21

Solutions

Solution 1 One-tail left test

It is of interest to ascertain whether there has been a decrease in strength, thus the test is the one-tail left alternative, whence:

H0 65: µ = H1 65: µ <

(i) The Z statistic is:

Z X X x= ( )/- s

N for the population is assumed large and the sample n = 9, therefore the standard error for the mean of the sample distribution is estimated from the population standard deviation σ:

σ σx n= Z X n= −( ) / ( )µ σ

Thus, for the sample:

Zsample = − = − =( ) / ( / ) /55 65 15 9 10 5 -2

It can be seen that the sample mean X occurs at a point Z = −2 standard deviations from the population mean.

(ii) Statisticians arbitrarily set an error limit on Z beyond which the sample is considered to be from another population distribution. For example, Ztables = + or − 1.64, repre-senting α = 5% in the tail area(s) or a 1 in 20 chance of error, is called 95% confidence level or significance.(Other confidence levels may be chosen appropriate to the risk assessment for the problem, for example, 99%, i.e. α = 1% in the tail allows the decision maker more discretion.)

(iii) In this example, the permitted mean concrete strength limit for quality control was set at a value giving Z = −1.64 (α = 5%), but the actual sample produces Z = −2 (i.e. 97.725% confidence) which is clearly further away from the population mean, i.e. Z = −2 > −1.64.

179

Modern Construction Management, Seventh Edition. Frank Harris and Ronald McCaffer with Francis Edum-Fotwe.© 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.

180 Self-learning exercises

Hence, the null hypothesis, Ho: µ > or = 65, is rejected and it is therefore concluded that based on a comparison of mean strengths only, the sample may have come from a different and weaker batch of concrete. But there remains a 2.275% (α = 2.275% in the tail) chance that the random sample of 9 cubes of concrete with an average strength of 55 N/mm2 may have come from concrete with a population µ = 65 N/mm2 or greater strength.

The action now would be to improve the quality-control measures and take a further sample, repeated as necessary until the null hypothesis is upheld and the mix complies with the required standard deviation set for the population (σ).

Solution 2 Two-tail t-test

In this case the concern is in over-loads or under-loads being more or less than the mean, thus, for the 95% confidence level, α = 2.5% is put in each tail (note: depending on what needs to be ascertained, other arrangements of the confidence interval (see Fig. 21.5) may be examined, for example, between 1% and 4% limits in the left and right tails respectively).

If the process is in control, then:

H0 1 1: .µ =

The alternative hypothesis is that the mean is not 1.1 m3, that is, it could be more or less than the mean.

H1 1 1: .µ not =

At the 95% confidence level with α = 2.5% in each tail, t tables and 8 degrees of freedom give:

(i) tX

S n=

–

/

µ

Since the data is drawn from a sample, the population standard deviation σ is estimated from the sample standard deviation S.

X = 1 075. S = 0 03.

t =−

= −1 075 1 1

0 03 92 5

. .

. /.

(ii) Degrees of freedom = n − 1 = 8 and with 2.5% in each tail, from ttables

t = + or – .2 31

(iii) t = >-2.5 – .2 31

The result negatively exceeds the t statistic and, thus, at the 95% confidence level, production is considered to be out of control.

Solutions 181

X1 = 30 σ1 = 2.5 n1 = 150X2 = 29.5 σ2 = 2.3 n2 = 200

Solution 3 Paired samples one-tail right t-test

It is of interest to ascertain whether there has been an increase in deliveries, thus the test is the one-tail right case, whence:

H0 2 1 0: µ µ− = H1 2 1 0: µ µ− >

(i) With α = 5% in the tail, from ttables and n − 1 = 4 degrees of freedom,

t tables = 2 13.

(ii) d is the difference between the before, and after, values:

Supplier d d2

A 1 1B 4 16C 1 1D 2 4E −1 1

Total 7 23

Mean of the sample differences = d = 7 / 5 = 1.4

t d S nd= ( ( ))/( / )2 1- -m m

The standard deviation of the population σ in the t-test statistic is estimated from the paired differences sample standard deviation (Sd).

