List of Practical Experiments (Official Syllabus)

1. Fitting of straight line by the method of least squares.
2. Fitting of parabola by the method of least squares.
3. Fitting of exponential curve of two types by the method of least squares.
4. Fitting of power curve by the method of least squares.
5. Computation of correlation coefficient and regression lines for ungrouped data.
6. Computation of correlation coefficient for bivariate frequency distribution.
7. Computation of correlation coefficient and forming regression lines for grouped data.
8. Computation of partial and multiple correlation coefficients.
9. Computation of Yule's coefficient of association and colligation.

Experiment 1 — Fitting a Straight Line

1. Problem

Fit a straight line \(Y = a + bX\) by the method of least squares to:

x	0	1	2	3	4
y	1	1.8	3.3	4.5	6.3

2. Aim

To fit a straight line to bivariate data by the method of least squares.

3. Formula

Applying it:

1. Form the columns \(x, y, xy, x^2\) and total them.
2. Compute the slope \(b\) and intercept \(a\) from the least-squares formulas.

4. Calculation

Blank working table (fill in xy and x², then total):

x	y	xy	x²
0	1
1	1.8
2	3.3
3	4.5
4	6.3
Σ

\(n = 5\); \(\sum x = 10,\ \sum y = 16.9,\ \sum xy = 47.1,\ \sum x^2 = 30\).

\(b = \dfrac{5(47.1) - (10)(16.9)}{5(30) - 10^2} = \dfrac{235.5 - 169}{150 - 100} = \dfrac{66.5}{50} = 1.33\).

\(a = \dfrac{16.9}{5} - 1.33(2) = 3.38 - 2.66 = 0.72\).

5. Result

The fitted least-squares line is \(Y = 0.72 + 1.33\,X\).

Experiment 2 — Fitting a Parabola

1. Problem

Fit a second-degree parabola \(Y = a + bX + cX^2\) by the method of least squares to:

x	1	2	3	4	5
y	2	5	10	17	26

2. Aim

To fit a second-degree polynomial (parabola) to data by the method of least squares.

3. Formula

Applying it:

1. Form the sums \(\sum x, \sum x^2, \sum x^3, \sum x^4, \sum y, \sum xy, \sum x^2 y\).
2. Write the three normal equations and solve for \(a, b, c\).

4. Calculation

Blank working table (fill in the product columns and total):

x	y	x²	x³	x⁴	xy	x²y
1	2
2	5
3	10
4	17
5	26
Σ	60

\(\sum x = 15,\ \sum x^2 = 55,\ \sum x^3 = 225,\ \sum x^4 = 979,\ \sum y = 60,\ \sum xy = 240,\ \sum x^2 y = 1034\).

Normal equations:

• \(60 = 5a + 15b + 55c\)
• \(240 = 15a + 55b + 225c\)
• \(1034 = 55a + 225b + 979c\)

Solving gives \(a = 1,\ b = 0,\ c = 1\) (the equations are satisfied exactly: \(5+55 = 60\), \(15+225 = 240\), \(55+979 = 1034\)).

5. Result

The fitted parabola is \(Y = 1 + X^2\) — an exact fit, since the data were generated by \(1 + X^2\).

Experiment 3 — Fitting Exponential Curves

1. Problem

Fit (Type I) \(Y = a\,b^X\) and (Type II) \(Y = a\,e^{bX}\) by least squares to:

x	1	2	3	4	5
y	3	9	27	81	243

2. Aim

To fit exponential curves by linearising them with logarithms and applying least squares.

3. Formula

Applying it:

1. Type I: take \(\log_{10} y\); then \(\log y = \log a + (\log b)\,X\) is linear in \(X\).
2. Fit the line: slope \(B = \log b\), intercept \(A = \log a\); recover \(a = 10^{A},\ b = 10^{B}\).
3. Type II: the same data give \(b = \ln(\text{growth factor})\) directly, since \(a\,e^{bX}\) with \(a = 1\).

4. Calculation

Blank working table (fill in log y and x·log y, then total):

x	y	log y	x · log y	x²
1	3			1
2	9			4
3	27			9
4	81			16
5	243			25
Σ				55

\(\log y\): 0.477, 0.954, 1.431, 1.908, 2.386; \(\sum \log y = 7.157\); \(\sum x = 15\), \(\sum x^2 = 55\).

