| Next two questions are related |
| 1 | You run a gizmo factory. The average output of the factory is 480 gizmos and the standard deviation of that output is 12 units per day. In repeated random samples of size n = 40 days, the expected value of sample mean is: |
| a | 12 |
| b | 480 |
| c | 75.89 |
| d | 48 |
| 2 | Using the population mean and standard deviation in question 1, what is the probability that the mean of a random sample of n = 40 days is between 478 and 482 gizmos: |
| a | 0.7062 |
| b | 0.8858 |
| c | 0.9232 |
| d | 0.9426 |
| Next three questions are related |
| 3 | In the population of employees in your company, 35 percent (π = 0.35) contribute to the annual United Way campaign. In repeated samples of size 200, what is the standard error of the sample proportion p̅, the proportion in the sample that contribute to United Way? |
| a | 0.0473 |
| b | 0.0415 |
| c | 0.0398 |
| d | 0.0337 |
| 4 | In the previous question, 95 percent of sample proportions from samples of size n = 200 deviate from the population proportion of 0.35 by no more that ±____ (or ____ percentage points). |
| a | 0.044 | (4.4 percentage points.) |
| b | 0.05 | (5 percentage points.) |
| c | 0.058 | (5.8 percentage points.) |
| d | 0.066 | (6.6 percentage points.) |
| 5 | In the previous question, to obtain a margin of error of ±0.03 (3 percentage points) for a 95% middle interval of sample proportions, what is the minimum sample size? |
| a | 972 |
| b | 1005 |
| c | 1046 |
| d | 1068 |
| Next two questions are related |
| 6 | As part of a statistics assignment in October to estimate the percentage of voters who would vote for a mayoral candidate, each of 500 students collects his or her own random sample of likely voters. There are 400 voters in each student’s random sample. |
| | Each student then constructs a 95 percent confidence interval for the population proportion who will vote for the candidate using his or her own random sample. Considering the 500 intervals constructed by the students, the expected number of intervals that will contain the population proportion who will vote for that candidate will be approximately: |
| a | 380 |
| b | 400 |
| c | 475 |
| d | 500 |
| 7 | Suppose the sample proportion of one of the students in the previous question, Beth's sample, is p̅ = 0.46. Beth's 95% interval estimate of proportion of the population of voters voting for the candidate is: |
| a | [0.41 , 0.51] |
| b | [0.40 , 0.52] |
| c | [0.39 , 0.53] |
| d | [0.38 , 0.54] |
| 8 | It is estimated that 80% of Americans go out to eat at least once per week, with a margin of error of 0.04 (for 95% confidence). A 95% confidence interval for the population proportion of Americans who go out to eat once per week or more is: |
| a | [0.722, 0.878] |
| b | [0.760, 0.840] |
| c | [0.771, 0.829] |
| d | [0.798, 0.802] |
| Next two questions are related |
| 9 | 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 8 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? |
| | | 450 | 470 | 430 | 420 |
| | | 440 | 460 | 420 | 500 |
| a | [433, 467] |
| b | [430, 470] |
| c | [426, 472] |
| d | [424, 476] |
| 10 | As the manager of the bank in the previous question, you want the 95% interval estimate to capture the population mean customers per day within ±10 customers. Using a planning value of σ̂ = 30, how many days should you include in the sample? |
| a | 68 |
| b | 59 |
| c | 48 |
| d | 35 |
| Next three questions are related |
| You work for a charitable organization and you want to test the hypothesis that the average age of people who donate to your organization is above 45 years. You get a random sample of n = 220 donors and the value of the sample mean is 47 years. The value of the sample standard deviation is 19 years. |
| 11 | Which of the following is the correct statement of the hypotheses for the test? |
| a | H₀: μ ≥ | 45 | H₁: μ < | 45 |
| b | H₀: μ ≤ | 45 | H₁: μ > | 45 |
| c | H₀: μ > | 45 | H₁: μ ≤ | 45 |
| d | H₀: μ < | 45 | H₁: μ ≥ | 45 |