Standard deviation of the paired differences sample is:

S d d n nd = − −{( ( ) / ) / ( )}Σ Σ2 2 1

Sd = − =( / ) / .23 7 5 4 1 8162

n = 5, thus:

t = − = =( . ) / . . / . .1 4 0 1 816 5 1 4 0 813 1 72

(iii) t = <1.72 2 13.There is no reason to suppose that delivery has improved, that is, the means of the two samples are not significantly different at the 95% confidence level.

Solution 4 Two-sample two-tail Z test

182 Self-learning exercises

The hypotheses are:

H0 1 2 0: µ µ− = H1 1 2: µ µnot =

i.e. two-tail test (α = 2.5%)

(i) Z tables = 1 96.

(ii) Z X X n n= {( ) ( )}/ ( / + / )1 2 1 2) 12

1 22

2- - - s sm m

Z = − − +

= + =

{( . ) } / ( . / . / )

. / . . .

30 29 5 0 2 5 150 2 3 200

0 5 0 0416 0 0265 1 30

2 2

44

(iii) Z = <1.304 1 96.

Thus, the null hypothesis is confirmed that the two firms are producing similarly, that is, the means are not significantly different at the 95% confidence level.

Solution 5 Two sample one-tail t-test

H0 1 2 0: µ µ− = H1 1 2 0: µ µ− >

A is µ1, B is µ2

i.e. one-tail test (α = 5%) with n – 2 degrees of freedom

= +n n1 2 2– = + =4 6 2 8–

(i) t ftables with 8 1 86deg . .=

(ii) t X X S n n= {( ) ( )}/ (1/ +1/ )1 2 1 2 1 2- - -m m

S n S n S n n= − + − + −{(( ) ( ) ) / ( )}1 12

2 22

1 21 1 2

S = × + × + −( ) ={( . . ) / } .3 0 14 5 0 18 4 6 2 0 406

t = − − + =(( . . ) ) / . ( / / ) .6 5 5 9 0 0 406 1 4 1 6 2 29(iii) t = >2.29 1 86.

There is reason to suppose that B is working quicker than A, that is, the means are significantly different at the 95% confidence level.

Solution 6 F test

One-tail test

S n12 0 18 1 5= − =. , degrees of freedom

S n22 0 14 1 3= − =. , degrees of freedom

Solutions 183

H0 1 2: σ σ= H1 1 2 0: σ σ− >

i.e. one-tail test (α = 5%)

(i) The σ22 estimate is the larger variance and thus the numerator.

F = × × = ={( . ) / } / {( . ) / } . / . .6 0 18 5 4 0 14 3 0 216 0 1867 1 157(ii) Thus, Ftable

53 is the comparator

Ftable5

3 9 01= .(iii) F = <1.157 9 01.

Therefore, the null hypothesis is proved and therefore σ1 can be considered equal to σ2, and the F test assumption is upheld.

Two-tail test

The two-tail test is used when there is no reason to suppose that one variance is greater than the other.

H1 1 2 0: σ σ− not equal to

The F statistics are therefore:

F = σ σ12

22/

or,

F = σ σ22

12/

α = 2.5% and use the larger variance (σ2) as the numerator.

Solution 7 Binomial distribution

n = 4 x = 3 p = 0 4. P( ) ( !/ ! !)( . ) ( . ) .3

3 14 3 1 0 4 0 6 0 1536= =

Solution 8 Two-sample two-tail proportion test

Two samples produce proportions x1/n1 and x2/n2 from two separate populations, for example, production processes, then:

184 Self-learning exercises

(i) The distribution of x1/n1 – x2/n2 is approximately normal if n1 and n2 are large(ii) Mean of the x1/n1 – x2/n2 distribution is p1 – p2

(iii) Variance of x1/n1 – x2/n2 distribution is {p1(1 − p1) / n1 + p2(1 − p2) / n2}

x n1 1 0 11/ .= x n2 2 0 06/ .= n n1 2150 125= =,

The problem is of the two-tail type, thus, an amount α = 2.5% is placed in each tail.The hypotheses are:

