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Vidaara.orgClass 12 · Mathematics
CodeVID-M12-15-CT
Linear Regression — Full Chapter Test
Chapter: Linear Regression
Topic: Complete chapter — all topics
Maximum Marks: 35
Time: 80 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
  • Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
  • Show all working for Sections B, C and D. Final answers are given in the Answer Key.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
If the regression coefficients are $b_{xy}=0.4$ and $b_{yx}=0.9$, the correlation coefficient $r$ is:
  • A.$0.36$
  • B.$0.65$
  • C.$0.6$
  • D.$0.5$
2.
The regression lines $x+2y=5$ and $2x+3y=8$ intersect at the point:
  • A.$(2,1)$
  • B.$(1,2)$
  • C.$(5,0)$
  • D.$(4,0)$
3.
Assertion (A): The regression coefficients $b_{xy}=1.2$ and $b_{yx}=0.9$ cannot belong to the same data set.
Reason (R): The product $b_{xy}\,b_{yx}=r^2$ cannot exceed $1$.
  • A.Both A and R are true and R is the correct explanation of A
  • B.Both A and R are true but R is not the correct explanation of A
  • C.A is true but R is false
  • D.A is false but R is true
4.
If $r=0.8$, $\sigma_x=4$ and $\sigma_y=6$, then the regression coefficient $b_{yx}$ is:
  • A.$0.53$
  • B.$0.8$
  • C.$1.2$
  • D.$3$
5.
If the two lines of regression coincide, the value of $r$ is:
  • A.$0$
  • B.$\pm\dfrac12$
  • C.$\pm1$
  • D.any value in $(-1,1)$
Section B — Short Answer I 4 × 2 = 8 marks
6.
If $b_{yx}=-1.5$ and $b_{xy}=-0.2$, find the correlation coefficient $r$.
7.
For a data set, $\bar x=10$, $\bar y=12$ and $b_{yx}=0.6$. Estimate $y$ when $x=15$.
8.
The two regression lines are $3x+2y=26$ and $6x+y=31$. Find $\bar x$ and $\bar y$.
9.
Why are there two regression lines? When do they coincide?
Section C — Short Answer II 4 × 3 = 12 marks
10.
The two regression lines of a data set are $4x-5y+33=0$ and $20x-9y=107$. Find $\bar x$, $\bar y$ and the correlation coefficient $r$.
11.
Five students scored the following marks: $x$ (test 1) $=1,2,3,4,5$ and $y$ (test 2) $=2,3,5,4,6$. Find the regression line of $y$ on $x$ and predict $y$ when $x=6$.
12.
Prove that the arithmetic mean of the two regression coefficients is at least the correlation coefficient in absolute value: $\dfrac{b_{xy}+b_{yx}}{2}\ge|r|$.
13.
Given $r=0.5$, $\sigma_x=3$ and $\sigma_y=4$, find both regression coefficients and verify that $b_{xy}\,b_{yx}=r^2$.
Section D — Long Answer / Case Based 2 × 5 = 10 marks
14.
The two regression lines are $8x-10y+66=0$ and $40x-18y=214$, and the variance of $x$ is $9$. Find (a) the means $\bar x$ and $\bar y$, [2] (b) the correlation coefficient $r$, [1] (c) the standard deviation of $y$. [2]
15.
Case study. A shop owner studies monthly advertising spend $x$ (₹ thousand) and sales $y$ (₹ lakh). The data give $\bar x=20$, $\bar y=50$, $\sigma_x=4$, $\sigma_y=6$ and $r=0.75$.
(a) Find the regression line of $y$ on $x$ and the regression line of $x$ on $y$. [3]
(b) Estimate the sales when the advertising spend is ₹28 thousand. [1]
(c) Estimate the advertising spend needed for sales of ₹62 lakh. [1]

Answer Key

Section A — Multiple Choice Questions
  1. (C) $0.6$
  2. (B) $(1,2)$
  3. (A) Both A and R are true and R is the correct explanation of A
  4. (C) $1.2$
  5. (C) $\pm1$
Section B — Short Answer I
  1. $r^2=b_{yx}\,b_{xy}=0.3$. Both regression coefficients are negative, so $r$ is negative: $r=-\sqrt{0.3}\approx-0.55$.
