5 Marks Questions

Question 15 Marks

Ten competitors in a beauty contest are ranked by three judges of the following order:

1^st judge	1	5	4	8	9	6	10	7	3	2
2^ndjudge	4	8	7	6	5	9	10	3	2	1
3^rd judge	6	7	8	1	5	10	9	2	3	4

Use Spearman's rank correlation coefficient to test which pair of judges has nearest approach to common tasks in beauty.

Answer

Construct the following table:

R₁	R₂	R₃	d₁₂=R₁-R₂	d₁₃=R₁-R₃	d₂₃=R₂-R₃	d₁₂²	d₁₃²	d₂₃²
1	4	6	-3	-5	-2	9	25	4
5	8	7	-3	-2	1	9	4	1
4	7	8	-3	-4	-1	9	16	1
8	6	1	2	7	5	4	49	25
9	5	5	4	4	0	16	16	0
6	9	10	-3	-4	-1	9	16	1
10	10	9	0	1	1	0	1	1
7	3	2	4	5	1	16	25	1
3	2	3	1	0	-1	1	0	1
2	1	4	1	-2	-3	1	4	9
Total						$\Sigma d_{12}^2=74$	$\Sigma d_{13}^2=156$	$\Sigma d_{23}^2=44$

Spearman's correlation of rank coefficient between ranking given by : $1^{\text {st }}$ and 2nd judges, $ \begin{aligned} r_{12} & =1-\frac{6 \Sigma d_{12}^2}{n\left(n^2-1\right)} \\ & =1-\frac{6 \times 74}{10\left(10^2-1\right)} \\ & =1-\frac{444}{990}=\frac{546}{990}=0.551 \\ r_{13} & =1-\frac{6 \Sigma d_{13}^2}{n\left(n^2-1\right)} \end{aligned} $
$ \begin{aligned} & =1-\frac{6 \times 156}{10\left(10^2-1\right)} \\ & =1-\frac{936}{990}=\frac{54}{990}=0.054 \\ r_{23} & =1-\frac{6 \Sigma d_{23}^2}{n\left(n^2-1\right)} \\ & =1-\frac{6 \times 44}{10\left(10^2-1\right)} \\ & =1-\frac{264}{990}=\frac{726}{990}=0.733 \end{aligned} $
Since, $r_{23}$ is maximum, we conclude that $2^{\text {nd }}$ and $3^{\text {rd }}$ judges have the nearest approach to common beauty test.

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Question 25 Marks

Find spearman's coefficient of rank correlation between marks obtained in History and Geography.

Student	A	B	C	D	E	F	G	H	I	J
History	30	20	40	50	30	20	30	50	10	0
G ugraphy	15	40	40	45	20	30	15	50	20	10

Answer

We construct the following table (giving rank 1 to the lowest):

Student	Marks in History	Rank R₁	Marks in Geography	Rank R₂	Difference d=R₁-R₂	d²
A	30	6	15	2.5	3.5	12.25
B	20	3.5	40	7.5	-4	16
C	40	8	40	7.5	0.5	0.25
D	50	9.5	45	9	0.5	0.25
E	30	6	20	4.5	1.5	2.25
F	20	3.5	30	6	-2.5	6.25
G	30	6	15	2.5	3.5	12.25
H	50	9.5	50	10	-0.5	0.25
I	10	2	20	4.5	-2.5	6.25
J	0	1	10	1	0	0
Total						$\Sigma d^2=56$

There are 3 ties in ranks $R_{1}$ two ties of 2 items and one tie of 3 items. Also, there are 3 ties in ranks $R_{2}$ - each of two items.
Hence,
$r_{ s }=1-\frac{6\left[\Sigma d^2+\Sigma \frac{1}{12}\left(m^3-m\right)\right]}{n\left(n^2-1\right)}$
$=1-\frac{6\left[56+\frac{1}{12}\left(2^3-2\right)+\frac{1}{12}\left(3^3-3\right)+\frac{1}{12}\left(2^3-2\right)+\frac{1}{12}\left(2^3-2\right)+\frac{1}{12}\left(2^3-2\right)+\frac{1}{12}\left(2^3-2\right)\right]}{10\left(10^2-1\right)}$
$=1-\frac{6(60.5)}{990}=1-0.367=0.633$.

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Question 35 Marks

Coefficient of correlation between $x$ and $y$ for 20 items is 0.4. The AM's and SD's of x and y are known to be 12 and 15 and 3 and 4, re Later on, it was found that the pair (2 wrongly taken as (15, 20). Find the correct value of the correlation coefficient.

