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Question
Suppose that you have just performed an experiment using the randomized complete block analysis of variance design and the output from Excel is shown as follows. The primary null hypothesis to be tested involves whether the means of the three groups are equal.ANOVA: Two-Factor Without Replication
| Summary | Count | Sum | Average | Variance |
| Block 1 | 3 | 450 | 150 | 468.9 |
| Block 2 | 3 | 210 | 70 | 72.9 |
| Block 3 | 3 | 610 | 203.3333 | 633.3333 |
| Block 4 | 3 | 700 | 233.3333 | 3333.333 |
| Group 1 | 4 | 560 | 140.0 | 5066.667 |
| Group 2 | 4 | 810 | 202.5 | 8425 |
| Group 3 | 4 | 600 | 150 | 3266.667 |
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | p-value | F-crit |
| Rows | 46158.33 | 3 | 15386.11 | 22.4251 | 0.001163 | 4.757055 |
| Columns | 9016.667 | 2 | 4508.333 | 6.57085 | 0.030797 | 5.143249 |
| Error | 4116.667 | 6 | 686.1111 | |||
| Total | 5929.67 | 11 |
Using a significance level of 0.05, what conclusions should be made about the three groups? Which groups have different means? Discuss.
Answer
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A) Since z = -0.423 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
B) Since z = -1.0614 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
C) z = -1.0231 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
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Assuming that a significance level of 0.05 is used, what conclusion should the governor reach based on these sample data?
A) Since z = 1.1594 < 1.645, do not reject the null hypothesis.
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The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.
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B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.
C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.
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B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error.
C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error.
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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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