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Question
Each year, Business Week publishes information and rankings of master of business administration (MBA) programs. The data file MBA Analysis contains data on several variables for eight reputable MBA programs as presented in the October 2, 2000, issue of Business Week. The variables include pre- and post-MBA salary, percentage salary increase, undergraduate GPA, average Graduate Management Admission Test (GMAT) score, annual tuition, and expected annual student cost. Compute the mean and median for each of the variables in the database.A)
| Mean | Median | |
| Pre-MBA Salary | 39077 | 43337.63 |
| Post-MBA Salary | 82203 | 98902 |
| Percentage Increase in Salary | 116 | 123 |
| Undergraduate GPA | 3.46 | 3.40 |
| GMAT Score | 635.00 | 631.13 |
| Annual Tuition | 13163.50 | 15967.50 |
| Expected Annual Student Cost | 23169.5 | 27980.75 |
B)
| Mean | Median | |
| Pre-MBA Salary | 46667.63 | 39077 |
| Post-MBA Salary | 98902 | 84403 |
| Percentage Increase in Salary | 113 | 116 |
| Undergraduate GPA | 3.40 | 3.56 |
| GMAT Score | 661.16 | 635.00 |
| Annual Tuition | 15967.50 | 19163.70 |
| Expected Annual Student Cost | 28980.75 | 23179.5 |
C)
| Mean | Median | |
| Pre-MBA Salary | 43337.63 | 39077 |
| Post-MBA Salary | 98902 | 82203 |
| Percentage Increase in Salary | 123 | 116 |
| Undergraduate GPA | 3.40 | 3.46 |
| GMAT Score | 631.13 | 635.00 |
| Annual Tuition | 15967.50 | 13163.50 |
| Expected Annual Student Cost | 27980.75 | 23169.5 |
D)
| Mean | Median | |
| Pre-MBA Salary | 39077 | 46667.63 |
| Post-MBA Salary | 84403 | 98902 |
| Percentage Increase in Salary | 116 | 113 |
| Undergraduate GPA | 3.56 | 3.40 |
| GMAT Score | 635.00 | 661.16 |
| Annual Tuition | 19163.70 | 15967.50 |
| Expected Annual Student Cost | 23179.5 | 28980.75 |
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
B) 0.0712
C) 0.0537
D) 0.1412
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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.
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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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
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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.
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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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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The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. Conduct the test using a p-value.
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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