H p p0 1 2: = H p p1 1 2: not equal to

(i) Z tables or= + – .1 96(ii) The Z statistic is:

( / / ) ( ) = { (1 ) / + (1 ) / }1 1 2 2 1 2 1 1 1 2 2 2x n x n p p Z p p n p p n- - - -–

If H0 hypothesis is true, then it can be assumed that p = p1 = p2, and substituting in the Z statistic,

Z x n x n p p n n= − − − +{( / / ) } / {( ( )( / / )}1 1 2 2 1 20 1 1 1

The unknown value of p can be estimated as p* = {n1 x1/n1 + n1 x1/n1} / (n1 + n2) = (x1 + x2) / (n1 + n2)

p* ( . . ) / ( ) .= × + × + =0 11 150 0 06 125 150 125 0 0873

and substituting p*

Z = − − × × +{( . . ) } / ( . . ) ( / / )0 11 0 06 0 0 0873 0 9127 1 150 1 125 Z = × =0 05 0 282 0 121 1 465. / ( . . ) .

(iii) Z = <1.465 1 96.

that is, not significant at the 95% confidence level, and therefore the idle time proportions of the two samples are not presumed to be different.

Solution 9 Poisson distribution probability of success

n p= = =300 2 1000 1 500, / /

Solutions 185

µ = = =np 300 500 0 6/ .

(i) P e x e ex( )

. ./ ! . / ! .00 6 0 0 60 6 0 0 5488= = = =− − −µµ

(ii) P e( ). . / .1

0 6 10 6 1 0 3293= =−

The remaining portion under the curve is

P( ) ( . . )2 1 0 5488 0 3293or more = − + = 0.1219

The probability that more than one worker will be seriously injured during the year is 0.1219, that is, approximately a 12% chance.

Solution 10 Chi squared goodness-of-fit test

H0 60 30 9 1: : : :Actual payment days conform to the ratio H1: Payments are in some other ratio

(i) From χ2 tables with r − 1 = 3 degrees of freedom (r is the number of possible outcomes, i.e. categories) at the 95% confidence level,

χ2 7 81tables value is .

(ii) Expected numbers of payments are:eA = 200 × 0.6 = 120eB = 200 × 0.3 = 60eC = 200 × 0.09 = 18eD = 200 × 0.01 = 2Total 200

Category Observed(f) Expected(φ) f – φ (f – φ)2 (f – φ)2 /φ

A 130 120 10 100 0.8333B 50 60 −10 100 1.6667C 15 18 −3 9 0.5000D 5 2 3 9 4.5000

χ2 = 7.5

(iii) χ2 = 7.5 < 7.81, hence, since the null hypothesis is just about upheld, the payments appear to be under control.

186 Self-learning exercises

Solution 11 Correlation analysis

n = 5

r = − − −

−

{ . } / [ ][ . . ]

/ (

5 166 250 2 8 5 15100 250 5 1 86 2 8

830 700 130 0

2 2* * * *

000 9 3 7 84

130 13 000 1 46 0 94

)( . . )

/ ( * . ) .

−= =

Frequency

In the situation of frequencies (f) for test results or observations, the r formula is modified to:

r n xy xf yf n x f xf n y f yfx y x x y y= − − −{ }{ ( )( )} [ ( ) ][ ( ) ]Σ Σ Σ Σ Σ Σ Σ2 2 2 2

Hypothesis test

The sample is assumed to derive from a bivariate normal distribution, and the sampling distribution of r around the population correlation coefficient (ρ) will vary according to the value of ρ. When ρ is not = 0, the sample distribution is skewed around ρ and transformation techniques are then required to test the confidence interval for r. Readers are referred to statistical texts for detailed examples such as Spearman Rank Correlation Coefficient, or use a suitable computer package such as Minitab.

