18W (1)

.pdf

School

University of Texas *

*We aren’t endorsed by this school

Course

121

Subject

Economics

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by ChiefBravery9827

MGEB12 – Quantitative Methods in Economics-II Winter 2018 Instructor: A. Mazaheri Test-1 (Solutions) Instructions: This is a closed book test. A formula sheet is attached. You are allowed the use of a calculator. Write your full name and your student ID # on both this test book as well as the Scantron paper. You have 90 Minutes. Good Luck! Last Name: First Name: ID FOR MARKERS ONLY: The University of Toronto's Code of Behaviour on Academic Matters applies to all University of Toronto Scarborough students. The Code prohibits all forms of academic dishonesty including, but not limited to, cheating, plagiarism, and the use of unauthorized aids. Students violating the Code may be subject to penalties up to and including suspension or expulsion from the University. MC Q#1 Q#2 Total Marks Earned Maximum Marks Possible 68 12 15 95 _ song 9966 22092996275

Page 2 of 28 Part A (Long Questions) Question-1 (12 Points) The administrator at your local hospital states that on weekends the average wait time for emergency room (ER) visits is 10 minutes. Based on discussions you have had with friends who have complained on how long they waited to be seen in the ER over a weekend, you believe it takes longer. You decide to test your hypothesis. Over the course of a few weekends you record the wait time for 40 randomly selected patients. The average wait time for these 40 patients is 11 minutes with a standard deviation of 3 minutes. a) [7 Points] State the null and alternative hypotheses. Using the p-value approach, test the hypothesis. Explain in some detail your conclusion. Solution: 𝐻𝐻 0 : 𝜇𝜇 = 10 𝐻𝐻 1 : 𝜇𝜇 > 10 𝑡𝑡 39 = 11 − 10 3 √ 40 = 2.11 P-Value: Since t statistics is in between t 0.025, 40 = 2.021 , t 0.01,40 = 2.423, the p-value will be between 1% and 2.5% Decision : There is strong evidence in rejecting the null as the Type-I error is less than 2.5%. _ song 9966 22092996275

Page 3 of 28 b) [5 Points] (Ignore Part a) A patient is to take a medical test for high blood pressure. Suppose any reading higher than 130 is considered high blood pressure. b1) What would be the null and the alternative in this case? (2.5 Points) 𝐻𝐻 0 : 𝜇𝜇 = 0 ( 𝜇𝜇 ≤ 0) 𝐻𝐻 1 : 𝜇𝜇 > 0 b2) Explain in words what are Type-I and Type-II errors are in this case? (Note: A generic definition is not enough) (2.5 Points) Type-I error: Conclude that the patient does have High blood pressure when he/she does not. Type-II error: Conclude the patient does not have high blood pressure when he/she does. _ song 9966 22092996275

Page 4 of 28 Question-2 (15 Points ) In 1990, 5.8% of job applicants who were tested for drugs failed the test. A researcher believes that the failure rate is now lower. She has taken a random of 1520 current job applicants which resulted in 58 failures. a) [6 Points] Express this claim by identifying the Null Hypothesis and the Alternative Hypothesis. Compute the p-value associated with this hypothesis test. Briefly explain what the p- value means. Using the calculated p-value and a significance level of 1%, would you reject the null hypothesis? Solution: 𝐻𝐻 0 : 𝑃𝑃 = 0.058 𝐻𝐻 1 : 𝑝𝑝 < 0.058 𝑝𝑝̅ = 58 1520 = 0.03816 𝑍𝑍 = 0.03816 − 0.058 � 0.058(1 − 0.058) 1520 = − 3.3096 2 Point for setting up the hypothesis p-value = P(Z < -3.3096) < (0.5-0.4990) = 0.001 2 Points for calculating the test statistics. 1 Points : P-value is the probability of obtaining a sample with proportion of 0.03816 or less if the null is correct. 1 Points: p-value is smaller than the significance level of 1% => reject the null. _ song 9966 22092996275

Page 5 of 28 b) [4 Points] Find the 99% confidence interval for the proportion. Briefly interpret the interval. Solution : 𝑝𝑝̅ ± 𝑍𝑍 𝛼𝛼 / 2 � 𝑝𝑝̅ (1 − 𝑝𝑝̅ ) 𝑛𝑛 = 0.03816 ± 2.565 × 0.004914 = 0.03816 ± 0.012604 3 Points for the CI 1 Point: In a repeated sampling, 99% of the calculated intervals will include the true population => we are 99% confident that the proportion of the population will be between 2.5554% to 5.0762%%. c) [5 Points] Suppose you wish to conduct the test as in part (a) at α = 2.5 %. Find the power of your test if the actual population proportion is 3%. Briefly interpret your results. Solution: 𝑝𝑝 ∗ �� > 0.058 − 1.96 � 0.058(1 − 0.058) 1520 = 0.046249 𝑍𝑍 𝛽𝛽 = 0.046249 − 0.03 � 0.03(1 − 0.03) 1520 = 3.7137 𝛽𝛽 − 𝑃𝑃�𝑍𝑍 > 𝑍𝑍 𝛽𝛽 � = 𝑃𝑃 ( 𝑍𝑍 > 3.7137) < 0.001 The power of the test is > 0.999. For this alternative, the test has a very high power. _ song 9966 22092996275

Page 6 of 28 Part B (Multiple Choice) Form-1 For the following 18 questions, choose the most correct answer and mark it on the SCANTRON form : 1. Identify your form number (Note: if you don’t answer this question correctly you will lose 5 points) A. 1 B. 2 C. 3 D. 4 2. Suppose our p-value is .037. What of the following statements is True? (D) A. We will reject Ho at alpha=.10, but not at alpha=.05 B. We will reject Ho at alpha=.10, .05, or .01 C. We will not reject Ho at alpha=.10, .05, or .01 D. We will reject Ho at alpha=.10 or .05, but not at alpha=.01 E. We will can never reject the null because the p-value is small. 3. The power curve provides the probability of (A) A. rejecting the null hypothesis when the alternative is correct B. accepting the null hypothesis when the alternative is correct C. accepting the null hypothesis when the alternative is correct D. rejecting the alternative when the null is correct E. correctly rejecting the alternative hypothesis 4. In hypothesis testing, we need to assume: (E) I. The sample size is equal or more than 30 II. Distribution of the population is symmetric III. The population variance is known A. I only B. II only C. III only D. I and II only E. None of the above 5. As a researcher you have already calculated the following three confidence intervals using the same data: 90%: (-.29, 6.29), _ song 9966 22092996275

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version