Research School of Finance, Actuarial Studies and Statistics PAST FINAL EXAMINATION 4 SOLUTIONS STAT7055 Introductory Statistics for Business and Finance Writing Time: 180 minutes Reading Time: 15 minutes Exam Conditions: Central examination. Students must return the examination paper at the end of the examination. This examination paper is not available to the ANU Library archives. Materials Permitted in the Exam Venue: (No electronic aids are permitted, e.g., laptops, phones). Calculator (non-programmable). Two A4 pages with notes on both sides. Unannotated paper-based dictionary (no approval required). Materials to be Supplied to Students: Script book. Scribble paper. Instructions to Students: Please write your student number in the space provided on the front of the script book. Attempt all 5 questions. Start your solution to each question on a new page and clearly label each solution with the corresponding question number. To ensure full marks show all the steps in working out your solutions. Marks may be deducted for failure to show working or formulae. Selected statistical tables are attached to the back of the examination paper. If a required degree of freedom is not listed in a statistical table, please use the closest degree of freedom. Unless otherwise stated, use a significance level of α = 5%. Round all numeric answers to 4 decimal places. Question: 1 2 3 4 5 Total Marks: 21 21 18 21 22 103 Question 1 [21 marks] There are many things which can affect the price of a second hand car. Data was collected on 105 second hand car sales. A multiple linear regression model was fitted with sale price as the dependent variable (Y ), and the odometer reading (X1), the odometer reading squared (X21 ), the age (X2) and an indicator of whether the car has an automatic transmission (Z = 1 if the car has an automatic transmission and Z = 0 otherwise) as the independent variables. That is, the following model was fitted: Y = β0 + β1X1 + β2X 2 1 + β3X2 + β4Z + Note that sale price was measured in thousands of dollars (e.g., Y = 32 corresponds to a sale price of 32 000 dollars), odometer reading was measured in thousands of kilometres (e.g., X1 = 19.1 corresponds to 19 100 kilometres) and age was measures in years. The regression output, which includes some missing entries, is displayed below: Predictor Coef SE Coef T p-value Intercept 57.5629 5.7896 9.94 0.0000 Odometer 2.03 0.0455 Odometer2 1.25 0.2136 Age 0.1216 0.1441 0.84 0.4009 Z 0.1707 0.6040 0.28 0.7780 Analysis of Variance Source DF SS MS F p-value Regression Residual Error Total 5783.875 (a) [4 marks] The adjusted R2 for the model is equal to 0.8544375. Test the overall significance of the model. Clearly state your hypotheses and use a significance level of α = 5%. Solution: The hypotheses are: H0 : β1 = β2 = β3 = β4 = 0 H1 : Not all coefficient parameters are equal to 0. In order to calculate the F -statistic, we need to know the SSR and the SSE. From the given values of the adjusted R2 = 0.8544375, SS(Total) = 5783.875, Past Final Examination 4 Page 2 of 22 STAT7055 n = 105 and k = 4, we can calculate the SSE: adj R2 = 1 SSE n k 1 SS(Total) n 1 0.8544375 = 1 SSE 105 4 1 5783.875 105 1 SSE = (1 0.8544375)× 5783.875 104 × 100 SSE = 809.5339 We can then calculate the SSR: SSR = SS(Total) SSE = 4974.3411 Now we can calculate our F -statistic: F = MSR MSE = SSR k SSE n k 1 = 4974.3411 4 809.5339 100 = 153.6174 We need to compare this to an F -distribution with k = 4 numerator degrees of freedom and n k 1 = 100 denominator degrees of freedom and reject H0 when F > F0.05,4,100 = 2.46. Since 153.6174 > 2.46, we reject H0 and conclude that