close to −1 so the data values show a strong to moderate negative correlation. A linear
regression model is suitable for these data.
7 a ‘tr’ should be interpreted as a trace, which means a small amount.
b r = −0.473 (3 s.f.), treating ‘tr’ values as zero.
c The data show a weak negative correlation so a linear model may not be best; there may be other
variables affecting the relationship or a different model might be a better fit.
Challenge
Take logs of the data in order to compute all of the required relationships:
x 3.1 5.6 7.1 8.6 9.4 10.7
y 3.2 4.8 5.7 6.5 6.9 7.6
log x 0.491 0.748 0.851 0.934 0.973 1.03
log y 0.505 0.681 0.756 0.813 0.839 0.881
Compute the PMCC for x and log y: r = 0.985 (3 s.f.).
Compute the PMCC for log x and log y: r = 1.00 (3 s.f.).
Therefore the data indicate that log x and log y have a strong positive linear relationship. From the
previous section, the data indicate a relationship of the form y = kxn.
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