Solution 12 Linear univariate regression analysis

(i)

The regression equation is determined from the sample data:n = 5 Σx = 52 Σy = 115Σxy = 1100 Σx2 = 558 Σy2 = 3225

slope b n xy x y n x x= − −{ ( )( )}/{ ( ) }Σ Σ Σ Σ Σ2 2

slope * * *b = − − = −( ) / ( ) .5 1100 52 115 5 558 52 5 5812

intercept *a y n b x n= −( ) / ( ) /Σ Σ a = − − =115 5 5 58 52 5 81 032/ ( . / ) .* y a bx xe = + = + −81 032 5 58. ( . )*

(ii)

x x= =0 11£ per tonne y a bx xe = + = + −81 032 5 58. ( . )* ye = + − = − =81 032 5 58 11 81 032 61 38 19 94. ( . ) . . .* tonnes

For £11 per tonne spent, the predicted emissions are 19.94 tonnes

Solutions 187

(iii)

Since the regression equation is an estimate of the population data and the sample is small, the t statistic is used to determine the confidence interval for the estimated ye at the chosen x0 = 11 (for example, the large circle in Fig. 21.10), and can be found expressed in statistical texts as:

t S n x x x x n y yyx e* * 1+1 / + ( mean ) / { ( ) / }] =02 2 2[ - S - S -

where y is the value at the tail position (large disc).Syx in the t statistic is an estimate of the population standard deviation σyx, namely:

S y a y b xy nyx = − − −{( ) / ( )}Σ Σ Σ2 2

The n–2 denominator is the degrees of freedom based on the two regression coefficients, a and b.

Substituting for values:

Syx = − − − −

= − +

{ . ( . )} / ( )

{ .

3225 81 032 115 5 58 1100 5 2

3225 9318 68 631

* *

88 3 8 66} / .=

For two degrees of freedom n–2, from ttables at 95% confidence level with 2.5% in each tail:

tn− =2 3 18. y n x x x x n= + + + − −19 94 3 18 8 66 1 1 0

2 2 2. . . / [( ) / { ( ) / }]* * mean Σ Σ y = + + + − −19 94 3 18 1 1 5 11 52 5 558 52 52 2. . / [( / ) / { / }]*

y = + + = +19 94 3 18 1 2 0 36 17 2 19 94 3 51. . . . / . . .*

or, for −t

y = −19 94 3 51. .

The 95% confidence interval around the expected value ye = £19.94 + or − 3.51

(iv)

If y and x are independent, then the slope b of the regression equation would be zero, that is, horizontal, and if not, a relationship exists. To test this possibility, the hypotheses are stated as:

H b0 0: = H b1 0: not =

The standard deviation of the slope b is estimated as:

Std dev of b S x n x

y a y b xy n

yx= −

= − − −

{ ( / )( ) }

{( ) / ( )}

Σ Σ

Σ Σ Σ

2 2

2

1

2 {{ ( ) / }

. . . / . .

Σ Σx x n2 2

8 66 17 2 8 66 4 147 2 09

−

= = =

188 Self-learning exercises

t statistic is:

t S byx* = – 0 t = =– . – / . – .0 252 0 0 046 5 48

for n − 2 = 3 degrees of freedomi.e. two-tail test (α = 2.5%)

t tables = – .3 182 t = >-5.48 – .3 182

Thus, the null hypothesis is rejected and thus the greenhouse-gas emissions reduce as spend-ing increases when considered at the 95% level of confidence.

(v)

Coefficient of correlation

r n xy x y n x x n y y= − − −{ ( )( )} [ ( ) ][ ( ) ]Σ Σ Σ Σ Σ Σ Σ2 2 2 2

r =

= =5 1100 52 115 5 558 2704 5 3225 13225

480 499 0 96

* * * *– [ – ][ – ]

– / – .

