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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 blocking?
A) Blocking was not effective because p-value = 2.66 is greater than 0.05.
B) Blocking was effective because p-value = 2.66E - 5 is less than 0.05.
C) Blocking was not effective because p-value = 1.6 is greater than 0.05.
D) Blocking was effective because p-value = 1.6E-14 is less than 0.05.
Answer
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According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008.
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A) 0.0872
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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.
D) z = 0.341 > -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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According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people.
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
B) 0.0712
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.
A) 0.9192
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D) 0.0314
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Because of the complex nature of the U.S. income tax system, many people have questions for the Internal Revenue Service (IRS). Yet, an article published by the Detroit Free Press entitled "Assistance: IRS Help Centers Give the Wrong Information" discusses the propensity of IRS staff employees to give incorrect tax information to tax-payers who call with questions. Then IRS Inspector General Pamela Gardiner told a Senate subcommittee that "the IRS employees at 400 taxpayer assistance centers nationwide encountered 8.5 million taxpayers face-to-face last year. The problem: When inspector general auditors posing as taxpayers asked them to answer tax questions, the answers were right 69% of the time."Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed.Using an = 0.05 level, based on the sample data, what conclusion should be reached about the IRS rate of correct tax returns?A) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.69 rate quoted in the Detroit Free Press articleB) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.58 rate quoted in the Detroit Free Press articleC) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z= -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.69 rate quoted in the Detroit Free Press article.D) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.58 rate quoted in the Detroit Free Press article.
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A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling.
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.
The sample data provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling.
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.
D) Since z = 2.1316 > 1.645, reject the null hypothesis.
The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.
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A) 2.7299
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C) 1.4919
D) -0.3421
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For the following hypothesis test: With n= 0.42 and p = 0.42, state the conclusion
A) Because the calculated value of the test statistic, t=0.4122, 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.
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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A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error.
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.
D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error.
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The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out-of-pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out-of-pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondents' 2012 out-of-pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out-of-pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test.
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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