the overall regression model is significant. (b) [2 marks] What do you conclude about the relationship between sale price and odometer reading squared Clearly state your hypotheses and use a significance level of α = 5%. Solution: The hypotheses are: H0 : β2 = 0 H1 : β2 6= 0 From the output, we see that the p-value for this test is 0.2136, which is greater than 0.05, so we fail to reject H0 and we conclude that β2 = 0. That is, once the other independent variables have been included in the model, there is no significant linear relationship between sale price and odometer reading squared. (c) [2 marks] Test whether a different intercept is needed for cars that have an au- tomatic transmission. Clearly state your hypotheses and use a significance level of α = 5%. Past Final Examination 4 Page 3 of 22 STAT7055 Solution: To see if a different value of the intercept is needed for cars that have an auto- matic transmission, we are testing: H0 : β4 = 0 H1 : β4 6= 0 From the output, we see that the p-value for this test is 0.7780, which is greater than 0.05, so we fail to reject H0 and we conclude that β4 = 0. That is, once the other independent variables have been included in the model, a different intercept is not needed for cars that have an automatic transmission. (d) [3 marks] Test whether the expected change in sale price when the age increases by 1 year (all other variables held constant) is less than +0.15 (that is, less than positive 150 dollars). Clearly state your hypotheses and use a significance level of α = 5%. Solution: The expected change in sale price when the age increases by 1 year (all other variables held constant) is, by definition, β3. So the hypotheses are: H0 : β3 = 0.15 H1 : β3 < 0.15 We can no longer use the p-value from the output to perform this test and we must manually calculate the T -statistic: T = β 3 c sβ 3 = 0.1216 0.15 0.1441 = 1.8848 We need to compare this to a t-distribution with n k 1 = 100 degrees of freedom and reject H0 when T < t0.05,100 = 1.660. Since 1.8848 < 1.660, we reject H0 and we conclude that the expected change in sale price when the age increases by 1 year (all other variables held constant) is less than +0.15 (e) [5 marks] Using the estimated regression model, the predicted sale price for a car that is 4 years old, travelled 32 000 km (X1 = 32) and has a manual transmission is y = 22.81569 and the predicted sale price for a car that is 7 years old, travelled 29 000 km (X1 = 29) and has an automatic transmission is y = 26.22919. Based on this information, calculate the estimates β 1 and β 2. Past Final Examination 4 Page 4 of 22 STAT7055 Solution: Using the estimated regression model, we get the following equations from the predicted sale prices for the two cars: 22.81569 = 57.5629 + β 1 × 32 + β 2 × 322 0.1216× 4 0.1707× 0 26.22919 = 57.5629 + β 1 × 29 + β 2 × 292 0.1216× 7 0.1707× 1 Some algebra reduces these equations to: 32β 1 + 1024β 2 = 34.2608 (1) 29β 1 + 841β 2 = 30.3118 (2) From the first equation we get: β 1 = 34.2608 1024β 2 32 (3) Substituting this into the second equation and solving for β 2 we get: β 2 = 0.0085 Substituting β 2 back into the third equation we get: β 1 = 0.7996 When people look to buy second hand cars, the odometer reading is generally the first thing they check. Given this, a simple linear regression was fitted with sale price (Y ) as the dependent variable and the odometer reading (X1) as the independent variable. The regression output is given below: Predictor Coef SE Coef T Intercept 63.3897 1.5936 39.78 Odometer 1.2906 0.0520 24.83 (f) [2 marks] Test the overall significance of the model. Clearly state