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Question
A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here:| ANOVA | ||||
| Source of Variation | SS | df | MS | F |
| Textbooks | 16,624 | |||
| Retailer | 2.4 | |||
| Error | ||||
| Total | 17,477.6 |
Complete the ANOVA table by filling in the missing sums of squares, the degrees of freedom for each source, the mean square, and the calculated F-test statistic for each possible hypothesis test.
A) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 40.05, Retailer df = 2, MSB = 1.2,
, SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 35B) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2,
, SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 35C) Textbooks df = 12, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2,
, SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 36D) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2,
, SSW = 831.2, Error df = 22, MSW = 38.7, Total df = 36 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.
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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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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.
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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
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