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First Name ________________ Last Name _______________ Student Number _______________ Acadia University Faculty of Arts DEPARTMENT OF ECONOMICS Course: Econ 2613FA01 - Empirical Analysis in Economics & Business Fall 2022 Instructor: C. Burç Kayahan ASSIGNMENT IV Notes: This assignment is due by 7:00PM ON WEDNESDAY, DECEMBER 7 th . Completed assignments, with your name and ID clearly shown, should be submitted to the Assignment 4 submission link on ACORN . I strongly recommend that you make a copy of your assignment before handing it in. Answers should be as complete as possible, SHOWING ALL STEPS AND ALGEBRAIC FORMULAS USED . Group work is encouraged, but you MUST ENSURE that the final, submitted version of the assignment is your OWN work. 1
First Name ________________ Last Name _______________ Student Number _______________ Question 1: Suppose that you are interested in estimating the average amount a customer spends for dinner at Paddy’s Pub in Kentville. Historical records show that the standard deviation of customer spending in Paddy’s Pub is $7.8. In order to estimate the average spending in the population, you collect a random sample of 30 expenditures, which are reported under the Excel sheet titled “ Paddy’s ” in ACORN. (12 pts) (a) The owner of Paddy’s Pub in Kentville claims that the average customer spending for dinner at his establishment is $30. At 10% level of significance , conduct a hypothesis test to determine whether the average customer spending in Paddy’s Pub is significantly different from $30 . Make sure to write down the null and alternative hypotheses , explain which test statistic is appropriate for this test, show your calculations for the critical and calculated test statistic , express the decision rule , provide a detailed & complete conclusion of your hypothesis test, and draw a graph that not only illustrates the rejection and non- rejection regions as well as the critical & the calculated test statistic values. Null Hypothesis: Mu =30 Alternative Hypothesis: Mu 30 Test: Statistic: Sigma known= z test Critical value of the test statistic: Ztest = norm.s.inv(0.05)= -1.64 norm.s.inv(0.95)= 1.64 Calculated value of the test statistic: (32.28-30)/ (7.8 30 ) = 1.6 Decision rule: |z calc| < |zcrit| 1.6< 1.64 Do not reject Hnull Conclusion: In conclusion, we cannot conclude that the average spending at paddy’s is much greater than 30$ with a 10% level of significance. (8%) (b) Calculate the p-value and show it on a graph. Confirm the results of our test from part (a) using the p-value approach. Would the results of your hypothesis test change if the level of significance of your test were 5% instead ? Explain. P-value: norm.s.dist = ( -1.60, true) = 0.055 *2 = 0.11 or 11% LOS- 10% 11%> 10% So we do not reject Hnull Draw your graph below: Reject 5% Reject 5% Do not reject 90% 1.60 1.64 Zcalc zcrit -1.64 zcrit 2
First Name ________________ Last Name _______________ Student Number _______________ Confirm the conclusion of the hypothesis test from part (a) using the p-value: P-value vs Level of Significance: The p value is larger than the level of significance, therefore, we shouldn’t reject Hnull Would the result change? Explain why or why not: No, they would not change because the p-value would remain the same (11%) and it would still be larger than the level of significance (5%) therefore we would still not reject Hnull. 3
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First Name ________________ Last Name _______________ Student Number _______________ Question 2: The average ambulance response time in Nova Scotia is believed to be around 26 minutes. The Minister of Health and Wellness in Nova Scotia would like to reduce the average ambulance response time, so the Ministry implemented a set of changes to existing procedures in order to reduce the response times. Historical records show that the standard deviation of the response time in the province is 5.7 minutes. As an analyst at the Ministry, you collect a sample of 81 randomly collected response times from the province to evaluate the effectiveness of the new policy. The average response time in your sample is calculated to be 24.2 minutes. (12pts) (a) At 5% level of significance , conduct a hypothesis test to determine whether the average ambulance response time in Nova Scotia is less than 26 minutes . Is the new policy effective? Explain. Hint: The average response time was initially believed to be 26 minutes. Hence, the new policy is effective if the average response time in the province is reduced to a value below 26 minutes following the changes to the existing procedures. Null Hypothesis: Mu >= 26 Alternative Hypothesis: Mu<26 Test: Statistic: sigma known = ztest = 5.7 Critical value of the test statistic: Z test = norm.s.inv (0.05) = -1.64 Calculated value of the test statistic: (24.2-26)/ (5.7/ 81 ) = -2.84 Decision rule: |z calc| > |zcrit | 2.84>1.6 Reject Hnull Conclusion: At 5% level of significance, we can conclude that the average ambulance time is significantly less than 26 mins. Is the new policy effective? Briefly explain: Yes the new policy would be effective because the average response time is less than 26 mins. (8%) (b) Calculate the p-value and show it on a graph. Confirm the results of our test from part (a) using the p-value approach. Would the results of your hypothesis test change if the level of significance of your test were 1% instead? Explain. P-value: = norm.s.dist(-2.84,true) = 0.023 0.23%< 5% Draw your graph below: 5% reject 95% do not reject -2.84 -1.64 Zcalc, zcrit 4