The proportion of explained variability by the coefficient of determination r2 = 92%, that is, high

(vi)

y a bx xe = + = + −81 032 5 58. ( . )* x = =x0 14

ye = 81.032 − 5.58 *14 = 2.912 tonnes of emissions for £14/tonne of spendingAt 95% confidence interval, tn-2 = 3.18The confidence interval around the population expected average value is derived from the

modified t statistic:

t S n x x x x n y yyx p e* * 1 / + ( mean ) / { ( ) / }]02 2 2[ - S - S -=

where yp is the population expected average value thus

y y t S n x x x x np e yx= + +or * * mean– / [( – ) /{ – ( ) / }]1 02 2 2Σ Σ

Students are recommended to derive yp for a range of x0 values and plot the two regression boundary lines.

Solutions 189

Solution 13 Time series

The data for a 3-order model are set out in Excel as follows:

A B C D E

Month(t) XT XT-1 XT-2 XT-3

1 8 0002 9 000 8 0003 10 000 9 000 8 0004 10 000 10 000 9 000 8 0005 9 000 10 000 10 000 9 0006 8 000 9 000 10 000 10 0007 7 000 8 000 9 000 10 0008 7 000 7 000 8 000 9 0009 7 000 7 000 7 000 8 000

10 8 000 7 000 7 000 7 00011 9 000 8 000 7 000 7 00012 9 000 9 000 8 000 7 00013 8 000 9 000 9 000 8 00014 7 000 8 000 9 000 9 00015 6 000 7 000 8 000 9 00016 6 000 6 000 7 000 8 00017 7 000 6 000 6 000 7 00018 7 000 7 000 6 000 6 00019 8 000 7 000 7 000 6 00020 9 000 8 000 7 000 7 00021 10 000 9 000 8 000 7 00022 10 000 10 000 9 000 8 00023 9 000 10 000 10 000 9 00024 8 000 9 000 10 000 10 000

25 ? 8 000 9 000 10 000

(1) Using Excel > tools > data analysis > regression with residuals ticked >OK, start by evalu-ating the first order auto-regressive model which tests cells B2:B24 and C2:C24, respec-tively columns XT and XT-1 columns (n = 23) to produce the following output results.

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.773399R Square 0.598146Adjusted R Square 0.57901Standard Error 821.9371Observations 23

190 Self-learning exercises

(i) As explained in the previous regression analysis problem, to test whether the slope of the regression line is a good predictor of XT, the null hypothesis is set at:

H b0 1 0: ( . . )= i e is not a significant predictor H b1 1 0: ( . . )not i e is a significant predictor=

The t statistic is:

t S b* = 1 0–

Hence, using the standard error and coefficient (b1 constant) shown in the auto-regression results, the calculated t value for variable XT-1 is:

t = =0 773 0 138. / . 5.59

(ii) From ‘t test’ tables for n − k − 1 = 23 − 2 = 21 degrees of freedom, where n is the number of items of data in the period month 2 to 24, k is the number of b constants and −1 is the number of a constants (i.e. a, b1), and 95% confidence level, the two-tailed critical t-test value is:

t tables or= + – . .2 08

(iii) Thus, t = 5.59 > 2.08

The null hypothesis is therefore rejected; hence b1 is considered a good predictor.Indeed, the P value 1.51E-05 is significantly less than α = 2.5 % or 0.025 probability in the tail.However, the value of the correlation coefficient R = 0.598 suggests the line of best fit is

barely a good estimator of XT.

(2) To test whether the second order equation with the inclusion of independent variable XT-2 produces a better solution, cells B3:B24 and C3:D24 are used to generate the fol-lowing output results (n = 22):

ANOVA

df SS MS F Significance F

Regression k = 1 21 117 156 21 117 156 31.25779 1.51E-05Residual n−k−1 = 21 14 187 192 675 580.6Total n−1 = 22 35 304 348

Coefficients Standard Error

t Stat P-value Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept a 1852.217 1143.633 1.61959 0.120242 −526.1 4230.533 −526.1 4230.533X Variable 1 0.773399 0.138333 5.590867 1.51E-05 0.485721 1.061077 0.485721 1.061077

Solutions 191

(i) For n − k − 1 = 22 − 2 − 1 = 19 degrees of freedom (a, b1, b2) the two-tailed critical t-test value is:

t tables or= + – . .2 093

(ii) The calculated t value for constant b2 is:

t = -6.13529

(iii) Thus, t = −6.1353 > −2.093

H0: b2 = 0 is rejected and hence the second-order model containing a, b1 and b2 is considered a good predictor. Indeed, the respective P coefficients 0.000246, 1.65E-09, 6.75E-06 are sig-nificant predictors.

