Multiple Choice Question 1. b 2. e 3. d 4. a 5. c 6. c Question 7 (7.a) (i) Stratified sampling. This is because the business owners have been stratified by business types. There are 5 strata; that is stratum 1 = retailers, stratum 2 = agriculture, stratum 3 = manufacturing, stratum 4 = financial services and stratum 5 = advertising. For each stratum we select a sample size of 6 owners. (ii) Simple random sampling. This is because it uses a random number generator to select the business owners from the state Directory of Businesses. (7.b) Running multiple t tests of each population mean against all the others separately will lead to a build-up of type I errors, i.e. for each test there is a probability of rejecting the null hypothesis of equality when in fact the null hypothesis is true (our chosen α value equals the type I error). Thus the probability of rejecting the null that all the population means are equal is much higher than α when doing these multiple tests. (7.c) The F-statistic is ππππππ 9600 1920 ππ 5 πΉπΉ = = = = 27.2 ππππππ 2400 70.59 ππ − ππ − 1 40 − 5 − 1 (7.d) True. When every observation is on the regression line, π¦π¦ππ = π¦π¦οΏ½ππ and thus the sum of squares for error ππππππ = ∑ππππ=1(π¦π¦ππ − π¦π¦οΏ½ππ )2 = 0. This means that the standard error of estimate ππππ = οΏ½ and thus π π 2 = 1 − ππππππ πππππ¦π¦ = 1. ππππππ ππ−ππ−1 =0 (7.e) (i) Since nA and nB are less than 10, the Wilcoxon Rank Sum test T = TA. It is a one-tailed test and πΌπΌ = 0.025. Thus the decision rule is reject H0 if ππ = πππ΄π΄ ≤ πππΏπΏ = 12. (ii) since nA and nB are more than 10, the Wilcoxon Rank Sum test T = z. It is a onetailed test and πΌπΌ = 0.05. Thus the decision rule is reject H0 if ππ = |π§π§| > π§π§πΌπΌ/2 = π§π§0.025 = 1.96. (7.f) We can remove the variation that we are not interested in. For example, in Week 4 we ask the question whether the new design tyres last longer, on average, than existing tyres. Independent samples can lead to more variability in our outcome as different drivers (for the new and existing tyres) may drive different ways. That is, some drivers drove in a way that extended the life of the tyre, while others drove faster and braked harder, resulting in shorter tyre lives. But, we are not interested in the variation among drivers, that is we are not interested in knowing whether drivers actually differ in the way they drive. We can eliminate that source of variation by designing a matched pairs experiment -- same drivers and cars were used in both new and existing tyres samples. This makes it easier to determine if tyre brand represented a real source of variation, and thus that one tyre design is superior to another tyre design. (7.g) (i) ππΜ = 0.55 and sampling error ππ = 0.03. So the 95% confidence interval is ππΜ ± ππ 0.55 ± 0.03 The lower confidence limit is 0.52 (0.55 minus 0.03) and the upper confidence limit is 0.58 (0.55 plus 0.03). (ii) The confidence interval is entirely above 50% because it is (0.52 , 0.58) , so it is reasonable to say that more than 50% of the population thinks their weight is about right. Question 8 (8.a) The test is called the Pooled-Variance-t statistic. This test is appropriate for quantitative data (i.e. total spending in dollars on hair care products), independent samples (i.e. 210 adults live in Melbourne and 250 adults live in Sydney) and assumed unknown equal population variances. (8.b) You could construct a histogram of the data in the samples and observe the pattern. If the histogram reveals a nice bell-shaped and symmetrical distribution, then we can be confident that the data is normally distributed. (8.c) Let ππ1 be the population mean for the spending on hair care products in Melbourne. Let ππ2 be the population mean for the spending on hair care products in Sydney. H0 : µ1 – µ2 = 0 HA : µ1 – µ2 < 0 (8.d) Pooled-Variance-t statistic (1) Construct the sample means and sample variances for Sydney π₯π₯Μ 2 , π π 22 and Melbourne π₯π₯Μ1 , π π 12 (2) Construct the Pooled-Variance-t test (π₯π₯Μ 1 − π₯π₯Μ 2 ) π‘π‘ = 1 1 οΏ½π π ππ2 οΏ½ + οΏ½ ππ1 ππ2 where π π ππ2 = (ππ1 − 1)π π 12 + (ππ2 − 1)π π 22 ππ1 + ππ2 − 2 Question 9 (9.a) The estimated logit model is οΏ½ ππππππππππππππ, ππππππππππππππππ) ππππ(π π π π π π π π π π = 1|ππππππ, ππππππππ, = πΉπΉ(1.656 − 0.016π΄π΄π΄π΄π΄π΄ − 0.111πΈπΈπΈπΈπΈπΈπΈπΈ − 0.004πΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆ − 0.466 π π π π π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘) (9.b) (i) The negative coefficient on AGE implies that the probability that an individual smokes is lower for older people, holding EDUC, CIGPRIC and RESTAURN fixed. (ii) The negative coefficient on EDUC implies that more educated people (more years of schooling) are less likely to smoke, holding AGE, CIGPRIC and RESTAURN fixed. (9.c) The negative coefficient on CIGPRIC implies that the probability that an individual smokes is lower for people in states where the price of cigarettes including taxes is higher, holding AGE, EDUC and RESTAURN fixed. Step 1 Step 2 π»π»0 : π½π½3 = 0 π»π»π΄π΄ : π½π½3 < 0 οΏ½ π½π½ 3 Test statistic is π‘π‘ = ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ The t-statistic has a standard normal distribution when ππ is large. Step 3 α = 0.05 Step 4 Decision rule: Reject H0 if t < -zα = -z0.05 = -1.645 Step 5 From the EViews output π‘π‘ = οΏ½3 π½π½ −0.0035 = = −0.23 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ 0.0155 Step 6 Since t = -0.23 > -1.645, we fail to reject the null hypothesis. The data does not support the hypothesis that increasing the cost (higher price including taxes) of smoking reduces the probability of people smoking. This analysis suggests that increasing cigarette taxes to discourage smoking may not be an effective government policy. (9.d) The probability that Ms. A smokes is οΏ½ ππππππππππππππ, ππππππππππππππππ) ππππ(π π π π π π π π π π = 1|ππππππ, ππππππππ, = πΉπΉ(1.656 − 0.016(40) − 0.111(13) − 0.004(60) − 0.466) = πΉπΉ(−1.133) 1 = = 0.24361 or 24.36% 1 + ππ −(−1.133) (9.e) The probability that Ms. B smokes is οΏ½ ππππππππππππππ, ππππππππππππππππ) ππππ(π π π π π π π π π π = 1|ππππππ, ππππππππ, = πΉπΉοΏ½1.656 − 0.016(40) − 0.111(13) − 0.004(60)οΏ½ = πΉπΉ(−0.667) 1 = 0.33917 or 33.92% = −(−0.667) 1 + ππ (9.f) Yes because Mr. A has a lower probability of smoking than Mr. B. Note that they are both identical in every way (ie same educ, cigpric and age) except the state in which they live. Question 10 (10.a) In 2005Q2, π‘π‘ = 122 (because (2004 − 1975 + 1) × 4 + 2 = 122) and so π¦π¦οΏ½122 = 4713.07 + 37.1(122) − 582.23 = 8657.0, which is $8657 millions. (10.b) Not enough. From the AR(1) model, we calculate the forecast for the period 2005Q2 as follows: οΏ½ 2005ππ2 = 48.04 − 0.26Δπ¦π¦2005ππ1 Δπ¦π¦ So to answer this question I need to construct Δπ¦π¦π‘π‘ = π¦π¦π‘π‘ − π¦π¦π‘π‘−1 for the final period 2005Q1. That is, I need to calculate Δπ¦π¦2005ππ1 = π¦π¦2005ππ1 − π¦π¦2004ππ4 . I know π¦π¦2005ππ1 = 9138 from the table. But the table does not provide a value for π¦π¦2004ππ4 . Thus, I cannot calculate Δπ¦π¦2005ππ1 and subsequently I cannot forecast the value of π¦π¦π‘π‘ for the period 2005Q2. In short, the table needs to provide a value for π¦π¦2004ππ4 . Question 11 (11.a) The estimated coefficient on LAND_SIZE equals 0.112. This tells us that a one metre squared increase in the size of the house block is associated with, on average, a $112 increase in house selling price, holding number of bedrooms and street number fixed. (11.b) The coefficient on SN4 equals -50.486. This implies that the average sale price of houses with the number “4" in their street number is $50,486 less than that for houses without a “4" in their street number, holding land size and number of bedrooms fixed. This negative sign is what we would expect, as the number “4" is supposed to bring bad luck, reducing demand for such houses and hence a lower house sale price. (11.c) The regression model is ππππππππππππ = π½π½0 + π½π½1 ππππππππ_π π π π π π π π ππ + π½π½2 πππππππππππππππ π ππ + π½π½3 ππππ4 + π½π½4 ππππ8 + ππππ Testing for ‘good luck’ according to the sales agent’s theory and so this is a right onetailed test. Let π½π½4 denote the coefficient for SN8. So we expect π½π½4 > 0 because street number containing “8” will have a higher sale price than house without the street number “8”. Step 1 Step 2 The test statistic is the t-statistic Step 3. πΌπΌ = 0.05 π»π»0 : π½π½4 = 0 π»π»π΄π΄ : π½π½4 > 0 π‘π‘ = π½π½Μ4 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ Because it is a one-tail