Chapter 2, Problem 93RE

Solution

(a)

To find: The linear regression model and scatter graph.

The model of the given data $y=57.0019+0.013x$

Given information

The table for the given data is below.

period	2	7	12	17	22	27	32	37	42	47	52	57	62
temperature	57.15	57.16	57.18	57.16	57.20	57.26	57.44	57.47	57.53	57.67	57.75	57.83	57.87

For 1952,x=2

Straight line equation is y=a+bx.

The normal equations are

∑y=an+b∑x

∑xy=a∑x+b∑x²

The values are calculated using the following table

x	y	x2	x·y
2	57.15	4	114.3
7	57.16	49	400.12
12	57.18	144	686.16
17	57.16	289	971.72
22	57.2	484	1258.4
27	57.26	729	1546.02
32	57.44	1024	1838.08
37	57.47	1369	2126.39
42	57.53	1764	2416.26
47	57.67	2209	2710.49
52	57.75	2704	3003
57	57.83	3249	3296.31
62	57.87	3844	3587.94
---	---	---	---
∑x=416	∑y=746.67	∑x2=17862	∑x·y=23955.19

Substituting these values in the normal equations

13a+416b=746.67

416a+17862b=23955.19

Solving these two equations using Elimination method,

13a+416b=746.67

and 416a+17862b=23955.19

13a+416b=746.67 →(1)

416a+17862b=23955.19 →(2)

equation(1)×32⇒416a+13312b=23893.44

equation(2)×1⇒416a+17862b=23955.19

Substracting ⇒-4550b=-61.75

⇒4550b=61.75

⇒b=61.754550

⇒b=0.0136

Putting b=0.0136 in equation (1), we have

13a+416(0.0136)=746.67

⇒13a=746.67-5.6457

⇒13a=741.0243

⇒a=741.024313

⇒a=57.0019

∴a=57.0019 and b=0.0136

Now substituting this values in the equation is y=a+bx, we get

  $y=57.0019+0.013x$

The scatter graph of above data is below.

  

(b)

To find: The quadratic regression model and scatter graph.

The model of the given data $y=57.1108+0.0032x+0.0002x$

Given information

The table for the given data is below.

period	2	7	12	17	22	27	32	37	42	47	52	57	62
temperature	57.15	57.16	57.18	57.16	57.20	57.26	57.44	57.47	57.53	57.67	57.75	57.83	57.87

The equation is y=a+bx+cx² and the normal equations are

  ${\displaystyle \sum y=an+b{\displaystyle \sum x+c{\displaystyle \sum x^2}}}$

  ${\displaystyle \sum xy=a{\displaystyle \sum x}+b{\displaystyle \sum x^2+c{\displaystyle \sum x^3}}}$

  ${\displaystyle \sum x^{2}y=a{\displaystyle \sum x^2}+b{\displaystyle \sum x^3+c{\displaystyle \sum x^4}}}$

The values are calculated using the following table

x	y	x2	x3	x4	x·y	x2·y
2	57.15	4	8	16	114.3	228.6
7	57.16	49	343	2401	400.12	2800.84
12	57.18	144	1728	20736	686.16	8233.92
17	57.16	289	4913	83521	971.72	16519.24
22	57.2	484	10648	234256	1258.4	27684.8
27	57.26	729	19683	531441	1546.02	41742.54
32	57.44	1024	32768	1048576	1838.08	58818.56
37	57.47	1369	50653	1874161	2126.39	78676.43
42	57.53	1764	74088	3111696	2416.26	101482.92
47	57.67	2209	103823	4879681	2710.49	127393.03
52	57.75	2704	140608	7311616	3003	156156
57	57.83	3249	185193	10556001	3296.31	187889.67
62	57.87	3844	238328	14776336	3587.94	222452.28
---	---	---	---	---	---	---
∑x=416	∑y=746.67	${\displaystyle \sum x^2}$ =17862	${\displaystyle \sum x^3}$ =862784	${\displaystyle \sum x^4}$ =44430438	∑x·y=23955.19	${\displaystyle \sum x^{2}y}$ =1030078.83

Substituting these values in the normal equations

13a+416b+17862c=746.67

416a+17862b+862784c=23955.19

17862a+862784b+44430438c=1030078.83

Solving these 3 equations,

Total Equations are 3

13a+416b+17862c=746.67→(1)

416a+17862b+862784c=23955.19→(2)

17862a+862784b+44430438c=1030078.83→(3)

Select the equations (1) and (2), and eliminate the variable a.

  

Select the equations (1) and (3), and eliminate the variable a.

  

Select the equations (4) and (5), and eliminate the variable b.

  

Now use back substitution method

From (6)

1251250c=202.25

⇒c=202.251251250=0.0002

From (4)

-4550b-291200c=-61.75

⇒-4550b-291200(0.0002)=-61.75

⇒-4550b-47.0691=-61.75

⇒-4550b=-61.75+47.0691=-14.6809

⇒b=-14.6809-4550=0.0032

From (1)

13a+416b+17862c=746.67

⇒13a+416(0.0032)+17862(0.0002)=746.67

⇒13a+4.2294=746.67

⇒13a=746.67-4.2294=742.4406

⇒a=742.440613=57.1108

Solution using Elimination method.

a=57.1108,b=0.0032,c=0.0002

Now substituting this values in the equation is y=a+bx+cx², we get

  $y=57.1108+0.0032x+0.0002x$

Scatter graph of the above data is given below

  

(c)

To find: Which model is best fit.

The quadratic model of the given data is best fit.

Given information

The quadratic model of the given data is best fit.

By observing the both graph it is seen that the data is more scatter to the quadratic graph and gives clear idea about how temperature goes up and down .Hence the quadratic model is best fit for the above data.