| 12 | The test statistic for your hypothesis test is ______. At a 5% level of significance, |
| a | 1.96 | Reject H₀. Conclude the mean age is above 45. |
| b | 1.64 | Do not reject H₀. Do not conclude the mean age is above 45. |
| c | 1.56 | Do not reject H₀. Do not conclude the mean age is above 45. |
| d | 1.35 | Reject H₀. Conclude the mean age is above 45. |
| 13 | In the previous question, |
| a | If H₀ were in fact false and you rejected it, you committed a Type II error |
| b | If H₀ were in fact false and you did not reject it, you committed a Type I error |
| c | If H₀ were in fact true and you rejected it, you committed a Type II error. |
| d | If H₀ were in fact true and you rejected it, you committed a Type I error. |
| Next two questions are related |
| 14 | According to the Census Bureau the sample proportion of American children without health insurance rose from 0.109 in 2005 to 0.117 in 2006. Using the Census data you test the hypothesis that the change in the population proportion of American children without health insurance is zero with a Probability of Type I error = 0.05. You reject the zero change hypothesis. |
| | The correct interpretation is: |
| a | Even if the change in the population of American children without health insurance really is zero, 5% of all repeated samples would have produced test statistics that cause you to reject the hypothesis that the change is zero. |
| b | The odds are 5 percent that if the sample of American children in 2006 had been expanded to include the entire population of American children, the change in the proportion of children without health insurance would still not be zero. |
| c | The test statistic for the null hypothesis that H₀: π = 0.109 is less than the critical value. |
| d | 95 percent of repeated samples would produce confidence intervals within ± 0.05 points of 0.117. |
| 15 | To test the hypothesis, at a 5% level of significance, that the proportion of American children without health insurance has increased from 0.109, a random sample of 1020 children revealed a sample proportion of 0.121. Compute the p-value. |
| a | 0.0556 | Conclude that proportion of children without health insurance has increased. |
| b | 0.0556 | Do not conclude that proportion of children without health insurance has increased. |
| c | 0.1093 | Conclude that proportion of children without health insurance has increased. |
| d | 0.1093 | Do not conclude that proportion of children without health insurance has increased. |
| Next 5 questions are based on the following regression problem: |
| You run a regression to analyze the relationship between the wage rate and years of education. The following is the regression output. The dependent variable, WAGE is in dollars per hour and the independent variable EDUC is the years of education. |
| | SUMMARY OUTPUT |
| | Regression Statistics |
| | Multiple R | | 0.4006 |
| | R Square |
| | Adjusted R Square | | 0.1593 |
| | Standard Error |
| | Observations | | 700 |
| | ANOVA |
| | | df | SS | MS | F | Significance F |
| | Regression | | | 25104.62 | 133.4634 | 2.26E-28 |
| | Residual | | 131294.59 |
| | Total | | 156399.21 |
| | | Coefficients | Std Error | t Stat | P-value | Lower 95% | Upper 95% |
| | Intercept | 3.179 | 2.2199316041 | 1.4319032388 | 1.53E-01 | -1.1798164292 | 7.5373 |
| | EDUC | 1.995 | | | 2.26E-28 |
| | Note: | ∑(x −x̅)² = | 6305 |
| 16 | The predicted WAGE when EDUC = 16 is: |
| a | 35.1 |
| b | 32.6 |
| c | 30.5 |
| d | 28.8 |
| 17 | The standard error of the estimate is, |
| a | 11.593 |
| b | 12.604 |
| c | 13.715 |
| d | 14.826 |
| 18 | The fraction of the variations in WAGE that is explained by years of education is: |
| a | 0.2331 |
| b | 0.2119 |
| c | 0.1926 |
| d | 0.1605 |
| 19 | The test statistic to perform the hypothesis test for the significance of the slope coefficient is: |
| a | 11.55 |
| b | 10.86 |
| c | 9.99 |
| d | 8.89 |
| 20 | The lower boundary of the 95% confidence interval for the slope coefficient is: |
| a | 1.33 |
| b | 1.66 |
| c | 1.74 |
| d | 1.83 |