ECON 203 Midterm 1

Assume that the distribution of final scores of golf players at the UC golf course is normal, with a standard deviation of 3 and a median of 78. Consider that the par is 72, what percentage of players score higher than the par?

97.5% REASON: In a normal distribution, the mean equals the median. We are interested in finding the percentage of players scoring higher than 72. Given that 72 is 2*σ to the left of the mean, we know that the area to the left of 72 will be (1-.95)/2 = .025. The area to the right of 72 therefore is 1-.025=.975 or 97.5%.

As a farmer in Champaign, you want to plant the corn that will give you the highest yield per acre. You come across four different seeds and your cousin tells you to "choose any because they are all the same". Not satisfied with this advice, you find the following statistics on yields for each type. Seed A: Nr plots: 42 Avg. Yield: 20.1 Standard dev.: 2.2 Seed B: Nr plots: 30 Avg. Yield: 18 Standard dev.: 1.5 Seed C: Nr plots: 20 Avg. Yield: 19.3 Standard dev.: 2 Seed D: Nr plots: 8 Avg. Yield: 16.9 Standard dev.: 4.4 Calculate the F statistic. a. 7.92 b. 8.02 c. 8.41 d. 39.20 e. There is not enough information to calculate the F Statistic.

A. 7.92 REASON: F = MST/MSE = 39.20/4.95 = 7.92. Where MSE = SSE / (n-k) - 475.21/96 = 4.95.

Which of the following statements is FALSE? a. The mode is always equal to the median. b. The mode is greater than the median when a distribution is skewed to the left c. The mode of a distribution that is symmetric is equal to the mean. d. The mode is less than the mean when a distribution is skewed to the right. e. The mode does not change when adding an extreme value to the right of the distribution.

A. The mode is always equal to the median. REASON: This is not true for symmetrical distributions

The workers at a factory are discussing whether or not to unionize. They hire you to evaluate whether unionized workers at other factories around the US earn higher wages than non-unionized workers. You collect the following data on salaries (in thousands of dollars per year) for unionized (population 1) and non-unionized (population 2) workers: Unionized: Sample size: 222 Average Salary: 73.1 Standard deviation: 10.1 Non-unionized Sample size: 412 Average salary: 72.5 Standard deviation: 4.2

A. h0: = 1 h1: REASON: The null hypothesis is that the variances are equal, hence that their ratio is 1; the alternative is that they are unequal, or that their ratio is different from 1.

Suppose you have a mound-shaped, symmetrical distribution, and now we add some extreme values on the lower tail of the distribution. Which of the following could be true? a. The distribution would be skewed to the right. b. The distribution would be skewed to the left. c. The mode would be smaller than the median. d. The mode would be smaller than the mean. e. The mean would be larger than the median.

B. The distribution would be skewed to the left. REASON: Starting from a mound-shaped, symmetrical distribution, adding some extreme values on the lower tail (left side) would produce a fatter left tail. Therefore, the distribution would be skewed to the left.

As a farmer in Champaign, you want to plant the corn that will give you the highest yield per acre. You come across four different seeds and your cousin tells you to "choose any because they are all the same". Not satisfied with this advice, you find the following statistics on yields for each type. Seed A: Nr plots: 42 Avg. Yield: 20.1 Standard dev.: 2.2 Seed B: Nr plots: 30 Avg. Yield: 18 Standard dev.: 1.5 Seed C: Nr plots: 20 Avg. Yield: 19.3 Standard dev.: 2 Seed D: Nr plots: 8 Avg. Yield: 16.9 Standard dev.: 4.4 Compute the mean of the squares for treatments: a. 19.05 b. 20.10 c. 39.20 d. 65.25 e. 117.61

C. 39.20 REASON: First compute the grand mean: 19.054. Then the sum of the required treatments is 39.20

You decide to go to London as an exchange student for a year. When you arrive you realize that everyone is crazy about soccer and you have to pick a winning local team to support. A friend advises you to choose Arsenal over Chealsea since they win more games. Just to make sure his advise is correct, you collect data on the number of games played and won by each team and find that Arsenal (population 1) won 45 of the last 60 games, and Chealsea won 32 of the last 50 games. What is appropriate conclusion of the test allowing a 10% chance of Type I error? z0.10 = 2.326 z0.025 = 1.960 z0.050 = 1.645 z0.100 = 1.282 t0.010,148 = 2.361 t0.025,148 = 1.982 t0.050,148 = 1.659 t0.100,148 = 1.289 a. reject the null hypothesis, Arsenal wins a higher proportion of games than Chealsea. b. Reject the null hypothesis, the teams win the same proportion of games. c. Do not reject the null hypothesis, there is insufficient evidence that Arsenal wins more than Chealsea. d. Do not reject the null hypothesis, the teams win the same proportion of games. e. More data is needed to determine whether we should reject the null hypothesis.

C. Do not reject the null hypothesis, there is insufficient evidence that Arsenal wins more than Chealsea. REASON: We are told that the significance level is of 10%. the critical value is hence z.1 = 1.282. Since our observed z- statistics of 1.254 is less extreme, we do not reject the null and conclude that there is insufficient evidence that Arsenal wins more than Chealsea.

