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Does the data indicate that the revised (one week) forecast is significantly more accurate than the first (one month) forecast? In order to find out whether the revised forecast is significantly more accurate than the first forecast, we can use t-test to

Extracts from this document...

Introduction

Question1

A company supplying parts to a large customer receives forecasts of the expected demand in advance of the delivery date. One forecast is received one month ahead of delivery and a revised forecast is received one week ahead of delivery (i.e. three weeks after the first forecast). Finally, the actual requirements are indicated electronically on the delivery date. The following table shows that the data sent to the supplier covering a period of six months (twenty four weeks).

image00.png

(i) Does the data indicate that the revised (one week) forecast is significantly more accurate than the first (one month) forecast?

In order to find out whether the revised forecast is significantly more accurate than the first forecast, we can use t-test to test whether the means are equal for two populations.

At first, we calculate the errors of both forecasts, which give us the following data:

Week

Error 1 month ahead

Error 1Week ahead

Difference between 2 errors

1

-129

-80

49

2

-508

-468

40

3

-1050

-989

61

4

-476

-366

110

5

-1800

-1615

185

6

-1041

-956

85

7

-949

-979

-30

8

-515

-630

-115

9

-1188

-1243

-55

10

-1068

-34

1034

11

-621

123

744

12

-579

-178

401

13

-956

258

1214

14

-890

-819

71

15

27

53

26

16

48

108

60

17

-780

-750

30

18

-1089

-1012

77

19

-1072

-1173

-101

20

-1758

-1769

-11

21

-1032

-984

48

22

-1444

-1229

215

23

-893

-639

254

24

-453

-247

206

In order to test whether the revised forecast is more accurate than the first one, the paired t-test could be used.  

Define the question: is the result from the revised forecast more accurate than that of the first forecast?

Identify the appropriate model: as the mean difference is required and samples are small and not independent, we choose paired t-test.

Difference could be in either way; therefore, it is a two-tailed test.

Formulate the Null hypothesis and the Alternative hypothesis (H0 and H1):

  H0: µ12

H1: µ1≤µ2

The level of significance is α = 5%, then α/2 = 2.5%.

...read more.

Middle

image49.png=image50.pngimage50.png=-3.348

 a =image51.pngimage51.png - bimage52.pngimage52.png =840/10-(-3.348)image53.pngimage53.png =99.066

Therefore, the investment=99.066+ (-3.348)image15.pngimage15.pngnumber of years.

An estimate of the error variance could be found out by the following calculations:

Source of Variance

Sum of Squares

Degrees of Freedom

Mean Square

Quantity estimated by Mean Square

Due to regression

image54.png

=image55.pngimage55.png=924.75

1

M1=924.75

image03.pngimage03.png + image56.pngimage56.png

About regression(error)

By difference

=59

n-2=8

M0=7.375

image03.png

Total

image04.png

=983.75

n-1=9

Therefore, the value of the error mean square image05.pngimage05.png = 7.375

 And the standard deviation image06.pngimage06.png =2.716

In order to test the likelihood of any real relationship existing between the two variables, we choose F-test to be the assessment method.

H0: image56.pngimage56.png =0 or image03.pngimage03.png + image56.pngimage56.png=0

H1: image56.pngimage56.pngimage33.pngimage33.png 0 or image03.pngimage03.png + image56.pngimage56.pngimage33.pngimage33.png0

It is a one-tailed test, because a value of image56.pngimage56.pngimage02.pngimage02.png 0 does not have any sense.

The F value is calculated as:  F =image59.pngimage59.png=image60.pngimage60.png=125.39

At the level of significance is 5%, with vG=1, vL=8 (from the Degrees of Freedom column in above table ), we find Fcrit is 5.318, from Statistical Table 6(b).

As Fcalcimage33.pngimage33.png Fcrit, we reject H0. Therefore, the conclusion is that there is a significant relationship between investment and times.  

(iii) Produce an equation to show the rate of decline over the ten year period.

As is showed clearly in the previous graph, the rate of decline might follow the rules of linear regression, therefore, we assume the equation is Y= a + bX, where x is number of years and y is the rate of decline.

Using Excel, the following data are transformed

X

Y

X2

Y2

XY

0

0

0

0

0

1

-0.0375

1

0.001406

-0.0375

2

-0.1125

4

0.012656

-0.225

3

-0.125

9

0.015625

-0.375

4

-0.1375

16

0.018906

-0.55

5

-0.125

25

0.015625

-0.625

6

-0.2

36

0.04

-1.2

7

-0.275

49

0.075625

-1.925

8

-0.2625

64

0.068906

-2.1

9

-0.325

81

0.105625

-2.925

45

-1.6

285

0.354375

-9.9625

b=image49.pngimage49.png=image61.pngimage61.png=-0.033

 a =image51.pngimage51.png - bimage52.pngimage52.png =(-1.6)/10-(-0.033)image53.pngimage53.png =-0.012

Therefore, the rate of decline=-0.012+ (-0.033)image15.pngimage15.pngnumber of years.