  2. $y-\bar y=b_{yx}(x-\bar x)\Rightarrow y-12=0.6(x-10)$. At $x=15$, $y=12+3=15$.
  3. Both lines pass through $(\bar x,\bar y)$. From the second, $y=31-6x$; in the first, $3x+62-12x=26\Rightarrow x=4$, so $y=7$. Hence $\bar x=4$, $\bar y=7$.
  4. The line of $y$ on $x$ minimises the sum of squared vertical deviations (used to predict $y$), while the line of $x$ on $y$ minimises the squared horizontal deviations (used to predict $x$); they are different lines in general. They coincide only when $r=\pm1$ (perfect linear correlation).
Section C — Short Answer II
  1. Solving the two lines simultaneously: $x=\dfrac{5y-33}{4}$ in the second gives $25y-165-9y=107\Rightarrow y=17$, $x=13$. So $\bar x=13$, $\bar y=17$. Taking $4x-5y+33=0$ as the line of $y$ on $x$: $b_{yx}=\dfrac45$; and $20x-9y=107$ as the line of $x$ on $y$: $b_{xy}=\dfrac{9}{20}$. Then $r^2=\dfrac45\cdot\dfrac9{20}=0.36\le1$ (consistent), so $r=0.6$ (positive, as both coefficients are positive).
  2. $\bar x=3$, $\bar y=4$. $\sum(x-\bar x)(y-\bar y)=4+1+0+0+4=9$ and $\sum(x-\bar x)^2=10$. So $b_{yx}=\dfrac9{10}=0.9$. The line is $y-4=0.9(x-3)$, i.e. $y=0.9x+1.3$. At $x=6$, $y=6.7$.
  3. Since $b_{xy}$ and $b_{yx}$ have the same sign as $r$, AM $\ge$ GM gives $\dfrac{|b_{xy}|+|b_{yx}|}{2}\ge\sqrt{|b_{xy}||b_{yx}|}=\sqrt{r^2}=|r|$. Hence the result.
  4. $b_{yx}=r\dfrac{\sigma_y}{\sigma_x}=0.5\times\dfrac43=\dfrac23$ and $b_{xy}=r\dfrac{\sigma_x}{\sigma_y}=0.5\times\dfrac34=\dfrac38$. Their product is $\dfrac23\times\dfrac38=\dfrac14=0.25=r^2$. Verified.
Section D — Long Answer / Case Based
  1. (a) Multiply the first by $5$: $40x-50y+330=0$. Subtract from the second: $32y=544\Rightarrow y=17$; then $8x=170-66\Rightarrow x=13$. So $\bar x=13$, $\bar y=17$. (b) Line of $y$ on $x$: $y=0.8x+6.6$, so $b_{yx}=0.8$. Line of $x$ on $y$: $x=0.45y+5.35$, so $b_{xy}=0.45$. $r^2=0.8\times0.45=0.36$, so $r=0.6$. (c) $b_{yx}=r\dfrac{\sigma_y}{\sigma_x}\Rightarrow0.8=0.6\cdot\dfrac{\sigma_y}{3}\Rightarrow\sigma_y=4$.
  2. (a) $b_{yx}=r\dfrac{\sigma_y}{\sigma_x}=0.75\times\dfrac64=1.125$, so $y-50=1.125(x-20)$, i.e. $y=1.125x+27.5$. $b_{xy}=r\dfrac{\sigma_x}{\sigma_y}=0.75\times\dfrac46=0.5$, so $x-20=0.5(y-50)$, i.e. $x=0.5y-5$. (b) At $x=28$: $y=1.125(28)+27.5=59$, so sales are about ₹59 lakh. (c) At $y=62$: $x=0.5(62)-5=26$, so the spend should be about ₹26 thousand.
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