your hypotheses and use a significance level of α = 5%. Solution: Past Final Examination 4 Page 5 of 22 STAT7055 The hypotheses are: H0 : β1 = 0 H1 : β1 6= 0 From the output, we see that the T -statistic for this test is equal to T = 24.83. We need to compare this to a t-distribution with n 2 = 103 degrees of freedom and reject H0 when T > t0.025,103 ≈ t0.025,100 = 1.984 or T < t0.025,103 ≈ 1.984. Since 24.83 < 1.984, we reject H0 and we conclude that the overall regression model is significant. (g) [3 marks] The following sample statistics for the odometer readings are given: Xˉ1 = 30.19046 and s 2 X1 = 28.60936. Calculate a 90% prediction interval for the sale price of a car that has travelled 39 000 km (X1 = 39) given that the standard error of estimate is s = 2.835572. Solution: A 90% prediction interval for the sale price of a car that has travelled 39 000 km is given by: ( y g ± tα 2 ,n 2 × s √ 1 + 1 n + (x1g Xˉ1)2 (n 1)s2X1 ) where y g = 13.0563, t0.05,103 ≈ t0.05,100 = 1.660, x1g = 39, Xˉ1 = 30.19046, n = 105, s2X1 = 28.60936 and s = 2.835572. Plugging this all in we get: (8.2662, 17.8464) Question 2 [21 marks] The 100 metre sprint is one of the most watched events in any Olympic Games. Sprinters will often go to great lengths to improve their times, looking into factors such as the equipment they use and the coaches they hire. Listed in the table below are the personal best times for the 100 metre sprint for 27 sprinters that are in training. The 27 sprinters were chosen as follows: For each of three shoe brands (the Fast, the Quick and the Speedy brand), 9 sprinters who used that particular brand were randomly selected. The sample variances of the times for each shoe brand are also listed in the table. (a) [6 marks] Test whether the mean time for sprinters using the speedy brand is more than 0.2 seconds faster than for sprinters using the quick brand. Clearly state your hypotheses and use a significance level of α = 5%. Clearly state any assumptions you have made (without testing them) when performing this test. Past Final Examination 4 Page 6 of 22 STAT7055 Shoe Brand Times s2 Fast 9.82 10.22 10.15 9.77 10.12 9.86 9.87 10.11 10.18 0.0313111 Quick 9.67 9.95 10.10 10.16 9.98 10.09 10.20 10.68 10.29 0.0753278 Speedy 9.69 9.75 9.49 10.03 9.66 10.24 10.04 9.87 10.01 0.0557528 Solution: Let μs and σ 2 s denote the population mean and variance, respectively, of times for the speedy brand and let μq and σ 2 q denote the corresponding values for the quick brand. We will use a T -statistic to compare the population means and the main assumption we are making is that the two population variances are equal, i.e., σ2s = σ 2 q . The hypotheses we want to test are: H0 : μq μs = 0.2 H1 : μq μs > 0.2 Letting Y represent times, from the table we can calculate Yˉs = 9.8644 and Yˉq = 10.1244. From the given sample variances for the speedy and quick brands, we can calculate the pooled sample variances to be s2p = 0.0655. The test statistic is therefore: T = (Yˉq Yˉs) D0√ s2p ( 1 nq + 1 ns ) = (10.1244 9.8644) 0.2√ 0.0655 ( 1 9 + 1 9 ) = 0.4972 We need to compare this to a t-distribution with nq + ns 2 = 16 degrees of freedom and reject H0 when T > t0.05,16 = 1.746. Since 0.4972 < 1.746, we fail to reject H0 and we conclude that the mean time for sprinters using the speedy brand is not more than 0.2 seconds faster than those using the quick brand. A one-way ANOVA was performed on this data, using shoe brand as the factor. The partially filled ANOVA table is provided below: Source Sum of squares Degrees of freedom Mean squares F Shoe Brand 0.3059 Error Total (b) [2 marks] Calculate the sum of