EBK STATISTICS FOR MANAGEMENT AND ECONO
EBK STATISTICS FOR MANAGEMENT AND ECONO
10th Edition
ISBN: 9780100546462
Author: KELLER
Publisher: YUZU
Question
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Chapter 14.2, Problem 49E

(a)

To determine

Compute the differences that exist between the burning times of the four brands of flares.

(a)

Expert Solution
Check Mark

Explanation of Solution

The null and alternative hypothesis can be written as follows:

H0:μ1=μ2=μ3=μ4H1:At least two means differ

x¯=10(61.6)+10(57.3)+10(61.8)+10(51.8)10+10+10+10=616+573+618+51840=2,32540=58.125

SST can be calculated as follows:

SST=nj(xj¯x¯)=10(61.658.13)2+10(57.358.13)2+10(61.858.13)+10(51.858.13)2=662.7

SSE can be calculated as follows:

SSE=(nj1)sj2=(101)80.49+(101)70.46+(101)22.18+(101)75.29=2,236

Thus, the value of SSE is 2,236.

Table-1 shows the ANOVA Table as follows:

Table-2
SourceDegree of freedomSum of squareMean squareF
Treatmentsk-1 = 3SST = 662.7220.93.56
Errorn- k = 36SSE = 2,23662.11 

Thus, the estimated F-value is 3.56 and p-value is 0.0236. Here, there is an adequate evidence to accomplish that differences exist between the flares with respect to burning times.

(b)

To determine

Compute theLSD with Bonferroni adjustment.

(b)

Expert Solution
Check Mark

Explanation of Solution

t-value can be written as follows:

C=4(3)2=6αE=0.05α=αECt(α2,nk)=2.794

LSD can be calculated as follows:

LSD=t(α2,nk)MSE(1ni+1nj)=2.79462.11(110+110)=9.45

Table-2 shows the means and difference as follows:

Table-2
 
TreatmentMeansDifference
i = 1, j =261.657.34.3
i = 1, j =361.661.8-2
i = 1, j =461.651.89.8
i =2, j =357.361.8-4.5
i =2, j =457.351.85.5
i = 3, j =461.851.810
 
    

Therefore, from the above calculation, the mean of flares C and D differs.

(c)

To determine

Compute the Tukey’s method.

(c)

Expert Solution
Check Mark

Explanation of Solution

qα(k,v)=q0.05(4,136)3.79ϖ=qα(k,v)MSEng=3.7962.1110=9.45

Table-2 shows the means and difference as follows:

Table-2
 
TreatmentMeansDifference
i = 1, j =261.657.34.3
i = 1, j =361.661.8-2
i = 1, j =461.651.89.8
i =2, j =357.361.8-4.5
i =2, j =457.351.85.5
i = 3, j =461.851.810

Therefore, from the above calculation, the mean of flares A and D, C and D differs.

