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Question
A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities. Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel:ANOVA: Single Factor
SUMMARY
| Groups | Count | Sum | Average | Variance |
| City 1 | 30 | 7897.179 | 236.2393 | 3334.11 |
| City 2 | 30 | 10322.1 | 344.0701 | 2201.818 |
| City 3 | 30 | 6045.102 | 201.5034 | 2215.919 |
ANOVA
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 306701.8 | 2 | 153350.9 | 59.3475 | 5.54E-17 | 3.101292 |
| Within Groups | 224803.5 | 87 | 2583.949 | |||
| Total | 531505.4 | 89 |
Based on the information provided, should we conclude that the three populations (cities) have equal mean spending during August? Test at the 0.05 level of significance.
Answer
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Q:
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A) Because t = -2.69 is less than -2.3456, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
B) Because t = -2.69 is less than -2.3456, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
C) Because t = -1.69 is less than -0.8712, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
D) Because t = -1.69 is less than -0.8712, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
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A) p-value = 0.0872 > 0.025; therefore do not reject H0
B) p-value = 0.0422 > 0.005; therefore do not reject H0
C) p-value = 0.0314 < 0.105; therefore reject H0
D) p-value = 0.0121< 0. 0805; therefore reject H0
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