squares for error for the one-way ANOVA. Past Final Examination 4 Page 7 of 22 STAT7055 Solution: The sum of squares for the one-way ANOVA can be calculated from the given sample variances: SSE = k∑ j=1 nj∑ i=1 (Yij Yˉj)2 = nf∑ i=1 (Yif Yˉf )2 + nq∑ i=1 (Yiq Yˉq)2 + ns∑ i=1 (Yis Yˉs)2 = (nf 1)× s2f + (nq 1)× s2q + (ns 1)× s2s = 8× (0.0313111 + 0.0753278 + 0.0557528) = 1.2991 (c) [3 marks] Test whether there is a difference in the mean times between the three shoe brands. Clearly state your hypotheses and use a significance level of α = 5%. Solution: Let μf , μq and μs denote the population mean times for the fast, quick and speedy shoe brands, respectively. The hypotheses are: H0 : μf = μq = μs H1 : At least two population means differ. From the ANOVA table we know that SST = 0.3059 and from part (b) we know that SSE = 1.2991. Since n = 27 and k = 3, we can calculate the F -statistic: F = MST MSE = SST k 1 SSE n k = 0.3059 2 1.2991 24 = 2.8256 We need to compare this to an F -distribution with k 1 = 2 numerator and n k = 24 denominator, degrees of freedom, and reject H0 when F > F0.05,2,24 = 3.40. Since 2.8256 < 3.40, we fail to reject H0 and we conclude that there is no difference in mean times between the three shoe brands. For each shoe brand, suppose the first 3 sprinters in the table were trained by coach Andy, the next 3 sprinters were trained by coach Bobby and the last 3 sprinters were trained by coach Carl. A two-way ANOVA was performed on the same data, using shoe brand and coach as factors. The partially filled ANOVA table is displayed below: (d) [3 marks] Test whether there is an interaction between coach and shoe brand. Clearly state your hypotheses and use a significance level of α = 5%. Past Final Examination 4 Page 8 of 22 STAT7055 Source Sum of squares Degrees of freedom Mean squares F Coach Shoe Brand Interaction 0.2970 Error 0.6781 Total Solution: The hypotheses are: H0 : There is no interaction between coach and shoe brand. H1 : There is an interaction. The SS(Int) and the SSE are given in the ANOVA table, and we can calculate df(Int) = (3 1) × (3 1) = 4 and df(E) = 27 3 × 3 = 18. Therefore, the F -statistic is: F = MS(Int) MSE = SS(Int) df(Int) SSE df(E) = 0.2970 4 0.6781 18 = 1.9709 We need to compare this to an F -distribution with 4 numerator and 18 denom- inator, degrees of freedom, and reject H0 when F > F0.05,4,18 = 2.93. Since 1.9709 < 2.93, we fail to reject H0 and we conclude that there is no interaction. (e) [3 marks] Test whether there is a difference in the mean times between the three shoe brands. Clearly state your hypotheses and use a significance level of α = 1%. Solution: The hypotheses are: H0 : μf = μq = μs H1 : At least two population means differ. The SS(Shoe) in the two-way ANOVA is the same as the SST from the one-way ANOVA of part (c) so the SS(Shoe) = 0.3059. The F -statistic is then equal to: F = MS(Shoe) MSE = SS(Shoe) df(Shoe) SSE df(E) = 0.3059 2 0.6781 18 = 4.0600 We need to compare this to an F -distribution with 2 numerator and 18 denom- inator, degrees of freedom, and reject H0 when F > F0.01,2,18 = 6.01. Since Past Final Examination 4 Page 9 of 22 STAT7055 4.0600 < 6.01, we fail to reject H0 and we conclude that there is no difference in mean times between the three shoe brands. (f) [4 marks] Test whether there is a difference in the mean times between the three coaches. Clearly state your hypotheses and use a significance level of α = 1%. Solution: Let μa, μb and μc denote the population mean times for coach Andy, Bobby and Carl, respectively. The hypotheses are: H0 : μa = μb = μc H1 : At least two population means differ
