Example Problems -simple linear regression - University Questions

Example

Determine the regression equation by finding the regression slope coefficient and the intercept value using the following data.
x  55 60  65 70  80
y  52 54  56 58  62

beta0=30

beta1=0.4

so the regression equation is  y=30+0.4x

Easy Method (Using Summations)

The regression equation is

$$\hat{y}=\theta_0+\theta_1x$$

where

$$\theta_1=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$$

and

$$\theta_0=\frac{\sum y-\theta_1\sum x}{n}$$

This method only requires the columns $x$, $y$, $x^2$, and $xy$.

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Step 1: Prepare the Table

x	y	x²	xy
55	52	3025	2860
60	54	3600	3240
65	56	4225	3640
70	58	4900	4060
80	62	6400	4960

Σ      330   282   22150   18760

Also,

• $n=5$
  
• $\sum x=330$
  
• $\sum y=282$
  
• $\sum x^2=22150$
  
• $\sum xy=18760$
  

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Step 2: Calculate the Slope

$$\theta_1= \frac{5(18760)-330(282)} {5(22150)-330^2}$$

Substitute the values:

$$= \frac{93800-93060} {110750-108900}$$ $$= \frac{740}{1850}$$ $$=0.4$$

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Step 3: Calculate the Intercept

$$\theta_0= \frac{\sum y-\theta_1\sum x}{n}$$

Substitute the values:

$$= \frac{282-(0.4)(330)}{5}$$ $$= \frac{282-132}{5}$$ $$= \frac{150}{5}$$ $$=30$$

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Step 4: Regression Equation

$$\boxed{\hat{y}=30+0.4x}$$

Example

The values of x and their corresponding values of y are shown in the table below. Find the least square regression lineEstimate the value of y when x=10

x=0  1  2  3  4

y=2  3  5  4  6

Create Calculation Table

$x$	$y$	$x^2$	$xy$
0	2	0	0
1	3	1	3
2	5	4	10
3	4	9	12
4	6	16	24

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🔸 Compute Sums

$$\sum x = 10,\quad \sum y = 20$$
$$\sum x^2 = 30,\quad \sum xy = 49$$
$$n = 5$$

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🔹 Formula for${\mathrm{ \beta }}_1$​ (Slope)

$$\beta_1 = \frac{n\sum xy - (\sum x)(\sum y)} {n\sum x^2 - (\sum x)^2}$$

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🔸 Substitute values

$$\beta_1 = \frac{5(49) - (10)(20)} {5(30) - (10)^2}$$ $$= \frac{245 - 200}{150 - 100} = \frac{45}{50} = 0.9$$

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🔹  Formula for${\beta }_{}$$0_{}$(Intercept)

$$\beta_0 = \frac{\sum y - \beta_1 \sum x}{n}$$

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🔸 Substitute values

$$\beta_0 = \frac{20 - 0.9(10)}{5}$$ $$= \frac{20 - 9}{5} = \frac{11}{5} = 2.2$$

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✅ Regression Line

$$\boxed{y = 0.9x + 2.2}$$

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✅ Estimate y when x=10

$$y = 0.9(10) + 2.2 = 9 + 2.2 = 11.2$$

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✅ Final Answer

• Regression line:

$$y = 0.9x + 2.2$$

• Estimated value at $x = 10$:

$$\boxed{11.2}$$

Example

Use the following data to construct a linear regression model for the auto insurance premium as a function of driving experience.

DrivingExp            |5 |2   |12|9  |15 |6  |25 |16

MonthlyPremium    |64|87|50|71|44 |56 |42 |60

Construct Calculation Table

$x$	$y$	$x^2$	$xy$
5	64	25	320
2	87	4	174
12	50	144	600
9	71	81	639
15	44	225	660
6	56	36	336
25	42	625	1050
16	60	256	960

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Compute Sums

$$\sum x = 90,\quad \sum y = 474$$
$$\sum x^2 = 1396,\quad \sum xy = 4739$$
$$n = 8$$

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 Compute Slope${\beta }_{}$$1_{}$

​

$$\beta_1 = \frac{n\sum xy - (\sum x)(\sum y)} {n\sum x^2 - (\sum x)^2}$$

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Substitute:

$$\beta_1 = \frac{8(4739) - (90)(474)} {8(1396) - (90)^2}$$$$\text{Denominator} = 11168 - 8100 = 3068$$

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Final:

$$\beta_1 = \frac{-4748}{3068} \approx -1.55$$

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Compute Intercept β0

$$\beta_0 = \frac{\sum y - \beta_1 \sum x}{n}$$

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Substitute:

$$\beta_0 = \frac{474 - (-1.55 \times 90)}{8}$$ $$= \frac{474 + 139.5}{8} = \frac{613.5}{8} \approx 76.69$$

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✅ Regression Model

$$\boxed{y = -1.55x + 76.69}$$

∣

Example

Predict the price of a 1000 square feet house using the regression model generated from the following data.

Square feet     Price(Lakhs)

500                     5

900                     10

1200                   13

1500                   18

2000                   25

2500                   32

2700                   35

Construct Table

$x$	$y$	$x^2$	$xy$
500	5	250000	2500
900	10	810000	9000
1200	13	1440000	15600
1500	18	2250000	27000
2000	25	4000000	50000
2500	32	6250000	80000
2700	35	7290000	94500

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Compute Sums

$$\sum x = 11300,\quad \sum y = 138$$
$$\sum x^2 = 22290000,\quad \sum xy = 278600$$
$$n = 7$$

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Compute Slopeβ1

​$$\beta_1 = \frac{n\sum xy - (\sum x)(\sum y)} {n\sum x^2 - (\sum x)^2}$$

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Substitute:

$$\beta_1 = \frac{7(278600) - (11300)(138)} {7(22290000) - (11300)^2}$$

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$$\mathrm{}$$

Final:

$$\beta_1 = \frac{390800}{28340000} \approx 0.01379$$

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Compute Intercept β0

$${\mathrm{<br>}}_{}$$

​ $$\beta_0 = \frac{\sum y - \beta_1 \sum x}{n}$$

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Substitute:

$$\beta_0 = \frac{138 - (0.01379 \times 11300)}{7}$$ $$= \frac{138 - 155.83}{7} = \frac{-17.83}{7} \approx -2.55$$

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✅ Regression Model

$$\boxed{y = 0.01379x - 2.55}$$

Predict Price for 1000 sq.ft

$$y = 0.01379(1000) - 2.55$$
$$= 13.79 - 2.55 = 11.24$$

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✅ Final Answer

• Regression line:

$$\boxed{y = 0.01379x - 2.55}$$

• Predicted price for 1000 sq.ft:

$$\boxed{11.24 \text{ Lakhs}}$$