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Question
A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different persons' income tax returns are done by each of the three software packages and the time is recorded for each. The computer results are shown below.| SUMMARY | Count | Sum | Average | Variance |
| 1 | 3 | 9 | 3 | 1 |
| 2 | 3 | 30 | 10 | 1 |
| 3 | 3 | 12 | 4 | 0 |
| 4 | 3 | 6.5 | 2.166667 | 0.583333 |
| 5 | 3 | 25 | 8.333333 | 2.333333 |
| 6 | 3 | 7 | 2.333333 | 1.083333 |
| 7 | 3 | 10 | 3.333333 | 0.333333 |
| 8 | 3 | 18 | 6 | 1 |
| 9 | 3 | 33.5 | 11.16667 | 0.583333 |
| 10 | 3 | 4.5 | 1.5 | 0.25 |
| Software A | 10 | 47.5 | 4.75 | 12.95833 |
| Software B | 10 | 47.5 | 4.75 | 10.79167 |
| Software C | 10 | 60.5 | 6.05 | 13.46944 |
ANOVA
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Rows | 329.9083 | 9 | 36.65648 | 130.227 | 1.6E-14 | 2.456281 |
| Columns | 11.26667 | 2 | 5.633333 | 20.01316 | 2.66E-05 | 3.554557 |
| Error | 5.066667 | 18 | 0.281481 | |||
| Total | 346.2417 | 29 |
Based on these results and using a 0.05 level of significance which is correct regarding the primary hypothesis?
A) The three software packages are not all the same because p-value = 1.6E-14 is less than 0.05.
B) The three software packages are all the same because p-value = 1.6 is greater than 0.05.
C) The three software packages are not all the same because p-value = 2.66E-5 is less than 0.05.
D) The three software packages are all the same because p-value = 2.66 is greater than 0.05.
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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Calculate the probability of committing a Type II error if the true population mean is 230 gallons. Assume that the population standard deviation is known to be 40 gallons.
A) 0.0331
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C) 0.0537
D) 0.1412
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You are given the following null and alternative hypotheses: If the true population mean is 4,345, determine the value of beta. Assume the population standard deviation is known to be 200 and the sample size is 100.
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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.
The sample data do not provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling.
B) Since z = 2.1316 > 1.645, reject the null hypothesis.
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C) Since z = 1.1594 < 1.645, do not reject the null hypothesis.
The sample data do not provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.
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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.
D) Because the calculated value of the test statistic, z = 0.3266, 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.
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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
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