You decide to go to London as an exchange student for a year. When you arrive you realize that everyone is crazy about soccer and you have to pick a winning local team to support. A friend advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his advise is correct, you collect data on the number of games played and won by each team and find that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50 games. What are the appropriate null and alternative hypotheses to test whether the proportion of games won by Arsenal is higher than by Chelsea? a. H0: p-hat1 - p-hat2 = 0 vs. p-hat1 - p-hat2 > 0 b. H0: p1 - p2 = 0 vs. H1: p1 - p2 < 0 c. H0: p1 - p2 =0 vs. H1: p1-p2> 0 d. H0: p1 - p2 > 0 vs. H1 : p1 - p2 = 0 e. H0: p-hat1 - p-hat2 = 0 vs. H1: p-hat1 - p-hat 2 ╪ 0

C. H0: p1 - p2 =0 vs. H1: p1-p2> 0 REASON: Null and alternative hypotheses always deal in population parameters, so the ones with p-hat are eliminated. You want to test whether the proportion of games won by Arsenal is higher than that of Chelsea.

You have just performed a hypothesis test with a 5% level of significance. You have not rejected the null hypothesis that the population mean is equal to 50 (H1: μ╪ 50). If you were to create a 90% confidence interval, what could you say about 50 being included or excluded from that confidence interval? a. The 90% confidence interval will include 50, since it is larger than the 95% confidence interval. b. The 90% confidence interval will include 50, since it is smaller than the 95% confidence interval. c. You are unsure if the 90% confidence interval includes 50, because the 95% confidence interval contains 50. d. The 90% confidence interval will not include 50, since it is smaller than the 95% confidence interval. e. You are unsure if the 90% confidence interval includes 50, because the 95% confidence interval does not contain 50.

C. You are unsure if the 90% confidence interval includes 50, because the 95% confidence interval contains 50. REASON: For the two-tailed test at the 5% level, we cannot reject the null; This means that 50 would be contained in a 95% confidence interval. However, since the 90% confidence interval is smaller than the 95% CI, it is uncertain whether 50 will be contained or not in it.

You decide to go to London as an exchange student for a year. When you arrive you realize that everyone is crazy about soccer and you have to pick a winning local team to support. A friend advises you to choose Arsenal over Chelsea since they win more games. Just to make sure his advise is correct, you collect data on the number of games played and won by each team and find that Arsenal (population 1) won 45 of the last 60 games, and Chelsea won 32 of the last 50 games. What is the value of the appropriate test statistic? a. 0.958 b. 0.987 c. 1.062 d. 1.254 e. 1.897

D. 1.254 REASON: We know that this is case 2 of difference in proportion since we are interested in testing whether the proportion of wins for one team is larger than for the other. Our test statistic will be z = [(p-hat1 - p-hat2)] / (SQRT(p-hat(1-p-hat)(1/n1+1/n2))

A hooligan walking erratically down the street tells you hat the proportion of games won by Arsenal is at least 15% higher than Chelsea. Which graph would best represent the p-value for testing the hooligan's claim.

D. A full graph with a little bit of area (15%) to the right not shaded. REASON: We want to test H1: p1- p2 > 0.15. Therefore, the test statistic would be: z = [(p-hat1 - p-hat2) - 0.15] / (SQRT(p-hat1(1-p-hat1)/n1 + p-hat2(1-p-hat2)n2)) = -.455. Since the test statistic is negative and we are interested in testing a right sided test, the appropriate graph will be D.

As a farmer in Champaign, you want to plant the corn that will give you the highest yield per acre. You come across four different seeds and your cousin tells you to "choose any because they are all the same". Not satisfied with this advice, you find the following statistics on yields for each type. Seed A: Nr plots: 42 Avg. Yield: 20.1 Standard dev.: 2.2 Seed B: Nr plots: 30 Avg. Yield: 18 Standard dev.: 1.5 Seed C: Nr plots: 20 Avg. Yield: 19.3 Standard dev.: 2 Seed D: Nr plots: 8 Avg. Yield: 16.9 Standard dev.: 4.4 Which of the following options is the correct critical value if you want to run a 5% significance level test on the equality of the yields between the four types of seeds? a. only F0.1,3,95 b. F0.025,4,100 and F0.975, 4, 100 c. Only F0.05,4,100 d. F0.025,3,96 and F0.975,3,96 e. Only F0.05,3,96

E. Only F0.05,3,96 REASON: The ANOVA test is always a one-sided greater than test, so we use ∝, not ∝/2. The degrees of freedom are k - 1 and n - k. Hence the correct choice is F0.05, 3, 96

Which of he following about the Chi-squared distribution is FALSE a. We use the Chi-squared distribution to perform a test on variance. b. The Chi-squared is not symmetrical. c. Only positive Chi-squared values are possible. d. The Chi-squared distribution is asymptotic to the horizontal axis to the right. e. 1 is an approximation to the center of the distribution.

e. 1 is an approximation to the center of the distribution REASON: The center of Chi-squared distribution is approximated by its degrees of freedom minus one.