\(\sum x\log y = 1(0.477) + 2(0.954) + 3(1.431) + 4(1.908) + 5(2.386) = 26.24\).

\(B = \dfrac{5(26.24) - 15(7.157)}{5(55) - 15^2} = \dfrac{131.2 - 107.36}{50} = 0.477\), so \(b = 10^{0.477} = 3\).

\(A = \dfrac{7.157}{5} - 0.477(3) = 1.431 - 1.431 = 0\), so \(a = 10^{0} = 1\).

Type II: \(b = \ln 3 = 1.0986\), \(a = 1\).

5. Result

Type I: \(Y = 1 \cdot 3^{X} = 3^{X}\). Type II: \(Y = e^{1.0986\,X}\) (the same curve written with base \(e\)).

Experiment 4 — Fitting a Power Curve

1. Problem

Fit a power curve \(Y = a\,X^{b}\) by least squares to:

x	1	2	3	4
y	2	16	54	128

2. Aim

To fit a power curve by taking logarithms of both variables and applying least squares.

3. Formula

Applying it:

1. Take \(\log x\) and \(\log y\); then \(\log y = \log a + b\log x\) is linear.
2. Fit the line: slope \(b\), intercept \(\log a\); recover \(a = 10^{\log a}\).

4. Calculation

Blank working table (fill in the log columns and products, then total):

x	y	log x	log y	(log x)²	log x · log y
1	2
2	16
3	54
4	128
Σ

\(\log x\): 0, 0.301, 0.477, 0.602; \(\log y\): 0.301, 1.204, 1.732, 2.107.

\(\sum \log x = 1.380,\ \sum \log y = 5.344,\ \sum(\log x)^2 = 0.681,\ \sum(\log x)(\log y) = 2.458\).

\(b = \dfrac{4(2.458) - (1.380)(5.344)}{4(0.681) - 1.380^2} = \dfrac{9.832 - 7.375}{2.724 - 1.904} = \dfrac{2.457}{0.820} = 3.0\).

\(\log a = \dfrac{5.344}{4} - 3\!\left(\dfrac{1.380}{4}\right) = 1.336 - 1.035 = 0.301\), so \(a = 10^{0.301} = 2\).

5. Result

The fitted power curve is \(Y = 2\,X^{3}\) — an exact fit, since the data were generated by \(2X^3\).

Experiment 5 — Correlation & Regression Lines (Ungrouped)

1. Problem

For the bivariate data below, compute the correlation coefficient and the two regression lines:

x	1	2	3	4	5
y	2	3	5	4	6

2. Aim

To compute Karl Pearson's correlation coefficient and the two regression lines for ungrouped data.

3. Formula

Applying it:

1. Form the sums \(\sum x, \sum y, \sum xy, \sum x^2, \sum y^2\).
2. Compute \(r\) and the regression coefficients \(b_{yx},\ b_{xy}\); check \(r = \sqrt{b_{yx}\,b_{xy}}\).

4. Calculation

Blank working table (fill in the product columns and total):

x	y	xy	x²	y²
1	2
2	3
3	5
4	4
5	6
Σ

\(n = 5\); \(\sum x = 15,\ \sum y = 20,\ \sum xy = 69,\ \sum x^2 = 55,\ \sum y^2 = 90\).

\(r = \dfrac{5(69) - (15)(20)}{\sqrt{[5(55) - 225][5(90) - 400]}} = \dfrac{345 - 300}{\sqrt{50 \times 50}} = \dfrac{45}{50} = 0.9\).

\(b_{yx} = 45/50 = 0.9\), \(b_{xy} = 45/50 = 0.9\); with \(\bar x = 3,\ \bar y = 4\):

• Regression of Y on X: \(Y - 4 = 0.9(X - 3) \Rightarrow Y = 0.9X + 1.3\).
• Regression of X on Y: \(X - 3 = 0.9(Y - 4) \Rightarrow X = 0.9Y - 0.6\).