test (and not a two-tail test), we cannot use the p-value approach. See Week 8 lecture slides. So we will use the critical value approach. Step 4. Decision Rule. Reject π»π»0 if π‘π‘ > π‘π‘πΌπΌ,ππ−ππ−1 = π‘π‘0.05,28−4−1 = 1.714 Step 5. The value of the t-test from the EViews output is π½π½Μ4 38.97 π‘π‘ = = = 1.51 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ 25.895 Step 6. Fail to reject the null hypothesis since t = 1.51 < 1.714 and so insufficient evidence in support of the agent's theory here. (11.d) The hypothesis that BEDROOMS has an effect on Price and so we will conduct a two-tail test since it says “an effect” without saying positive effect or negative effect. Step 1 π»π»0 : π½π½2 = 0 π»π»π΄π΄ : π½π½2 ≠ 0 Step 2 The test statistic is the t-statistic π‘π‘ = π½π½Μ2 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ Critical value can be obtained from the student t-table. Step 3. πΌπΌ = 0.05 Because we are testing the null hypothesis π½π½2 equals to zero and it is a two-tail test, we can either use the critical value approach or the p-value approach. See Week 8 lecture slide. However the EViews output does not give us the p-value and so we will use the critical value approach. Step 4. Decision Rule. Reject π»π»0 if |π‘π‘| > π‘π‘πΌπΌ/2,ππ−ππ−1 = π‘π‘0.05/2,28−4−1 = 2.069 Step 5. We can calculate the value of the t-test as π½π½Μ2 86.39 π‘π‘ = = = 12.99 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ 6.65 Step 6. Reject the null hypothesis since t =12.99 >2.069 and so sufficient evidence that BEDROOMS has an effect on PRICE. (11.e) Step 1 π»π»0 : π½π½1 = π½π½2 = π½π½3 = π½π½4 = 0 π»π»π΄π΄ : π΄π΄π΄π΄ ππππππππππ ππππππ π½π½ππ (ππ = 1,2,3,4) ππππ ππππππ ππππππππππ π‘π‘π‘π‘ π§π§π§π§π§π§π§π§ Step 2 The test statistic is the F-statistic ππππππ/ππ ππππππ/(ππ − ππ − 1) Critical value can be obtained from the F-table. πΉπΉ = Step 3. πΌπΌ = 0.05 Step 4. Decision Rule. Reject π»π»0 if πΉπΉ > πΉπΉπΌπΌ,ππ,ππ−ππ−1 = πΉπΉ0.05,4,28−4−1 = 2.8 Step 5. Since F-test is not given, we can however calculate the F-test by using information from the π π 2 . Recall from Tutorial 8 (Week 9) question B2. To compute the F-test, we need to know the values of SSR and SSE. From the EViews output, SSE = 38847.96 (which is the Sum squared resid) To calculate SSR, we note from the formula sheet that π π 2 is πππππΈπΈ π π 2 = 1 − πππππ¦π¦ From the EViews output π π 2 = 0.904. ππππππ πππππ¦π¦ 38847.96 0.904 = 1 − πππππ¦π¦ π π 2 = 1 − After re-arranging, we have πππππ¦π¦ = We can now calculate SSR as 38847.96 = 404666.25 1 − 0.904 πππππ¦π¦ = ππππππ + ππππππ 404666.25 = ππππππ + 38847.96 After re-arranging, we have πππππ π = 365818.29 From the EViews output, n = 28 and k = 4. Now the F-test can be calculated as ππππππ/ππ 365818.29/4 πΉπΉ = = = 54.15 ππππππ/(ππ − ππ − 1) 38847.96/23 Step 6. Reject the null hypothesis since F = 54.15 > 2.8 and so sufficient evidence to infer that the model is useful. (11.f) The standard error of regression is denoted as π π ππ , which is the S.E. of regression in the EViews output. See Tutorial 7 (Week 8) or Week 7 lecture slides. From the formula sheet, ππππππ 38847.96 =οΏ½ = 41.098 π π ππ = οΏ½ ππ − ππ − 1 28 − 4 − 1 The standard error of regression is $41,098 (as stated in Week 8 lecture slides π π ππ is measured in the units of dependent variable and in this case the dependent variable is measured in thousands of dollars). This means that the standard deviation of the residuals ππΜππ is $41,098. (11.g) Step 1 π»π»0 : π½π½1 = 0 π»π»π΄π΄ : π½π½1 > 0 Step 2 The test statistic is the t-statistic π‘π‘ = π½π½Μ2 ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ Critical value can be obtained from the student t-table. Step 3. πΌπΌ = 0.05 We will use the critical value approach. Step 4. (same as question 11.c above) Decision Rule. Reject π»π»0 if π‘π‘ > π‘π‘πΌπΌ,ππ−ππ−1 = π‘π‘0.05,28−4−1 = 1.714 Step 5. We cannot calculate the t-test because 0.11 π½π½Μ1 = = π’π’π’π’π’π’π’π’π’π’π’π’π’π’ π‘π‘ = ππππππ. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ ππππππ πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ “Not enough”. To answer this question, we need to know the Std. Error in order to calculate the t-test. Using the t-test we can determine whether we reject π»π»0 or not.