. In order to determine the SS(Coach), we need to find the total sum of squares SS(Total). Since we are using the same data, the SS(Total) for the two-way ANOVA is the same as the SS(Total) for the one-way ANOVA. Hence from the one-way ANOVA: SS(Total) = SST + SSE = 0.3059 + 1.2991 = 1.6050 From the SS(Total), we can now determine the SS(Coach): SS(Coach) = SS(Total) SS(Shoe) SS(Int) SSE = 1.6050 0.3059 0.2970 0.6781 = 0.3240 Hence we calculate the F -statistic to be: F = MS(Coach) MSE = SS(Coach) df(Coach) SSE df(E) = 0.3240 2 0.6781 18 = 4.3003 We need to compare this to an F -distribution with 2 numerator and 18 denom- inator, degrees of freedom, and reject H0 when F > F0.01,2,18 = 6.01. Since 4.3003 < 6.01, we fail to reject H0 and we conclude that there is no difference in mean times between the three coaches. Question 3 [18 marks] We have a room of 6 people, all of who are celebrating their birthday today. Each person’s age (the age that they turned today) is listed below: 31, 18, 21, 21, 18, 29 (a) [3 marks] A study has shown that 70% of people younger than 22 are likely to Past Final Examination 4 Page 10 of 22 STAT7055 hold a party on their birthday, whereas only 22% of people older than 22 are likely to hold a party on their birthday. For a person randomly selected from the room of 6 people, find the probability that they will be holding a birthday party today. Solution: Let BP denote the event that the selected person will hold a birthday party today and let the age of the selected person be denoted by the random variable X. We can use the law of total probability by considering the events {X < 22}, {X = 22} and {X > 22}. For a person randomly selected from the room, we know that P (X < 22) = 2 3 , P (X = 22) = 0 and P (X > 22) = 1 3 . So we get: P (BP ) = P (BP ∩ {X < 22}) + P (BP ∩ {X = 22}) + P (BP ∩ {X > 22}) = P (BP |{X < 22})P (X < 22) + P (BP |{X = 22})P (X = 22) + P (BP |{X > 22})P (X > 22) = 0.7× 2 3 + P (BP |{X = 22})× 0 + 0.22× 1 3 = 27 50 = 0.54 Suppose that we select a sample of 2 people from this room (without replacement). (b) [2 marks] Find the probability that the sample of size 2 contains the oldest person in the room. Solution: Let SO denote the event that the sample contains the oldest person in the room and let Xi denote the age of the ith person selected in the sample. Since 31 is the oldest age in the room, the event SO can be denoted as {X1 = 31}∪{X2 = 31}. Therefore: P (SO) = P ({X1 = 31} ∪ {X2 = 31}) = P (X1 = 31) + P (X2 = 31) P ({X1 = 31} ∩ {X2 = 31}) = 1 6 + 1 6 0 = 1 3 = 0.3333 (c) [2 marks] Find the probability that the oldest person in the sample of size 2 is younger than 24 years old. Solution: Past Final Examination 4 Page 11 of 22 STAT7055 Let Xi be defined as in part (b). For the oldest person in the sample to be younger than 24, both the first person selected and the second person selected must be younger than 24. Therefore we want to find the following probability: P ({X1 < 24} ∩ {X2 < 24}) = P ({X2 < 24}|{X1 < 24})P (X1 < 24) = 3 5 × 4 6 = 2 5 = 0.4 (d) [4 marks] Determine the sampling distribution of the age of the oldest person in the sample of size 2. Solution: Let Y denote the age of the oldest person in the sample. The following table lists all possible samples of size 2 selected without replacement, along with the age of the oldest person in each sample: Sample y Sample y (31, 18) 31 (18,29) 29 (31, 21) 31 (21, 21) 21 (31, 21) 31 (21, 18) 21 (31, 18) 31 (21, 29) 29 (31, 29) 31 (21, 18) 21 (18, 21) 21 (21, 29) 29 (18, 21) 21 (18, 29) 29 (18, 18) 18 Each sample is equally likely to be selected, so we obtain the sampling distri- bution of Y by