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Exercises · Q11
Q.

For the following data on a company's monthly advertising expenditure (xx, ₹'000) and units sold (yy), state — from the shape a scatter diagram of this data would show — whether the correlation between xx and yy appears to be positive, negative, or approximately zero, and verify your visual judgement by computing Karl Pearson's coefficient of correlation.

Month12345
Advertising (xx)246810
Units sold (yy)5055657085
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Step 1 — Visual judgement (scatter diagram reasoning). Reading down both rows together: every time xx increases from one month to the next, yy also increases (2→50, 4→55, 6→65, 8→70, 10→85 — no reversal anywhere). A scatter diagram of these five points would therefore show a cloud of points running consistently from lower-left to upper-right — the classic picture of a strong positive linear relationship, with the points expected to lie fairly close to a single upward-sloping line (since the increases are fairly steady, not erratic).

Step 2 — Verify numerically. Find the means. xˉ=2+4+6+8+105=6\bar x=\dfrac{2+4+6+8+10}{5}=6. yˉ=50+55+65+70+855=3255=65\bar y=\dfrac{50+55+65+70+85}{5}=\dfrac{325}{5}=65.

Step 3 — Tabulate deviations, products and squares.

xxyyx−xˉx-\bar xy−yˉy-\bar y(x−xˉ)(y−yˉ)(x-\bar x)(y-\bar y)(x−xˉ)2(x-\bar x)^2(y−yˉ)2(y-\bar y)^2
250−4−156016225
455−2−10204100
66500000
8702510425
10854208016280
Total17040630

Step 4 — Apply Pearson's formula.

r=17040630=17025200≈170158.75≈1.07r = \dfrac{170}{\sqrt{40}\sqrt{630}} = \dfrac{170}{\sqrt{25200}} \approx \dfrac{170}{158.75} \approx 1.07

Since a valid correlation coefficient can never exceed 11 in magnitude, this signals an arithmetic slip — re-checking the cross-product column: (−4)(−15)=60(-4)(-15)=60, (−2)(−10)=20(-2)(-10)=20, (0)(0)=0(0)(0)=0, (2)(5)=10(2)(5)=10, (4)(20)=80(4)(20)=80; total =60+20+0+10+80=170=60+20+0+10+80=170, which is correct as tabulated, so the error must be in a squared term. Re-checking ∑(y−yˉ)2\sum(y-\bar y)^2: 225+100+0+25+400=750225+100+0+25+400=750, not 630630 as first written (the last term, 202=40020^2=400, not 280280) — correcting this: ∑(y−yˉ)2=750\sum(y-\bar y)^2=750.

Step 5 — Recompute with the corrected sum.

r=17040750=17030000≈170173.2≈0.981r = \dfrac{170}{\sqrt{40}\sqrt{750}} = \dfrac{170}{\sqrt{30000}} \approx \dfrac{170}{173.2} \approx 0.981 …

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