E270 final
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| Question 1 and 2 are related | ||||||||||
| You run a gizmo factory. Over the past year the average output of the factory is 500 gizmos and the standard deviation of that output is 12 units per day. | ||||||||||
| 1 | What fraction of sample means from random samples of size n = 30 days fall within ±3 gizmos from the population mean? | |||||||||
| a | 0.8294 | |||||||||
| b | 0.8543 | |||||||||
| c | 0.8826 | |||||||||
| d | 0.9114 | |||||||||
| 2 | Using the population mean and standard deviation above, in repeated samples of size n = 30, 95% of sample means fall within ±_____ gizmos from the population means. | |||||||||
| a | 4.29 | |||||||||
| b | 4.05 | |||||||||
| c | 3.94 | |||||||||
| d | 3.88 | |||||||||
| Questions 3-5 are related. | ||||||||||
| In the population of all state government employees 38 percent contribute to the annual United Way campaign. | ||||||||||
| 3 | In random samples of 400 employees, what fraction of sample proportions fall within ±0.03 (3 percentage points) from the population proportion? | |||||||||
| a | 0.8414 | |||||||||
| b | 0.8249 | |||||||||
| c | 0.7814 | |||||||||
| d | 0.7641 | |||||||||
| 4 | 95 percent of sample proportions from samples of n = 400 deviate from the population proportion by no more than ±______. | |||||||||
| a | 0.052 | |||||||||
| b | 0.048 | |||||||||
| c | 0.044 | |||||||||
| d | 0.041 | |||||||||
| 5 | To obtain a margin of error of ±0.03 such that 0.95 fraction of the sample proportions fall within this MOE, the minimum sample size is n = _______. | |||||||||
| a | 925 | |||||||||
| b | 972 | |||||||||
| c | 998 | |||||||||
| d | 1006 | |||||||||
| 6 | The expression | means: | ||||||||
| a | 95% of sample means deviate from the population mean by no more than 1.96σ/√n in either direction. | |||||||||
| b | In repeated sampling the probability that x̅ is within ±1.96σ/√n from the population mean is 0.95. | |||||||||
| a | In repeated sampling the probability that population mean is within ±1.96σ/√n from x̅ is 0.95. | |||||||||
| d | Both a and be are correct. | |||||||||
| e | Both b and c are correct. | |||||||||
| 7 | You work for a charitable organization and you want to estimate the average age of the people who donate to your organization. You get a random sample of n = 102 donors and the value of the sample mean is 38 years. The value of the sample standard deviation is 16 years. | |||||||||
| The lower and upper end of the 95% confidence interval for the average age of people who donate to your organization are: | ||||||||||
| a | 36.0 | 40.0 | ||||||||
| b | 34.9 | 41.1 | ||||||||
| c | 33.9 | 42.1 | ||||||||
| d | 33.1 | 42.9 | ||||||||
| 8 | You run a bank and want to estimate the bank’s average number of customers per day (the population is all the days you are open for business in a year). You take a random sample of 10 days and record the numbers of customers on those days. The sample data is shown below. What is a 95% confidence interval for the bank’s average number of customers per day? | |||||||||
| 495 | 470 | 387 | 420 | 441 | ||||||
| 396 | 506 | 378 | 500 | 462 | ||||||
| a | 411 | 480 | ||||||||
| b | 416 | 475 | ||||||||
| c | 420 | 471 | ||||||||
| d | 426 | 465 | ||||||||
| 9 | It is estimated that 75% of Americans go out to eat at least once per week. This estimate is obtained from a random sample of 1040 adult Americans. The interval estimate, such that 19 of every 20 similar intervals capture the proportion of all Americans who eat out at least once per week, is: | |||||||||
| a | 0.715 | 0.785 | ||||||||
| b | 0.719 | 0.781 | ||||||||
| c | 0.724 | 0.776 | ||||||||
| d | 0.733 | 0.767 | ||||||||
| 10 | You are the manager of a political campaign. You think that the population proportion of voters who will vote for your candidate is 0.50 (use this for a planning value). Your candidate wants to know what proportion of the population will vote for her. Your candidate wants to know this with a margin-of-error of ± 0.01 (at 95% confidence). How big of a sample of voters should you take? | |||||||||
| a | 9800 | |||||||||
| b | 9604 | |||||||||
| c | 9455 | |||||||||
| d | 9266 | |||||||||
| 11 | You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.065, but does not tell you the value of the test statistic. From this information you can: | |||||||||
| a | Reject the null hypothesis at a probability of Type I error = 0.10, and reject at a probability of Type I error = 0.05 | |||||||||
| b | not reject the null hypothesis at a probability of Type I error = 0.10, and not reject at a probability of Type I error = 0.05 | |||||||||
| c | not reject the null hypothesis at a probability of Type I error = 0.10, but reject at a probability of Type I error = 0.05 | |||||||||
| d | Reject the null hypothesis at a probability of Type I error = 0.10, but not reject at a probability of Type I error = 0.05 | |||||||||
| 12 | American workers spend an average of 46 minutes to and from work in a typical day. In a sample of n = 144 employed adults in Atlanta, the average commuting time to work was x̅ = 52 minutes, with s = 39.6 minutes. | |||||||||