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Chapter 14 Solutions

EBK STATISTICS FOR MANAGEMENT AND ECONO

Ch. 14.1 - Prob. 11ECh. 14.1 - Prob. 12ECh. 14.1 - Prob. 13ECh. 14.1 - Prob. 14ECh. 14.1 - Prob. 15ECh. 14.1 - Prob. 16ECh. 14.1 - Prob. 17ECh. 14.1 - Prob. 18ECh. 14.1 - Prob. 19ECh. 14.1 - Prob. 20ECh. 14.1 - Prob. 21ECh. 14.1 - Prob. 22ECh. 14.1 - Prob. 23ECh. 14.1 - Prob. 24ECh. 14.1 - Prob. 25ECh. 14.1 - Prob. 26ECh. 14.1 - Prob. 27ECh. 14.1 - Prob. 28ECh. 14.1 - Prob. 29ECh. 14.1 - Prob. 30ECh. 14.1 - Prob. 31ECh. 14.1 - Prob. 32ECh. 14.1 - Prob. 33ECh. 14.1 - Prob. 34ECh. 14.1 - Prob. 35ECh. 14.1 - Prob. 36ECh. 14.1 - Prob. 37ECh. 14.1 - Prob. 38ECh. 14.1 - Prob. 39ECh. 14.1 - Prob. 40ECh. 14.1 - Prob. 41ECh. 14.1 - Prob. 42ECh. 14.2 - Prob. 43ECh. 14.2 - Prob. 44ECh. 14.2 - Prob. 45ECh. 14.2 - Prob. 46ECh. 14.2 - Prob. 47ECh. 14.2 - Prob. 48ECh. 14.2 - Prob. 49ECh. 14.2 - Prob. 50ECh. 14.2 - Prob. 51ECh. 14.2 - Prob. 52ECh. 14.2 - Prob. 53ECh. 14.2 - Prob. 54ECh. 14.2 - Prob. 55ECh. 14.2 - Prob. 56ECh. 14.2 - Prob. 57ECh. 14.2 - Prob. 58ECh. 14.2 - Prob. 59ECh. 14.2 - Prob. 60ECh. 14.2 - Prob. 61ECh. 14.2 - Prob. 62ECh. 14.2 - Prob. 63ECh. 14.2 - Prob. 64ECh. 14.2 - Prob. 65ECh. 14.2 - Prob. 66ECh. 14.2 - Prob. 67ECh. 14.2 - Prob. 68ECh. 14.4 - Prob. 69ECh. 14.4 - Prob. 70ECh. 14.4 - Prob. 71ECh. 14.4 - Prob. 72ECh. 14.4 - Prob. 73ECh. 14.4 - Prob. 74ECh. 14.4 - Prob. 75ECh. 14.4 - Prob. 76ECh. 14.4 - Prob. 77ECh. 14.4 - Prob. 78ECh. 14.4 - Prob. 79ECh. 14.4 - Prob. 80ECh. 14.4 - Prob. 81ECh. 14.4 - Prob. 82ECh. 14.4 - Prob. 83ECh. 14.4 - Prob. 86ECh. 14.4 - Prob. 87ECh. 14.4 - Prob. 88ECh. 14.4 - Prob. 89ECh. 14.4 - Prob. 90ECh. 14.4 - Prob. 91ECh. 14.4 - Prob. 92ECh. 14.4 - Prob. 93ECh. 14.4 - Prob. 94ECh. 14.4 - Prob. 95ECh. 14.4 - Prob. 96ECh. 14.4 - Prob. 97ECh. 14.4 - Prob. 98ECh. 14.4 - Prob. 99ECh. 14.5 - Prob. 100ECh. 14.5 - Prob. 101ECh. 14.5 - Prob. 102ECh. 14.5 - Prob. 103ECh. 14.5 - Prob. 104ECh. 14.5 - Prob. 105ECh. 14.5 - Prob. 106ECh. 14.5 - Prob. 107ECh. 14.5 - Prob. 108ECh. 14.5 - Prob. 109ECh. 14.6 - Prob. 110ECh. 14.6 - Prob. 111ECh. 14.6 - Prob. 112ECh. 14.A - Prob. 1ECh. 14.A - Prob. 2ECh. 14.A - Prob. 3ECh. 14.A - Prob. 4ECh. 14.A - Prob. 5ECh. 14.A - Prob. 6ECh. 14.A - Prob. 7ECh. 14.A - Prob. 8ECh. 14.A - Prob. 9ECh. 14.A - Prob. 10ECh. 14.A - Prob. 11ECh. 14.A - Prob. 12ECh. 14.A - Prob. 13ECh. 14.A - Prob. 14ECh. 14.A - Prob. 15ECh. 14.A - Prob. 16ECh. 14.A - Prob. 17ECh. 14.A - Prob. 18ECh. 14 - Prob. 113CECh. 14 - Prob. 114CECh. 14 - Prob. 115CECh. 14 - Prob. 116CECh. 14 - Prob. 117CECh. 14 - Prob. 118CECh. 14 - Prob. 119CECh. 14 - Prob. 120CECh. 14 - Prob. 121CECh. 14 - Prob. 122CE
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