Also, the correlation coefficient R = 0.87 suggests the line of best fit is an improved estima-tor of XT, that is, this model is an improvement over the previous first-order model.

(3) To test inclusion of the third order independent variable XT-3, Cells B4:B24 and C4:E24 (n = 21) are used to generate the following output results:

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.933694R Square 0.871785Adjusted R Square 0.858288Standard Error 483.1415Observations 22

ANOVA

df SS MS F Significance F

Regression k = 2 30155821 15077911 64.59406 3.35E-09Residual n − k −1 = 19 4435088 233425.7Total n − 1 = 21 34590909

Coefficients Standard Error

t Stat P-value Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept 3209.357 713.3052 4.499276 0.000246 1716.391 4702.322 1716.391 4702.322X Variable 1 1.412865 0.131581 10.73762 1.65E-09 1.137463 1.688268 1.137463 1.688268X Variable 2 −0.8152 0.132871 −6.13529 6.75E-06 −1.09331 −0.5371 −1.09331 −0.5371

192 Self-learning exercises

(i) For n = 21 − 4 = 17 degrees of freedom (a, b1, b2, b3 ), the two-tailed critical t-test value is:

t tables or= + – .2 11

(ii) The calculated t value for variable XT-3 is:

t = -0.85748

(iii) Thus, t = −0.85748 < −2.11, hence:

H b0 3 0: = is upheld

Thus, the inclusion of b3 does not suggest improved prediction.

Conclusion

The second-order solution apparently produces the line of best fit represented in the following time-series equation:

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.93543R Square 0.875029Adjusted R Square 0.852975Standard Error 477.0105Observations 21

ANOVA

df SS MS F Significance F

Regression 3 27084217 9028072 39.67702 6.8E-08Residual 17 3868164 227539.1Total 20 30952381

Coefficients Standard Error

t Stat P-value Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept 3895.833 1042.656 3.736452 0.001643 1696.019 6095.648 1696.019 6095.648X Variable 1 1.208822 0.228122 5.299024 5.89E-05 0.727527 1.690117 0.727527 1.690117X Variable 2 −0.49772 0.351316 −1.41673 0.174626 −1.23894 0.243492 −1.23894 0.243492X Variable 3 −0.20361 0.237455 −0.85748 0.403106 −0.7046 0.297373 −0.7046 0.297373

Solutions 193

X a b X b XT T T= + +− −1 1 2 2

X X XT T T= + −− −3209 357 1 412865 0 81521 2. . ( ) . ( ))

(4) Thus, predicted values for months 24, 25 and 26 are respectively:

X24 3209 357 1 412865 9000 0 8152 10 000 7772= + × − × =. . ( ) . ( ) (compare acctual of 8000)X25 3209 357 1 412865 8000 0 8152 9000 7175= + × − × =. . ( ) . ( )X26 3209 1 412865 7175 0 8152 8000 6824= + × − × =. ( ) . ( )

Alternative first-order model

X XT= + ×−1852 217 0 7733991. ( . )

The predicted values for months 24,25 and 26 are respectively:

X24 1852 217 0 773399 9000 8012 8000= + × =. . ( )compare actual ofX25 1852 217 0 773399 8000 8044= + × =. .X26 1852 217 0 773399 8044 8073= + × =. .

The final interpretation between the models is largely subjective and depends upon the knowledge and experience of the analyst with the conditions pertaining to the particular kind of data.