counting the possible outcomes. y 18 21 29 31 p(y) 1 15 5 15 = 1 3 4 15 5 15 = 1 3 (e) [3 marks] Calculate the expected value and variance of the sampling distribution of the age of the oldest person in the sample of size 2. Solution: Past Final Examination 4 Page 12 of 22 STAT7055 E(Y ) = ∑ all y (y × p(y)) = 18× 1 15 + 21× 1 3 + 29× 4 15 + 31× 1 3 = 394 15 = 26.2667 V (Y ) = E(Y 2) (E(Y ))2 = ∑ all y ( y2 × p(y)) (394 15 )2 = 182 × 1 15 + 212 × 1 3 + 292 × 4 15 + 312 × 1 3 ( 394 15 )2 = 5234 225 = 23.2622 (f) [4 marks] Suppose we took 10 samples of size 2, without replacement, from the room of 6 people (but before each new sample was taken, the previous sample was returned to the room). Find the probability that in more than 7 samples the oldest person in the sample was older than 30. Solution: Let W denote the number of the 10 samples for which the oldest person in the sample was older than 30. Then W ～ Bin(n = 10, p = P (Y > 30) = 1 3 ). Therefore we can calculate: P (W > 7) = P (W = 8) + P (W = 9) + P (W = 10) = 10! 8!× 2! × ( 1 3 )8 × ( 2 3 )2 + 10× ( 1 3 )9 × 2 3 + ( 1 3 )10 = 0.0034 Question 4 [21 marks] Let X and Y be independent continuous random variables with the following probability density functions: f(x) = 4 9 , 0 ≤ x < 1 2 9 x+ 6 9 , 1 ≤ x < 2 2 9 , 2 ≤ x ≤ 3 and f(y) = { y + 1, 0 ≤ y < 1 y 1, 1 ≤ y ≤ 2 (a) [3 marks] Find the probability that X is between 1.2 and 2.2. Past Final Examination 4 Page 13 of 22 STAT7055 Solution: x f(x ) 0 1 1.2 2 2.2 3 0 2/ 9 4/ 9 We want to find the red shaded area, which we can break into regions (either a triangle and two rectangles or a trapezium and a rectangle): P (1.2 < X < 2.2) = P (1.2 < X < 2) + P (2 < X < 2.2) = Area of trapezium + Area of rectangle = ( 1 2 × (Height at x = 1.2 + Height at x = 2)× (2 1.2) ) + ( 2 9 × (2.2 2) ) = ( 1 2 × ( 2 5 + 2 9 ) × 0.8 ) + ( 2 45 ) = 22 75 = 0.2933 (b) [3 marks] Find the probability that X is between 1.2 and 2.2, given that it is larger than 1.5. Solution: Past Final Examination 4 Page 14 of 22 STAT7055 xf(x ) 0 1 1.2 1.5 2 2.2 3 0 2/ 9 4/ 9 We want to find the red and blue shaded area, then divide that by the blue shaded area: P ({1.2 < X < 2.2}|{X > 1.5}) = P ({1.2 < X < 2.2} ∩ {X > 1.5}) P (X > 1.5) = P (1.5 < X < 2.2) P (X > 1.5) = P (1.5 < X < 2) + P (2 < X < 2.2) P (1.5 < X < 2) + P (X > 2) = ( 1 2 × (1 3 + 2 9 )× 0.5)+ ( 2 45 )( 1 2 × (1 3 + 2 9 )× 0.5)+ 1× 2 9 = 33 65 = 0.5077 (c) [3 marks] Find the probability that Y is between 0.7 and 1.1 or between 1.2 and 1.9. Solution: Past Final Examination 4 Page 15 of 22 STAT7055 yf(y ) 0 0.7 1 1.1 1.2 1.9 2 0 1 We want to find the red and blue shaded areas: P ({0.7 < Y < 1.1} ∪ {1.2 < Y < 1.9}) = P (0.7 < Y < 1.1) + P (1.2 < Y < 1.9) = P (0.7 < Y < 1) + P (1 < Y < 1.1) + P (1.2 < Y < 1.9) = 1 2 × 0.3× 0.3 + 1 2 × 0.1× 0.1 + 1 2 × (0.2 + 0.9)× 0.7 = 87 200 = 0.435 (d) [3 marks] Find the probability that X is between 1.2 and 2.2 or Y is between 0.7 and 1.1. Solution: Since we are told that X and Y are independent, using the Addition Rule we get: P ({1.2 < X < 2.2} ∪ {0.7 < Y < 1.1}) = P (1.2 < X < 2.2) + P (0.7 < Y < 1.1) P (1.2 < X < 2.2)× P (0.7 < Y < 1.1) = 22 75 + 1 20 22 75 × 1 20 = 493 1500 = 0.3287 (e) [2 marks] Find the probability that X is between 1 and 1.9, given that Y is between 0.7 and 1.1. Past Final Examination 4 Page 16 of 22 STAT7055 Solution: We want to find P ({1 < X < 1.9}|{0.7 < Y < 1.1}). However, since X and Y are independent, this probability is equal to P (1 < X < 1.9). x f(x ) 0 1 1.9 2 3 0 2/ 9 4/ 9 Therefore, we want to find the red shaded area, which we can calculate as the area of a trapezium: P (1 < X < 1.9) = ( 