(iv)  Calculate the 95% confidence limits for rate of change

Furthermore, we can find out an estimate of the error variance, as shown below

Source of Variance

Sum of Squares

Degrees of Freedom

Mean Square

Quantity estimated by Mean Square

Due to regression

image54.png

=image55.pngimage55.png=0.0898

1

M1=0.0898

image03.pngimage03.png + image56.pngimage56.png

About regression(error)

By difference

=0.0085

n-2=8

M0=0.0011

image03.png

Total

image04.png

=0.0983

n-1=9

Therefore, the value of the error mean square image05.pngimage05.png = 0.0011

 And the standard deviation image06.pngimage06.png =0.033

The standard error of b is given by

  S.E.(b)=image07.pngimage07.png=image08.pngimage08.png=image09.pngimage09.png=0.0037

The 95% confidence limits gives us image10.pngimage10.png=5%,image10.pngimage10.png/2=2.5%, v=8, from Statistical Table 5,

Therefore, image11.pngimage11.png=bimage12.pngimage12.pngimage14.pngimage14.pngimage15.pngimage15.png S.E.(b)

           =-0.033image12.pngimage12.png2.306image15.pngimage15.png0.0037

image16.pngimage16.png -0.042 to -0.024

(b)

A chemical filtration process was investigated to find the effect of different filter media on the speed of processing. Samples were collected from “Fast” filters and from “Slow” filters and the amount of a certain chemical ingredient was measured.  The results are shown below.  Does it seem that the amount of the chemical is significantly different between the two types of filter?

“Fast” Filters

“Slow” Filters

8.4%

9.0%

9.8%

10.2%

12.2%

9.6%

12.6%

4.4%

13.0%

7.0%

9.2%

9.2%

13.6%

...read more.

Conclusion

F-test:

Define the question: is there a difference in variance between the outputs of the two methods of analyzing the discharges?

Identify the appropriate model: comparison of two variances image13.pngimage13.png we use F-test

Because difference could be in either direction, image17.pngimage17.pngwe use two-tailed test.

Formulate the Null hypothesis and the Alternative hypothesis (H0 and H1)

H0: image18.pngimage18.png12 = image18.pngimage18.png22

H1:  image18.pngimage18.png12 image18.pngimage18.png22

We set the level of significance image10.pngimage10.png=5%, therefore, image10.pngimage10.png/2 =2.5%. image19.pngimage19.pngG =nG-1 =9-1=8, image19.pngimage19.pngL=nL-1=9image43.pngimage43.pngFrom Statistical Table 6(c), Fcrit =4.433

Because image21.png=image22.pngimage22.pngimage23.pngimage24.pngimage24.png2 / n         

image13.pngimage13.pngimage21.png1 = ∑image25.png / n  ; image21.png2 = ∑image26.png / n

Then we calculate the image23.pngA and image23.pngB.

Because image23.png = ∑x / n,

image13.pngimage13.pngimage23.pngA = ∑xA / n = 303/9=33.67

image23.pngB = ∑xB / n = 351/9=39

Therefore,

image21.png1 = ∑image25.png / n =149

image21.png2 = ∑image26.png / n=131.75

image17.pngimage17.png Fcalc= image21.png1/image21.png2=149/131.75=1.13

image13.pngimage13.png Fcalcimage02.pngimage02.pngFcritc

image13.pngimage13.png Accept H0

Therefore, there is no evidence of a significant difference in variance, which leads us to use the conventional t-test for difference between mean values.

The following table contains useful data from Excel Data Analysis:

F-Test Two-Sample for Variances

Variable 1

Variable 2

Mean

33.66667

39

Variance

149

131.75

Observations

9

9

df

8

8

F

1.13093

P(F<=f) one-tail

0.43305

F Critical one-tail

4.43326

T-test:

Define the question: is the mean of outputs using firm’s method lower than that of the EHA method?

Identify the appropriate model: because the population variance is not known and the mean is required, we use t-test.

As “underestimate” is a defined direction, we choose to use one-tailed test

Formulate the Null hypothesis and the Alternative hypothesis (H0 and H1)

H0: μ1 = μ2

H1: μ1image02.pngimage02.png μ2

We set the level of significance image10.pngimage10.png=5%, as it is a one-tailed test.

From Statistical Table 5, with v = 8 + 8 = 16, tcrit=1.746

As calculated previously, image23.pngA=33.67, image23.pngB=39, and pooled standard deviation of sample equals to  image28.pngimage28.png =image45.pngimage45.png=11.84

tcalc = image30.png

tcalc =image46.pngimage46.png = - 0.95

Therefore,  tcalc < tcritimage47.pngimage47.pngH0

There is no difference between the outputs of the two methods, and no evidence shows that the firm’s method underestimates the level of pollution.

The following table of data is abstracted from Excel Data Analysis.

t-Test: Two-Sample Assuming Equal Variances

Variable 1

Variable 2

Mean

33.66666667

39

Variance

149

131.75

Observations

9

9

Pooled Variance

140.375

Hypothesized Mean Difference

0

df

16

t Stat

-0.954904852

P(T<=t) one-tail

0.176915808

t Critical one-tail

1.745883669

P(T<=t) two-tail

0.353831617

t Critical two-tail

2.119905285

...read more.

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