Solution 14 Analysis of variance (ANOVA)

The hypotheses are:

H0 1 2 3: µ µ µ= = H i1: µ are not equal

As illustrated in Problem 4, sample means may be compared by the t-test, however, the data may also be evaluated using analysis of variance by assuming that the three samples (groups) represent three populations, which possess a constant variance σ2 as follows:

(i) ‘Within’ groups

An estimate of the population variance (σ2) may be estimated by the pooled variance of the samples (Sp

2) calculated from the sum of the squares of the ‘within’ groups (i.e. variation within the whole data), given by:

‘ ’ ( – ) /( – ) /( – ) .Within S x x n ap ij i2 2 1600 15 3 133 33= = =ΣΣ

Note: (n–a) degrees of freedom

(ii) ‘Between’ groups

194 Self-learning exercises

If it is assumed that µ1 = µ2 = µ3 and so form part of a single population, the population variance (σ2) may also be estimated from the sum of the squares of the ‘between’ groups (i.e. variation between the groups of data) given by:

‘ ’ ( – ) /( – ) /Between S n x x ad i i2 2 1 4000 2 2000= = =Σ

Note: (a – 1) degrees of freedom treats the three groups as a single sample

Application of the F test

If the null hypothesis were not true, the implication is that the ‘between’ means for the groups (i.e. subcontractor samples) are not equal and therefore would differ significantly from the overall data mean x. Conversely, the ‘within’ groups’ variance estimator would obviously not be affected.

Thus, the more the F ratio (F = ‘Between’ S2 /‘Within’ Sp2) exceeds unity, the greater the

variation between the groups and so the population means µi.

Result

(i) F S Sp= = =Between Within2 2 2000 133 33 15/ / . .(ii) There are respectively (3 − 1) and (15 − 3) degrees of freedom, hence from F tables for

the one-tail case (α = 5% or 0.05 probability) gives a critical value of F212 tables = 3.88.

(iii) Thus, F = 15 > 3.88 and the null hypothesis is rejected, inferring that the three samples are NOT from the same population and therefore the three means are unequal, that is, the performance of at least one of the subcontractors is different from the others.

Summary: using Microsoft Excel tools data, the Anova: Single Factor model produces cor-responding results as follows:

ANOVA

Source of Variation SS df MSq F P-value F crit

Between Groups 4000 2 2000 15 0.000544 3.88529Within Groups 1600 12 133.3333Total 5600 14

Solution 15 Two-way ANOVA (two factors)

The solution may be derived in similar manner as previously but with the inclusion of the additional factor B. The analysis now requires calculation of the ‘between groups’ variance for B variable also. Using Microsoft Excel tools data analysis program for two-way Anova without replication model:

Solutions 195

Subcontractor

A(x1j) A(x2j) A(x3j)

B(xi2) 30 40 70 140(xi1 = 46.67)B(xi2) 70 40 80 190(xi2 = 63.33)B(xi3) 40 30 70 140(xi3 = 46.67)B(xi4) 60 20 60 140(xi4 = 46.67)B(xi5) 50 20 70 140(xi5 = 46.67)Total 250(x1j = 50) 150(x2j = 30) 350(x3j = 70)

Anova: Two-factor without replication

SUMMARY Count Sum Average Variance

Row 1 3 140 46.66667 433.3333Row 2 3 190 63.33333 433.3333Row 3 3 140 46.66667 433.3333Row 4 3 140 46.66667 533.3333Row 5 3 140 46.66667 633.3333Column 1 5 250 50 250Column 2 5 150 30 100Column 3 5 350 70 50

ANOVA

Source of Variation SS df MSq F P-value F crit

Rows(B) 666.6667 4 166.6667 1.428571 0.308771 3.837854Columns(A) 4000 2 2000 17.14286 0.001281 4.458968Error 933.3333 8 116.6667Total 5600 14

Degress of freedom:

Rows ( )B b= − = − =1 5 1 4 Columns ( )A a= − = − =1 3 1 2 Error = − × − = × =( ) ( )a b1 1 4 2 8

(1) A variable

(i) F = =2000 116 67 17 14/ . .