1 2 × ( 4 9 + 11 45 ) × 0.9 ) = 31 100 = 0.31 For parts (f) and (g), we can assume that μX = E(X) = 34 27 and σ2X = V (X) = 0.69204. (f) [3 marks] Find the probability that, from a sample of size n = 50 taken from the distribution of X, the sample mean lies between 1.2 and 1.5. Solution: From the CLT, we know that Xˉ ～ N ( μXˉ = μX = 34 27 , σ2 Xˉ = σ2X n = 0.69204 50 ) . P (1.2 < Xˉ < 1.5) = P 1.2 3427√ 0.69204 50 < Xˉ μX√ σ2X n < 1.5 34 27√ 0.69204 50 = P ( 0.50 < Z < 2.05) = P (Z < 2.05) P (Z < 0.50) = 0.9798 0.3085 = 0.6713 (g) [4 marks] Suppose a sample of size 50 taken from the distribution of Y produced the following sample statistics: ∑50 i=1 Yi = 44 and ∑50 i=1 ( Yi Yˉ )2 = 45.08. Based Past Final Examination 4 Page 17 of 22 STAT7055 on this data, test whether the expected value of Y is less than the expected value of X. Clearly state your hypotheses and use a significance level of α = 5%. Solution: We want to test the following hypotheses: H0 : μY = 34 27 H1 : μY < 34 27 From the data we can calculate Yˉ = 44 50 = 0.88 and s2Y = 45.08 49 = 0.92. We use a T -statistic to test these hypotheses: T = Yˉ μ0√ s2Y n = 0.88 34 27√ 0.92 50 = 2.7959 We need to compare this to a t-distribution with n 1 = 49 degrees of freedom and reject H0 when T < t0.05,49 ≈ t0.05,50 = 1.676. Since 2.7959 < 1.676, we reject H0 and we conclude that the expected value of Y is less than the expected value of X. Question 5 [22 marks] Suppose X, the amount of time a person stays in bed after their alarm goes off, is uniformly distributed between 5 and a minutes. Also, suppose Y , the number of minutes they are late to work is uniformly distributed between 0 and b minutes. Over 100 days, how long they slept past their alarm (X) and how late they were to work (Y ) were recorded (in minutes). However, only the number of days for which X and Y fell within certain ranges was reported in the table below: 5 < X < 7 7 < X < 8 8 < X < 10 Totals 0 < Y < 2 18 9 12 2 < Y < 3 9 4 12 3 < Y < 5 13 9 14 Totals 100 (a) [5 marks] Based on the data given above, test whether a is equal to 10. Clearly state your hypotheses and use a significance level of α = 5%. Solution: Past Final Examination 4 Page 18 of 22 STAT7055 The given ranges of X values can be thought of as a categorical variable with 3 categories. If X was uniformly distributed between 5 and a = 10 minutes, then the observed count in each range should be close to the expected count we would see under the given uniform distribution. Since we are comparing observed counts to expected counts (for a single categorical variable), we should be performing a Chi-squared goodness-of-fit test. The hypotheses are: H0 : a = 10 (i.e., X is uniformly distributed between 5 and 10) H1 : a 6= 10 (i.e., X is not uniformly distributed between 5 and 10) We need to calculate the probability of falling into each category under the assumption that a = 10. We use the uniform probability density function to calculate the probabilities: p1 = P (5 < X < 7) = 2 5 , p2 = P (7 < X < 8) = 1 5 , p3 = P (8 < X < 10) = 2 5 So now we can also state our hypotheses as: H0 : p1 = 2 5 , p2 = 1 5 , p3 = 2 5 H1 : The probabilities do not match that given above. From H0, we calculate the expected counts to be e1 = 40, e2 = 20 and e3 = 40. The observed counts are obtained from the column totals and are equal to f1 = 40, f2 = 22, f3 = 38. The χ 2-statistic is therefore equal to: χ2 = 3∑ i=1 (fi ei)2 ei = 0.3 We need to compare this to a chi-squared distribution with 5 1 = 4 degrees of freedom and reject H0 when χ 2 > χ20.05,2 = 5.99. Since 0.3 < 5.99, we fail to reject H0 and we conclude that a = 10. (b) [3 