Answer

$\Rightarrow \Sigma x^2 =3060$
$\text { Similarly, } \sigma_y^2 =16$
$\Rightarrow \frac{\Sigma y^2}{N}-(\bar{y})^2 =16$
$\Rightarrow \frac{\Sigma y^2}{20}-(15)^2 =16$
$\Rightarrow \Sigma y^2 =4820$
$ \begin{aligned}\text { Thus, corrected } \quad\Sigma x & =n \bar{x}-\text { wrong } x \text { value } \\& + \text { correct } x \text { value } \\& =20 \times 12-15+20 \\& =245\end{aligned}$
Similarly corrected $\Sigma y=n \bar{y}$ - wrong $y$ value + correct $y$ value
$
=20 \times 15-20+15
$
$
=295
$
$
\begin{array}{ll}
\text { Corrected } & \Sigma x^2=3060-15^2+20^2=3235 \\
\text { Corrected } & \Sigma y^2=4820-20^2+15^2=4645
\end{array}
$
Thus, corrected value of the correlation coefficient is obtained as
$\begin{aligned} r & =\frac{\Sigma x y-\frac{\Sigma x \Sigma y}{N}}{\sqrt{\Sigma x^2-\frac{(\Sigma x)^2}{N}} \sqrt{\Sigma y^2-\frac{(\Sigma y)^2}{N}}} \\ & =\frac{3696-\frac{245 \times 295}{20}}{\sqrt{3235-\frac{(245)^2}{20}} \sqrt{4645-\frac{(295)^2}{20}}} \\ & =\frac{3696-3613.75}{\sqrt{3235-3001.25} \sqrt{4645-4351.25}} \\ & =\frac{82.25}{\sqrt{233.75} \sqrt{293.75}} \\ & =\frac{82.25}{15.28 \times 17.13} \\ & =\frac{82.25}{261.7464} \\ & =0.3142 \approx 0.31 .\end{aligned}$

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Question 45 Marks

Find the correlation coefficient between the heights of husbands and wives based on the following data (given in inches) and interpret the result.

Answer

We use assumed mean A = 70 B = 66 and use the formula
$r=\frac{\Sigma u v-\frac{1}{N} \Sigma u . \Sigma v}{\sqrt{\Sigma u^2-\frac{(\Sigma u)^2}{N}} \sqrt{\Sigma v^2-\frac{(\Sigma v)^2}{N}}}$

Couple	x	u=x-A =x-70	u²	y	v=y-B =y-66	v²	uv
1	76	6	36	71	5	25	30
2	75	5	25	70	4	16	20
3	75	5	25	70	4	16	20
4	72	2	4	67	1	1	2
5	72	2	4	71	5	25	10
6	71	1	1	65	-1	1	-1
7	71	1	1	65	-1	1	-1
8	70	0	0	67	1	1	0
9	68	-2	4	64	-2	4	4
10	68	-2	4	65	-1	1	2
11	68	-2	4	65	-1	1	2
12	68	-2	4	66	0	0	0
13	67	-3	9	63	-3	9	9
14	67	-3	9	65	-1	1	3
15	62	-8	64	61	-5	25	40
Total		0	194		5	127	140

Now, using (i),
$\begin{aligned} r & =\frac{140-\frac{(0)(5)}{15}}{\sqrt{194-\frac{(0)^2}{15}} \sqrt{127-\frac{(5)^2}{15}}} \\ & =\frac{140}{\sqrt{194} \sqrt{127-1.66}} \\ & =\frac{140}{\sqrt{194} \sqrt{125.34}}\end{aligned}$
$\begin{array}{l}=\frac{140}{13.93 \times 11.19} \\ =\frac{140}{155.87} \\ =0.90 \text { (Approx) }\end{array}$
which is a strong positive correlation. This short women (called assertive matching) shows that tall men usually marry tall women and short men marry

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Question 55 Marks

Examine whether there is any correlation between age and blindness on the basis of the following data:

Answer

Let us denote the mid-value of age in years as x and the number of blind persons per lakh as y. Then, to compute coefficient of correlation between x and y, we construct the following table:

Age in years	Mid-values	No. of persons	No. of Blind persons	*No. of blind per lakh y = B / P 1 lakh**	xy	x²	y²
0-10	5	90	10	11	55	25	121
10-20	15	120	15	12	180	225	144
20-30	25	140	18	13	325	625	169
30-40	35	100	20	20	700	1225	400
40-50	45	80	15	19	855	2025	361
50-60	55	60	12	20	1100	3025	400
60-70	65	40	10	25	1625	4225	625
70-80	75	20	6	30	2250	5625	900
Total	320	-	-	150	7090	17000	3120

The correlation coefficient between age and blindness is given by
$\begin{aligned} r & =\frac{\Sigma x y-\frac{\Sigma x \Sigma y}{N}}{\sqrt{\Sigma x^2-\frac{1}{N}(\Sigma x)^2} \sqrt{\Sigma y^2-\frac{1}{N}(\Sigma y)^2}} \\ & =\frac{7090-\frac{320 \times 150}{8}}{\sqrt{17000-\frac{(320)^2}{8}} \sqrt{3120-\frac{(150)^2}{8}}} \\ & =\frac{7090-6000}{\sqrt{17000-12800} \sqrt{3120-2812.5}}\end{aligned}$
$\begin{array}{l}=\frac{1090}{\sqrt{4200} \sqrt{307.5}}=\frac{1090}{64.80 \times 17.535} \\ =\frac{1090}{1136.268} \\ =0.959=0.96\end{array}$
Which exhibits a very high degree of positive correlation between age and blindness.