5. Result

\(r = 0.9\) (strong positive correlation). Regression lines: \(Y = 0.9X + 1.3\) and \(X = 0.9Y - 0.6\). Check: \(\sqrt{b_{yx}\,b_{xy}} = \sqrt{0.81} = 0.9 = r\) ✓.

Experiment 6 — Correlation for a Bivariate Frequency Distribution

1. Problem

Bivariate frequency table (X = age in years, Y = weight in kg):

X \ Y	40–50	50–60	60–70	Total
20–30	2	5	3	10
30–40	1	10	9	20
40–50	0	5	15	20
Total	3	20	27	50

Compute the correlation coefficient.

2. Aim

To compute Karl Pearson's correlation coefficient for a bivariate frequency distribution using step deviations.

3. Formula

Applying it:

1. Take the class mid-points and step deviations \(u = (x - A)/h_x,\ v = (y - B)/h_y\).
2. For each cell compute \(fu, fv, fu^2, fv^2, fuv\) and total them.
3. Substitute into the step-deviation correlation formula.

4. Calculation

Blank working table (mid-points \(x = 25, 35, 45\Rightarrow u=-1,0,1\); \(y = 45, 55, 65\Rightarrow v=-1,0,1\); fill in the product columns and total):

u	v	f	fu	fv	fu²	fv²	fuv
−1	−1	2
−1	0	5
−1	1	3
0	−1	1
0	0	10
0	1	9
1	−1	0
1	0	5
1	1	15
Σ		50

u	v	f	fu	fv	fu²	fv²	fuv
−1	−1	2	−2	−2	2	2	2
−1	0	5	−5	0	5	0	0
−1	1	3	−3	3	3	3	−3
0	−1	1	0	−1	0	1	0
0	0	10	0	0	0	0	0
0	1	9	0	9	0	9	0
1	−1	0	0	0	0	0	0
1	0	5	5	0	5	0	0
1	1	15	15	15	15	15	15
Σ		50	10	24	30	30	14

\(r = \dfrac{50(14) - (10)(24)}{\sqrt{[50(30) - 10^2][50(30) - 24^2]}} = \dfrac{700 - 240}{\sqrt{(1400)(924)}} = \dfrac{460}{1137.4} = 0.40\).

5. Result

\(r = 0.40\) — a moderate positive correlation between age and weight.

Experiment 7 — Correlation & Regression Lines for Grouped Data

1. Problem

For the bivariate frequency table of Experiment 6, compute the means, standard deviations and the two regression lines.

2. Aim

To obtain the regression lines of a bivariate frequency distribution from the correlation coefficient and the marginal standard deviations.

3. Formula

Applying it:

1. From the step-deviation sums (Experiment 6), compute \(\bar x,\ \bar y,\ \sigma_x,\ \sigma_y\).
2. Compute the regression coefficients \(b_{yx} = r\,\sigma_y/\sigma_x\) and \(b_{xy} = r\,\sigma_x/\sigma_y\).
3. Write the two regression lines through \((\bar x, \bar y)\).

4. Calculation

Blank working table (fill in each quantity from the Experiment 6 sums):

Quantity	Value
\(\bar x\), \(\bar y\)
\(\sigma_x\), \(\sigma_y\)
\(r\)
\(b_{yx}\), \(b_{xy}\)

With \(A = 35,\ B = 55,\ h_x = h_y = 10,\ N = 50\) and the sums \(\sum fu = 10,\ \sum fv = 24,\ \sum fu^2 = 30,\ \sum fv^2 = 30\):

\(\bar x = 35 + 10(10/50) = 37\); \(\bar y = 55 + 10(24/50) = 59.8\).

\(\sigma_x = 10\sqrt{30/50 - (10/50)^2} = 10\sqrt{0.56} = 7.48\); \(\sigma_y = 10\sqrt{30/50 - (24/50)^2} = 10\sqrt{0.3696} = 6.08\).

With \(r = 0.40\): \(b_{yx} = 0.40(6.08/7.48) = 0.33\); \(b_{xy} = 0.40(7.48/6.08) = 0.50\).

• Regression of Y on X: \(Y - 59.8 = 0.33(X - 37)\).
• Regression of X on Y: \(X - 37 = 0.50(Y - 59.8)\).

