EBK STATISTICS FOR MANAGEMENT AND ECONO
EBK STATISTICS FOR MANAGEMENT AND ECONO
10th Edition
ISBN: 8220102958609
Author: KELLER
Publisher: YUZU
Question
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Chapter 14.1, Problem 18E

(a)

To determine

Derive the data.

(a)

Expert Solution
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Explanation of Solution

The null and alternative hypothesis can be written as follows:

H0:μ1=μ2=μ3H1:At least two means differ

Age Rejection region can be written as follows:

F>Fα,k-1,n-k=F0.05,2,1472.61

SST can be calculated as follows:

SST=nj(xj¯x¯)=(63(31.336.23)2+81(34.4236.23)2+40(37.3836.23)2+111(39.9336.23)2)=3,366

SSE can be calculated as follows:

SSE=(nj1)sj2=(631)28.34+(811)23.2+(401)31.16+(1111)72.03=12,752

Thus, the value of SSE is 12,752

Table -1 shows the ANOVA Table as follows:

Table -1

SourceDegree of freedomSum of squareMean squareF
Treatmentsk-1 = 36SST = 3,3661,12225.60
Errorn- k = 291SSE = 12,75243.82 

Thus, the estimated F- value is 25.6 and p-value is 0. Here, there is an adequate evidence to infer that the age of the four group of cereal buyers differ.

(b)

To determine

Explain the required conditions for the test conducted.

(b)

Expert Solution
Check Mark

Explanation of Solution

The null and alternative hypothesis can be written as follows:

H0:μ1=μ2=μ3H1:At least two means differ

Income rejection region can be written as follows:

F>Fα,k-1,n-k=F0.05,3,2912.61

SST can be calculated as follows:

SST=nj(xj¯x¯)=(63(37.2239.97)2+81(38.9139.97)2+40(41.4839.97)2+111(41.7539.97)2)=1,008

SSE can be calculated as follows:

SSE=(nj1)sj2=(631)39.8+(811)40.85+(401)61.38+(1111)46.59=13,256

Thus, the value of SSE is 13,256

Table -1 shows the ANOVA Table as follows:

Table -2
SourceDegree of freedomSum of squareMean squareF
Treatmentsk-1 = 3SST = 1,0083367.37
Errorn- k = 291SSE = 13,25645.55 

Thus, the estimated F- value is 7.37 and p-value is 0.0001. Here, there is an adequate evidence to conclude that theincome differ between he four groups of cereal buyers.

(c)

To determine

Explain the required conditions for the test conducted.

(c)

Expert Solution
Check Mark

Explanation of Solution

The null and alternative hypothesis can be written as follows:

H0:μ1=μ2=μ3H1:At least two means differ

Education rejection region can be written as follows:

F>Fα,k-1,n-k=F0.05,3,2912.61

SST can be calculated as follows:

SST=nj(xj¯x¯)=(63(11.7511.98)2+81(12.4111.98)2+40(11.7311.98)2+111(11.8911.98)2)=21.71

SSE can be calculated as follows:

SSE=(nj1)sj2=(631)3.93+(811)3.39+(401)4.26+(1111)4.3=1,154

Thus, the value of SSE is 1,154

Table -1 shows the ANOVA Table as follows:

Table -2
SourceDegree of freedomSum of squareMean squareF
Treatmentsk-1 = 3SST = 21.717.241.82
Errorn- k = 291SSE = 1,1543.697 

Thus, the estimated F- value is 1.82 and p-value is 0.1428. Here, there is not an adequate evidence to conclude that the educationdiffer between he four groups of cereal buyers.

(d)

To determine

Summarize the findings.

(d)

Expert Solution
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Explanation of Solution

Using F – tests and descriptive statistics, the mean age and the mean household income are in ascending order. Example, sweet smacks buyers are younger and earn less than the buyers of the other three cereals. Thus, individual ‘C’ purchases are older and earn the most.

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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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