Chapter 1.7, Problem 51E

Solution

(a)

To find: The scatter graph of given data.

The graph is given

Given information:

Time(min)	Temperature	Time (L2)	Temperature
1	184.3	11	140
2	178.5	12	136.1
3	173.5	13	133.5
4	168.6	14	130.5
5	164.0	15	127.9
6	159.2	16	125
7	155.1	17	122.8
8	151.8	18	119.9
9	147	19	117.2
10	143.7	20	115.2

Graph of time from L1 and temperature from L2 is given below.

  

(b)

To find: Store L2-72 and find L3.

The table is given

Given information:

L3 Time	temperature
21	68
22	64.1
23	61.5
24	68.5
25	55.9
26	53
27	50.8
28	47.9
29	45.2
30	43.2

(c)

To find: Find the exponential regression for L3.

The exponential regression for L3 is $y=73.71\cdot {\left(0.9485\right)}^x$

Given information:

L3 Time	temperature
21	68
22	64.1
23	61.5
24	68.5
25	55.9
26	53
27	50.8
28	47.9
29	45.2
30	43.2

The curve to be fitted is $y=a{b}^x$ taking logarithm on both sides, we get

  ${\log }_{10}(y)={\log }_{10}(a)+x{\log }_{10}(b)$

  $y=a+bx$ where $Y={\log }_{10}(y),A={\log }_{10}(a),B={\log }_{10}(b)$

which linear in Y,x

So the corresponding normal equations are

  ${\displaystyle \sum Y=nA+B{\displaystyle \sum x}}$

  ${{\displaystyle \sum xY=A{\displaystyle \sum x}+B{\displaystyle \sum x}}}^2$

The values are calculated using the following table

x	y	Y=log10(y)	x2	x·Y
1	68	1.8325	1	1.8325
2	64.1	1.8069	4	3.6137
3	61.5	1.7889	9	5.3666
4	68.5	1.8357	16	7.3428
5	55.9	1.7474	25	8.7371
6	53	1.7243	36	10.3457
7	50.8	1.7059	49	11.941
8	47.9	1.6803	64	13.4427
9	45.2	1.6551	81	14.8962
10	43.2	1.6355	100	16.3548
---	---	---	---	---
∑x=55	∑y=558.1	∑Y=17.4124	∑x2=385	∑x· Y=93.8731

Substituting these values in the normal equations

10A+55B=17.4124

55A+385B=93.8731

Use elimination method we obtain A=1.8676,B=-0.023

∴a=antilog10(A)=antilog10(1.8676)=73.7191 and b=antilog10(B)=antilog10(-0.023)=0.9485

Now substituting this values in the equation is $y=a{b}^x$ , we get $y=73.71\cdot {\left(0.9485\right)}^x$