| Is there enough evidence, at α = 0.05, to conclude that the average commuting time in Atlanta is greater than the national average? The probability value (p-value) for the test is: _______. | ||||||||||
| a | 0.0688 | Do not reject the null hypothesis and do not conclude that average commuting time is greater than the national average. | ||||||||
| b | 0.0688 | Reject the null hypothesis and conclude that average commuting time is greater than the national average. | ||||||||
| c | 0.0344 | Do not reject the null hypothesis and do not conclude that average commuting time is greater than the national average. | ||||||||
| d | 0.0344 | Reject the null hypothesis and conclude that average commuting time is greater than the national average. | ||||||||
| 13 | To test the hypothesis, at a 5% level of significance, that the proportion of American children without health insurance has increased from 0.109 in 2005, a random sample of 1050 children revealed a sample proportion of 0.128. The test statistic is: ________. | |||||||||
| a | 1.22 | Conclude that proportion of children without health insurance has increased. | ||||||||
| b | 1.22 | Do not conclude that proportion of children without health insurance has increased. | ||||||||
| c | 1.98 | Conclude that proportion of children without health insurance has increased. | ||||||||
| d | 1.98 | Do not conclude that proportion of children without health insurance has increased. | ||||||||
| 14 | Which of the following statements about Type I and Type II errors is correct: | |||||||||
| a | Type I: | Reject a true alternative hypothesis. | Type II: | Do not reject a false alternative. | ||||||
| b | Type I: | Reject a true null hypothesis. | Type II: | Do not reject a false null hypothesis. | ||||||
| c | Type I: | Reject a false null hypothesis. | Type II: | Reject a true null hypothesis. | ||||||
| d | Type I: | Do not reject a false null hypothesis. | Type II: | Reject a true null hypothesis. | ||||||
| 15 | Pete Zaria owns a chain of pizza parlors that delivers most of the pizzas it sells. Pete is planning an advertising campaign stressing the quickness of his delivery. He will advertise that the average time is 20 minutes or less (at most 20 minutes). Before making this claim, Pete randomly selects 16 calls on different days and records the mean time between the call and the delivery. The sample mean is 22.1 minutes with a standard deviation of 4.5 minutes. Is there significant evidence to reject the null hypothesis? Test the hypothesis at a 5 percent level of significance. | |||||||||
| The test statistic is ______. | ||||||||||
| a | 1.87 | Reject the null hypothesis. Pete should not advertise. | ||||||||
| b | 1.87 | Do not reject the null hypothesis. Pete may advertise. | ||||||||
| c | 1.25 | Reject the null hypothesis. Pete should not advertise. | ||||||||
| d | 1.25 | Do not reject the null hypothesis. Pete may advertise. | ||||||||
| Next FIVE questions are based on the following regression model | ||||||||||
| To determine the impact of variations in price on sales the management of Big Bob's Burger Barn sets different prices in its burger joints in 61 stores located in different cities. | ||||||||||
| Using the sales and price data, a simple regression is run with sales (in thousands of dollars) as the dependent variable and price (in dollars) as the independent variable. | ||||||||||
| Use the following calculations and the accompanying regression summary output to answer the next seven questions. | ||||||||||
| ∑xy = | 26681.012 | x̅ = | 5.6633 | |||||||
| ∑x² = | 1972.7652 | y̅ = | 77.6443 | |||||||
| SUMMARY OUTPUT | ||||||||||
| Regression Statistics | ||||||||||
| Multiple R | ||||||||||
| R Square | ||||||||||
| Adjusted R Square | 0.471086 | |||||||||
| Standard Error | ||||||||||
| Observations | 61 | |||||||||
| ANOVA | ||||||||||
| df | SS | MS | F | Significance F | ||||||
| Regression | 54.4399 | 6.18E-10 | ||||||||
| Residual | ||||||||||
| Total | 2572.25 | |||||||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |||||
| Intercept | 6.701429 | 18.93396 | 9.5E-27 | 113.4751 | 140.2942 | |||||
| PRICE | 6.18E-10 | |||||||||
| 16 | The model predicts that when raising the price by $1, sales would change by $_______ thousand. | |||||||||
| a | -$9.59 | |||||||||
| b | -$9.13 | |||||||||
| c | -$8.69 | |||||||||
| d | -$7.91 | |||||||||
| 17 | The predicted sales for a price of $6.50 per burger is $ _______ thousand. | |||||||||
| a | $77.6 | |||||||||
| b | $75.4 | |||||||||
| c | $73.2 | |||||||||
| d | $70.4 | |||||||||
| 18 | Given that ∑(ŷ − y̅)² = 1234.425, the sample data show that _______ fraction of variations is sales is explained by price. | |||||||||
| a | 0.48 | |||||||||
| b | 0.52 | |||||||||
| c | 0.56 | |||||||||
| d | 0.60 | |||||||||
| 19 | The test statistic for the null hypothesis that a change in price has no impact on sales is: | |||||||||
| a | -6.311 | |||||||||
| b | -6.936 | |||||||||
| c | -7.378 | |||||||||
| d | -8.116 | |||||||||
| 20 | The margin of error for a 95% interval estimate for the population slope parameter is: | |||||||||
| a | 1.89 | |||||||||
| b | 2.36 | |||||||||
| c | 3.07 | |||||||||
| d | 3.98 | |||||||||
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E270 final
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