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Question 65 Marks

Calculate, the PR for the score 35 and 55 for the distribution of scores give in the following table :

Scores	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
f	2	4	5	10	35	20	13	8	3

Answer

Scores	f	c.f.
9.5-19.5	2	2
19.5-29.5	4	6
29.5-39.5	5	11
39.5-49.5	10	21
49.5-59.5	35	56
59.5-69.5	20	76
69.5-79.5	13	89
79.5-89.5	8	97
89.5-99.5	3	100
	n = 100

We use formula, $P R=\frac{100}{N}+\left(\right.$ c.f. $\left.+\frac{X-l}{i} \times f\right)$
(i) For $x=35$
$X$ lies between the class $29.5-39.5$
So, c.f. $=6, l=29.5, f=5, i=10, N=100$
$\therefore \quad P R=\frac{100}{100}+\left(6+\frac{35-29.5}{10} \times 5\right)$
$\begin{array}{l}=1+\left(6+\frac{5.5}{2}\right) \\ =1+(6+2.75)\end{array}$
\[=9.75 \]
Thus, percentile ranking of 35 is 9.75
(ii) For $X=55$,
$X$ lies between the class $49.5-59.5$
So, c.f. $=21, l=49.5, f=35, i=10, N=100$
$\therefore \quad P R=\frac{100}{100}+\left(21+\frac{(55-49.5)}{10} \times 35\right)$
$\begin{array}{l}=1+(21+19.25) \\ =41.25\end{array}$
Thus, percentile ranking of 55 is 41.25 .

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Question 75 Marks

Compute the correlation coefficient between x and y from the following data:
$\begin{array}{l}N=10, \Sigma x y=220, \Sigma x^2=200, \Sigma y^2=262, \Sigma x=40 \\ \text { and } \Sigma y=50 .\end{array}$

Answer

From the given data, we have by applying
$\begin{aligned} r & =\frac{\Sigma x y-\frac{\Sigma x \Sigma y}{N}}{\sqrt{\Sigma x^2-\frac{(\Sigma x)^2}{N}} \times \sqrt{\Sigma y^2-\frac{(\Sigma y)^2}{N}}} \\ & =\frac{220-(40 \times 50) / 10}{\sqrt{200-\frac{(40)^2}{10}} \times \sqrt{262-\frac{(50)^2}{10}}} \\ & =\frac{220-200}{\sqrt{200-160} \times \sqrt{262-250}} \\ & =\frac{20}{\sqrt{40} \times \sqrt{12}}\end{aligned}$
$\begin{array}{l}=\frac{20}{2 \sqrt{10} \times 2 \sqrt{3}}=\frac{20}{4 \sqrt{10} \times \sqrt{3}}=\frac{5}{\sqrt{10} \times \sqrt{3}} \\ =\frac{5}{10 \times 3} \times \sqrt{10} \times \sqrt{3} \\ =\frac{3.162 \times 1.732}{6} \\ =0.527 \times 1.732 \\ =0.91 . \text { (Approx) }\end{array}$
Alternate Solution
As given,
$\begin{aligned} \bar{x} & =\frac{\Sigma x}{N}=\frac{40}{10}=4 \\ \bar{y} & =\frac{\Sigma y}{N}=\frac{50}{10}=5 \\ \operatorname{Cov}(x, y) & =\frac{\Sigma x y}{N}-\overline{x y} \\ & =\frac{220}{10}-4 \times 5=22-20=2 \\ \sigma_x & =\sqrt{\frac{\Sigma x^2}{N}-(\bar{x})^2}=\sqrt{\frac{200}{10}-(4)^2} \\ & =\sqrt{20-16}=\sqrt{4}=2 \\ \sigma_y & =\sqrt{\frac{\Sigma y^2}{N}-(\bar{y})^2}=\sqrt{\frac{262}{10}-(5)^2} \\ & =\sqrt{26.20-25} \\ & =1.0954 \end{aligned} $
Thus applying the formula,
$\begin{aligned} r & =\frac{\operatorname{Cov}(x, y)}{\sigma_{x .}{ }^\sigma y} \\ & =\frac{2}{2 \times 1.0954} \\ & =0.91 \text { (Approx) } \end{aligned} $
As before, we draw the same conclusion.

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