(ii) From Ftables for the one-tail test (α = 5% or 0.05 probability), the critical value is:

F 28 4 46tables = .

(iii) F = 17.14 > 4.46, thus the null hypothesis is rejected and it can be inferred that the three samples are not from the same population. The three means are not equal and therefore the performance of the three contractors varies.

196 Self-learning exercises

(2) B variable(i) F = =166 67 116 67 1 43. / . .(ii) From Ftables for the one-tail test (α = 5% or 0.05 probability), the critical

value is:

F 48 3 84tables = .

(iii) F = 1.43 < 3.84, thus the null hypothesis is upheld and it can be inferred that the means of five different sources of materials are from the same population and therefore their variability is not significant.

It can therefore be inferred that most of the variability in the data is due to the performance of the subcontractors and not much affected by the sources of materials used.

Note:The alternative method of ‘Pairwise’ tests may be undertaken for pairs of groups in turn using the t-test.

Solution 16 Two-way ANOVA (interaction of the variables)

Null hypothesis for each variable:

H0 1 2 3: µ µ µ= = H i1: µ are not all equal

Anova: Two-factor with replication

SUMMARY D1 D2 D3 Total

C1Count 2 2 2 6Sum 145 135 120 400Average 72.5 67.5 60 66.66667Variance 12.5 12.5 0 36.66667

C2Count 2 2 2 6Sum 155 140 125 420Average 77.5 70 62.5 70Variance 12.5 0 12.5 50

TotalCount 4 4 4Sum 300 275 245Average 75 68.75 61.25Variance 16.66667 6.25 6.25

Solutions 197

ANOVA

Source of Variation SS df MS F P-value F crit

Sample(C) 33.33333 1 33.33333 4 0.092426 5.987374Columns(D) 379.1667 2 189.5833 22.75 0.001581 5.143249Interaction(C–D) 4.166667 2 2.083333 0.25 0.786527 5.143249Within 50 6 8.333333

Total 466.6667 11

(i) Variable C, construction workers

The calculated F = 4The critical value F1

6tables = 5.99

F = <4 5 987.

The null hypothesis is accepted and thus there is not considered to be a significant difference between performance results for before, and after, training.

(ii) Variable D, incentive scheme

The calculated F = 22.75The critical value F1

6tables = 5.143

F = >22.75 5 143.

The null hypothesis is rejected and thus there is considered to be a significant difference between performance results of the construction workers operating under the three incentive schemes.

(iii) Variable C–D, interaction between the two variables

The calculated F = 0.25The critical value F1

6tables = 5.143

F = <0.25 5 143.

The null hypothesis is accepted and thus there is not considered to be a significant effect on performance of the construction workers by the interaction of incentive scheme and training.

Solution 17 Bayes’ theorem or law

To compute this value P(A|B), the following need to be determined:

198 Self-learning exercises

• P(A), or the probability that the company fails regardless of any other information. Since the selected company is random and the fraction of failed firms among the total is 5%, this probability equals 0.05.

• P(A′), or the probability that the company is solvent regardless of any other information (A′) is the complementary event to (A). This is 95%, or 0.95.

• P(B|A), or the probability that company is assessed as insolvent(B) and actually fails(A). This is analysed from previous data to be 0.85.

• P(B|A′), or the probability that company is assessed as insolvent(B) but was solvent(A′). This is analysed from previous data to be 0.1.

P(B), or the probability that a (randomly selected) company is assessed as insolvent (B) regardless of any other information.

P B P B A P A P B A P A( ) ( ) ( ) ( ) ( ), . . . . .= + ′ ′ × + × =| | this is 0 85 0 05 0 1 0 95 0 13375

Thus, the probability that a company fails (A), given that it was assessed as insolvent (B), can be computed by substituting these values in the formula:

P A B P B A P A P B( ) ( ) ( ) / ( ) . . / . .| |= × = × =0 85 0 05 0 1375 0 309

That is: only 31% of companies fail from those assessed as insolvent.