First Name ________________ Last Name _______________ Student Number _______________ Reject Hnull Confirm the conclusion of the hypothesis test from part (a) using the p-value: P-value vs Level of Significance: We would sill be able to conclude that the average ambulance response time is less than 26 mins. Would the result change? Explain why or why not: No the result would not change because the level of significance (1%) is still greater than the p Value (0.23%) 5
First Name ________________ Last Name _______________ Student Number _______________ Question 3: RCMP detachments are concerned about traffic speeds in school zones, hence they use speed- measuring machines on roads to measure how fast the motorists are driving. Excel sheet titled “Speed” presents measurements (in kms/h) that were reported by a radar placement at the school-zone in New Minas. (12%) (a) At the 1% level of significance , conduct a hypothesis test to determine whether the average speed of motorists at the school-zone in New Minas is greater than 30km/h at the school zone in New Minas. Make sure to write down the null and alternative hypotheses , explain which test statistic is appropriate for this test, show your calculations for the critical and calculated test statistic , express the decision rule , provide a detailed & complete conclusion of your hypothesis test, and draw a graph that not only illustrates the rejection and non-rejection regions as well as the critical & the calculated test statistic values. Null Hypothesis: mu>=30 Alternative Hypothesis: mu<30 Test: Statistic: Sigma unknown = t test Critical value of the test statistic: =t.inv(0.99, 35) = 2.44 Calculated value of the test statistic: (31.3 – 30)/ (3.25/ 36 ) = 2.4 Decision rule: T calc < T crit 2.4 < 2.44 Do not reject Hnull Conclusion: In conclusion, we cannot conclude that the speed limit of motorists is greater than 30 km/h with a 1% level of significance. (8%) (b) Calculate the p-value and show it on a graph. Confirm the results of our test from part (a) using the p-value approach. Would the results of your hypothesis test change if the level of significance of your test were 5% instead ? Explain. P-value: =t.dist(-2.4, 35, true)= 1.09% 1.09% > 1% Do not reject Hnull Draw your graph below: Reject 1% 99% do not reject 2.4 2.44 T calc Tcrit 6
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First Name ________________ Last Name _______________ Student Number _______________ Confirm the conclusion of the hypothesis test from part (a) using the p-value: P-value vs Level of Significance: Do not reject Hnull Would the result change? Explain why or why not: The result would change, we would reject the Hnull because the p value is lower than the level of significance. 1.09%< 5% Question 4: The local diner in a small New Brunswick community is up for sale. You are interested in potentially making a bid to purchase the diner. You want to estimate the average number of customers that visits the diner before making an offer. The Excel sheet titled Diner presents a random sample of 35 days where the number of customers that visited the diner is recorded. (a) (12pts) At the 5% level of significance , conduct a hypothesis test to determine whether the average number of customers that visits the diner is significantly different from 80 people ? Make sure to write down the null and alternative hypotheses , explain which test statistic is appropriate for this test, show your calculations for the critical and calculated test statistic , express the decision rule , provide a detailed & complete conclusion of your hypothesis test, and draw a graph that not only illustrates the rejection and non- rejection regions as well as the critical & the calculated test statistic values. Null Hypothesis: mu=80 Alternative Hypothesis: mu 80 Test: Statistic: sigma unknown= t test Critical value of the test statistic: =t.inv(0.025,34)= -2.03 Calculated value of the test statistic: (84.03 – 80)/(10.26/ 35 ) = 2.322 Decision rule: T calc > t crit 2.32> 2.03 Reject Hnull Conclusion: with 95% level of significance, we can conclude that the average number of customers that visits the diner is significantly different than 80. Draw your graph below: Do not reject 95% Reject 2.5% Reject 2.5% 7
First Name ________________ Last Name _______________ Student Number _______________ (4%) (b) Calculate the p-value Would the results of your hypothesis test change if the level of significance of your test were 1% instead ? Explain. P-value: =t.dist.2t(2, 32,34) = 0.0265 2.65%< 5% Reject Hnull Would the result change? Explain why or why not: Yes the results would change because now the p value is larger than the level of significance 2.65%> 1% Do not reject Hnull (4%) (a) At the 99% level of confidence , construct a confidence interval for the average number of customers that visits the diner. Does your interval support your conclusion from part (a)? Explain. =t.inv.2t(0.01,34)= 2.73 84.03 +&- (2.73)(1.735) 84.03 + 4.736=88.766 84.03- 4.736= 79.294 My confidence interval shows that 99% of the average number of customers that visit the diner are between 79.3 and 88.8, this supports the conclusion in part a. 8