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

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
SS | df | MS | F | |
Treatments (T) | 3 | 555111578.33 | 67.05 | |
Error (E) | 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
Sum of Squares among Treatments (SST) is the measures the variation between the sample means and the grand mean, given by,
Sum of Squares for Error (SSE) is the measures the variation in the sample data resulting from the variability within each sample,
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
Mean Square for Treatments (MST) found by dividing the sum of squares among treatments by its degrees of freedom, given by
Mean Square for Error (MSE) found by dividing the sum of squares for error by its degrees of freedom, given by
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
Rejection Region for ANOVA Tests: Reject the null hypothesis,
ANOVA Table is formed as,
SS | df | MS | F | |
Treatments (T) | SST | DFT | ||
Error (E) | SSE | DFE | ||
Total | SST+SSE | DFT+DFE |
Calculation:
The grand Mean is,
From the table, substitute the following values as,
Sum of Squares among Treatments (SST) is given as,
From the table, substitute the following values as,
Sum of Squares for Error (SSE) is given as,
Substitute the following values as,
The total Variation is given as,
Substitute the following values as,
The mean Square for Treatments (MST) is,
with degree of freedom as
The mean Square for Error (MSE) is,
with degree of freedom as
The test Statistic for an ANOVA Test:
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
Let the null and alternative hypothesis for this test is
H1: At least one mean revenue is different from others
Level of significance is 0.05, thus
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Chapter 11 Solutions
Beginning Statistics, 2nd Edition
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