marks] Calculate a 90% confidence interval for the probability (population pro- portion) that the person sleeps in between 5 and 7 minutes and is between 0 and 2 minutes late for work. Solution: Let p denote the probability that a person sleeps in between 5 and 7 minutes and is between 0 and 2 minutes late for work, i.e.: p = P ({5 < X < 7} ∩ {0 < Y < 2}) Past Final Examination 4 Page 19 of 22 STAT7055 From the table of counts, we calculate the sample proportion to be p = 18 100 = 0.18. Hence a 90% confidence interval for p is given by: p ± zα 2 × √ p (1 p ) n = 0.18± 1.645× √ 0.18× 0.82 100 = (0.1168, 0.2432) (c) [4 marks] Test whether the probability (population proportion) that the person is between 2 and 3 minutes late for work is greater than 0.18. Clearly state your hypotheses and use a significance level of α = 1%. Solution: Let p denote the probability that the person is between 2 and 3 minutes late for work. The hypotheses are: H0 : p = 0.18 H1 : p > 0.18 From the data we can calculate the sample proportion, p = 25 100 = 0.25. The test statistic is given by: Z = p p0√ p0(1 p0) n = 0.25 0.18√ 0.18×(1 0.18) 100 = 1.8220 As this is an upper-tailed test, the rejection region is given by Z > z0.01 = 2.33. Since 1.8220 < 2.33, we fail to reject H0 and we conclude that the probability is not greater than 0.18. (d) [5 marks] Based on the data above, test whether X and Y are independent (that is, whether the rows and columns are independent). Clearly state your hypotheses and use a significance level of α = 5%. Solution: We should perform a Chi-squared test of a contingency table. The hypotheses are: H0 : X and Y are independent. H1 : X and Y are not independent. Under the assumption of independence, the expected counts are calculated using the row and column totals and are given in the table below: Past Final Examination 4 Page 20 of 22 STAT7055 5 < X < 7 7 < X < 8 8 < X < 10 0 < Y < 2 15.6 8.58 14.82 2 < Y < 3 10 5.5 9.5 3 < Y < 5 14.4 7.92 13.68 The test statistic is given by: χ2 = 3∑ i=1 3∑ j=1 (fij eij)2 eij = 2.3842 We need to compare this to a chi-squared distribution with (3 1)× (3 1) = 4 degrees of freedom and reject H0 when χ 2 > χ20.05,4 = 9.49. Since 2.3842 < 9.49, we fail to reject H0 and we conclude that X and Y are independent based on the data. The following sample statistics were produced from the sample of 100 sleep-in times and the sample of 100 late-for-work times: Xˉ = 7.51092, s2X = 1.894077, Yˉ = 2.488363 and s2Y = 1.999385. For part (e), assume that the samples of sleep-in times and late-for-work times are independent, and that σ2X = σ 2 Y . (e) [5 marks] Test whether a exceeds b by more than 3 minutes. Clearly state your hypotheses and use a significance level of α = 5%. Solution: The hypotheses are: H0 : a b = 3 H1 : a b > 3 However, since the data given are the sample means, we should express the hypotheses in terms of the population means, μX and μY . As X and Y are uniformly distributed, we know that μX = a+5 2 and μY = b 2 . Rearranging these equations we obtain a = 2μX 5 and b = 2μY , which means a b = 2(μX μY ) 5. Therefore we can also express the hypotheses as: H0 : μX μY = 4 H1 : μX μY > 4 Now we calculate a T -statistic: T = (Xˉ Yˉ ) D0√ s2p ( 1 nX + 1 nY ) = (7.51092 2.488363) 4√ 1.9467 ( 1 100 + 1 100 ) = 5.1823 Past Final Examination 4 Page 21 of 22 STAT7055 We need to compare this to a t-distribution with nX + nY 2 = 198 degrees of freedom and reject H0 when T > t0.05,198 ≈ t0.05,200 = 1.653. Since 5.1823 > 1.653, we reject H0 and we conclude that a exceeds b by more than 3 minutes. END OF EXAMINATION Past Final Examination 4 Page 22 of 22 STAT7055