Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 11.CR, Problem 4CR
To determine

Hypothesis Testing using Analysis of Variance i.e. ANOVA.

Expert Solution & Answer
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Answer to Problem 4CR

Solution:

The null hypothesis is rejected and it is concluded that at least one of the tables is not generating the same mean revenue, at the 0.05 level of significance and the required ANOVA table is,

SS df MS F
Treatments (T) 166534735 3 555111578.33 67.05
Error (E) 13245702 16 827856.38
Total 179780437 19

Explanation of Solution

Given:

The amount of revenue (in whole dollars) received during one particular three-hour period is recorded at each table each night for five nights.

Revenue (in Dollars)
Table 1 Table 2 Table 3 Table 4
8534 9821 10, 542 7367
7845 8997 9982 8021
8901 7905 8934 9034
9371 8923 9344 6703
7782 6675 8745 8432

Formula Used:

Grand Mean is the weighted mean of the k sample means, one from each of the k populations, equivalent to the mean of all the sample data combined, given by

x¯¯=i=1knix¯ii=1kni

Sum of Squares among Treatments (SST) is the measures the variation between the sample means and the grand mean, given by,

SST=i=1kni(x¯ix¯¯)2

Sum of Squares for Error (SSE) is the measures the variation in the sample data resulting from the variability within each sample,

SSE=j=1n1(x1jx¯1)2+j=1n2(x2jx¯2)2+...+j=1nk(xkjx¯k)2

Total Variation, it is the sum of the squared deviations from the grand mean for all of the data values in each sample, given by

Total Variation=j=1n1(x1jx¯ ¯)2+j=1n2(x2jx¯ ¯)2+...+j=1nk(xkjx¯ ¯)2=SST + SSE

Mean Square for Treatments (MST) found by dividing the sum of squares among treatments by its degrees of freedom, given by

MST=SSTDFTwith DFT=k1

Mean Square for Error (MSE) found by dividing the sum of squares for error by its degrees of freedom, given by

MSE=SSEDFEwith DFE=nTk

Test Statistic for an ANOVA Test: Used when independent, simple random samples are taken from populations with variances that are unknown and assumed to be equal, where all of the k population distributions are approximately normal, given by,

F=MSTMSEwith df1=DFT=k1 and df2=DFE=nTk 

Rejection Region for ANOVA Tests: Reject the null hypothesis, H0 when FFα.

ANOVA Table is formed as,

SS df MS F
Treatments (T) SST DFT SSTDFT MSTMSE
Error (E) SSE DFE SSEDFE
Total SST+SSE DFT+DFE

Calculation:

The grand Mean is,

x¯¯=i=1knix¯ii=1kni

From the table, substitute the following values as,

x¯¯=((5×42433)+(5×42321)+(5×47547)+(5×39557))(5+5+5+5)=42964.5

Sum of Squares among Treatments (SST) is given as,

SST=i=1kni(x¯ix¯¯)2

From the table, substitute the following values as,

SST=[5×(4243342964.5)2+...+5×(3955742964.5)2]=166534735

Sum of Squares for Error (SSE) is given as,

SSE=j=1n1(x1jx¯1)2+j=1n2(x2jx¯2)2+...+j=1nk(xkjx¯k)2

Substitute the following values as,

SSE=[(85348486.6)2+...+(77828486.6)2]+...+[(73677911.4)2+...+(84327911.4)2]=13245702

The total Variation is given as,

Total Variation=SST + SSE

Substitute the following values as,

Total Variation=SST + SSE=166534735+13245702=179780437

The mean Square for Treatments (MST) is,

MST =1665347353=55511578.33

with degree of freedom as 41 or 3.

The mean Square for Error (MSE) is,

MSE =1324570216=827856.38

with degree of freedom as 204 or 16.

The test Statistic for an ANOVA Test:

F=55511578.33827856.38=67.05 with df1=DFT=3 and df2=DFE=16 

with degree of freedoms as 3 and 16.

Let Table 1, Table 2, and Table 3 and Table 4 be the population 1, 2, 3 and 4 respectively and let μ1, μ2, μ3 and μ4 be the mean revenue for tables 1, 2, 3 and 4 respectively.

Let the null and alternative hypothesis for this test is

H0: μ1= μ2= μ3=μ4

H1: At least one mean revenue is different from others

Level of significance is 0.05, thus α = 0.05. The tabulated F-value for (3,16) degrees of freedom is 3.24. Since F>Fα i.e. 67.05>3.24, null hypothesis is rejected. Thus, it is concluded that at least one of the tables is not generating the same mean revenue.

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Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
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