5. Result

\(\bar x = 37,\ \bar y = 59.8,\ \sigma_x = 7.48,\ \sigma_y = 6.08,\ r = 0.40\). The regression lines are \(Y - 59.8 = 0.33(X - 37)\) and \(X - 37 = 0.50(Y - 59.8)\).

Experiment 8 — Partial & Multiple Correlation

1. Problem

Given the simple (zero-order) correlations \(r_{12} = 0.8,\ r_{13} = 0.5,\ r_{23} = 0.4\), compute the partial correlations \(r_{12.3},\ r_{13.2}\) and the multiple correlation \(R_{1.23}\).

2. Aim

To compute partial and multiple correlation coefficients from the zero-order correlations.

3. Formula

Applying it:

1. Substitute the given correlations into the partial-correlation formula for each pair.
2. Compute the coefficient of multiple determination \(R_{1.23}^2\) and take its square root.

4. Calculation

Blank working table (fill in each coefficient):

Coefficient	Value
\(r_{12.3}\)
\(r_{13.2}\)
\(R_{1.23}\)

\(r_{12.3} = \dfrac{0.8 - (0.5)(0.4)}{\sqrt{(1 - 0.25)(1 - 0.16)}} = \dfrac{0.6}{\sqrt{0.75 \times 0.84}} = \dfrac{0.6}{0.7937} = 0.756\).

\(r_{13.2} = \dfrac{0.5 - (0.8)(0.4)}{\sqrt{(1 - 0.64)(1 - 0.16)}} = \dfrac{0.18}{\sqrt{0.36 \times 0.84}} = \dfrac{0.18}{0.5499} = 0.327\).

\(R_{1.23}^2 = \dfrac{0.64 + 0.25 - 2(0.8)(0.5)(0.4)}{1 - 0.16} = \dfrac{0.89 - 0.32}{0.84} = 0.679\), so \(R_{1.23} = 0.824\).

5. Result

\(r_{12.3} = 0.756,\ r_{13.2} = 0.327,\ R_{1.23} = 0.824\). The two predictors together explain about 67.9 % of the variation in \(X_1\).

Experiment 9 — Yule's Coefficient of Association & Colligation

1. Problem

Distribution by smoking and lung cancer in 200 individuals:

	Cancer (B)	No Cancer (β)	Total
Smoker (A)	60	40	100
Non-smoker (α)	20	80	100
Total	80	120	200

Compute Yule's coefficient of association \(Q\) and the coefficient of colligation \(\omega\).

2. Aim

To measure the association between two attributes using Yule's \(Q\) and the coefficient of colligation.

3. Formula

Applying it:

1. Read off the four cell frequencies \((AB), (A\beta), (\alpha B), (\alpha\beta)\).
2. Compute \(Q\) and \(\omega\); verify the relation \(Q = 2\omega/(1 + \omega^2)\).

4. Calculation

Blank working table (fill in the products and coefficients):

Quantity	Value
\((AB)(\alpha\beta)\)
\((A\beta)(\alpha B)\)
\(Q\)
\(\omega\)

\((AB) = 60,\ (A\beta) = 40,\ (\alpha B) = 20,\ (\alpha\beta) = 80\); so \((AB)(\alpha\beta) = 4800\) and \((A\beta)(\alpha B) = 800\).

\(Q = \dfrac{4800 - 800}{4800 + 800} = \dfrac{4000}{5600} = 0.714\).

\(\omega = \dfrac{\sqrt{4800} - \sqrt{800}}{\sqrt{4800} + \sqrt{800}} = \dfrac{69.28 - 28.28}{69.28 + 28.28} = \dfrac{41.00}{97.56} = 0.420\).

Check: \(2\omega/(1 + \omega^2) = 0.840/1.176 = 0.714 = Q\) ✓.

5. Result

\(Q = 0.714\) and \(\omega = 0.420\) — a strong positive association between smoking and lung cancer.

Lab Record Format (to be followed for every experiment)

1. Problem — the data and what is to be found.
2. Aim — the objective of the experiment.
3. Formula — the formula(e) used, then the steps that apply them.
4. Calculation — the completed working table and the substitution of values.
5. Result — the final value(s) with a brief interpretation.