Questions and Answers #4 Data Distributions

Questions and Answers Sheet 4 Data Distributions Question #1 Following data is the raw Data Pertaining to Total Time Hours Worked by Machinists 90 94 03 93 88 4 84 93 93 Which of the following is the probability that the machinists work for 88 hours? 2 1. 2. 3. 4. 15 1 5 4 15 3 5 Answer: Step 1 Probability : Probability is a branch of mathematics that deals with the occurrence of a random event. For example, when a coin is tossed in the air, the possible outcomes are Head and Tail. The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of outcomes and the total number of outcomes. Assume an event E can occur in r ways out of a sum of n probable or possible equally likely ways. Then 𝑟 the probability of happening of the event or its success is expressed as ; 𝑃(𝐸) = 𝑛 Step 2 Given data is the raw Data Pertaining to Total Time Hours Worked by Machinists . Here, number of machinists work for hours = 15 machinists work for 88 hours in raw data = 4 4 Probability that the machinists work for 88 hours = . 15 Question #2 Find the slope of the regression line for the following data. x y 2 1 3 3 4 2 We have written the regression line in the form 𝑦 = 𝑏0 + 𝑏1 𝑥 . So in this problem you only need to find the value of 𝑏1 . Round your answer to two decimal digits after the decimal point. The answer might be negative. So, your answer could be a number like −2.07, or 0.58, etc. Answer: Step 1 x y 2 1 3 3 4 2 Step 2 The formula for 𝑏1 is as follows: 𝑏1 = 𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦) 𝑛(∑ 𝑥 2 )−(∑ 𝑥)2 Calculate the required values are as follows: x 2 3 4 x =9 y 1 3 2 y=6 xy 2 9 8  xy = 19 x2 4 9 16  x 2 = 29 Step 3 Calculate the value of 𝑏1 is as follows: 𝑏1 = = = = 𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦) 𝑛(∑ 𝑥 2 )−(∑ 𝑥)2 9(19)−(9)(6) 3(29)−(9)2 57−54 87−81 3 6 = 0.5 Therefore, the value of 𝑏1 is 0.5. Question #3 A credit card manager wishes to determine the proportion of customer calls that result in a dissatisfied customer. Based on some preliminary data, she estimates the percentage to be 10%(𝑝 = 0.10). A precision of 15% and confidence level of 90% are desired. What is the sample size? Answer: Step 1 From the given information, the percentage to be 10%(𝑝 = 0.10) Given precision of 15% and confidence interval of 90% are desired. Margin of Error is, 𝑀𝐸 = 15% of p = 0.15 ⋅ 0.10 = 0.015 Significance level is 100% − 90% = 10% . From the standard normal table, the critical value corresponding with 10% is 1.645. Step 2 Sample size: 𝑛=( =( 𝑧𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2 𝑀𝐸 1.645 2 0.015 ) 𝑝(1 − 𝑝) ) 0.10(1 − 0.10) = 1082.41 ≈ 1083 Question #4 You fit a CAPM that regresses the excess return of Coca-cola on the excess market return using 20 years of monthly data. You estimate 𝛼 = 0,72, 𝛽 = 1,37, 𝑆 2 = 20,38, 𝜎𝑋2 = 19,82 and 𝜇𝑥 = 0,71 . What are the standar errors of 𝛼 and 𝛽 ? Answer: Step 1 We know that 𝑣(𝑝) = 𝜎𝑥2 𝑠2 2 1 𝑥 𝑛 𝑠2 𝑉𝑎𝑟(𝛼) = 𝜎 2 ( + ) Standard error is given as 𝜎𝑥2 Standard error of 𝛽 = √ 𝑠2 2 1 𝑥 𝑛 𝑠2 Standard error of 𝛼 = √𝜎 2 ( + ) Step 2 given: 𝛼 = 0.72, 𝛽 = 1.37 𝑠 2 = 20.38, 𝜎𝑥2 = 19.82, 𝑛 = 20 𝜇𝑥 = 𝑥 = 0.71 Now standard error of 𝛼 = √19.82 ( = √19.82 ( 1 20 + 0.5041 20.38 1 20 + (0.71)2 20.38 ) ) = √1.48124829 = 1.2171 19.82 Standard error of 𝛽 = √ 20.38 = √0.97252208 = 0.9862 Question #5 You fit a CAPM that regresses the excess return of Coca-cola on the excess market return using 20 years of monthly data. You estimate 𝛼 = 0,72, 𝛽 = 1,37, 𝑆 2 = 20,38, 𝜎𝑋2 = 19,82 and 𝜇𝑥 = 0,71 . What is the 99% confidence interval for 𝛽 ? Answer: Step 1 Sample Statistic = Slope of Regression i.e. 𝛽 𝛽 = 1.37 , 99% confidence interval, 𝑛 = 20 Confidence interval 𝛼 =1−( ) 100 99 = 1−( ) 100 Degree of freedom (𝑑𝑓) = 𝑛 − 2 = 20 − 2 = 18 As per the t distribution table, the critical value at df 18 is 2.552. Standard Error or Standard Deviation = √19.82 = 4.45197 Margin of Error = Critical value × Standard Error = 2.552 × 4.45197 = 11.36142 Step 2 Confidence Interval = Sample Statistic ± Margin of Error = 1.37 ± 11.36142 Upper Interval = 1.37 + 11.36412 = 12.73142 Lower Interval = 1.37 − 11.36412 = −9.99412 Therefore, 99% confidence interval is −9.99412 to 12.73142 Question #6 A freelance computer consultant keeps a database of her clients, which contains the names 𝑆 = {𝐴𝑐𝑚𝑒, 𝐵𝑟𝑜𝑡ℎ𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑖𝑜𝑛, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐹𝑙𝑜𝑦𝑑, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑙𝑏𝑒𝑟𝑡} . The following clients owe her money: 𝐴 = {𝐴𝑐𝑚𝑒, 𝐶𝑜𝑟𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} . The following clients have done at least $40,000 worth of business with her: 𝐵 = {𝐴𝑐𝑚𝑒, 𝐵𝑟𝑜𝑡ℎ𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑖𝑜𝑛} . The following clients have employed her in the last year: 𝐶 = {𝐴𝑐𝑚𝑒, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑖𝑜𝑛, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑙𝑏𝑒𝑟𝑡} . A subset of clients is described that the consultant could find using her database. the clients who owe her money or have done at least $40,000 worth of business with her Write the subset in terms of A, B, and C. Answer: Step 1 Given that, 𝐴 = Set of clients who owe her money 𝐵 = Set of clients who have done at least $40,000 worth of business with her 𝐶 = Set of clients have employed her in the last year Step 2 We have to find the set of clients who owe her money or have done at least $40,000 worth of business with her That is given by A or B =𝐴 ∪𝐵 = {𝐴𝑐𝑚𝑒, 𝐶𝑜𝑟𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} ∪ {𝐴𝑐𝑚𝑒, 𝐵𝑟𝑜𝑡ℎ𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑖𝑜𝑛} = {𝐴𝑐𝑚𝑒, 𝐵𝑟𝑜𝑡ℎ𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑖𝑜𝑛, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} Answer: 𝐴 ∪ 𝐵 (in terms of A and B) Question #7 Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of 𝑥 = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is 𝜎𝑥 = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is 𝜎𝑦 = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is 𝜇𝑥 pounds, while the true (unknown) mean of 1999 AMs weights is 𝜇𝑦 pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data( X ) 77 189.030 14.04 1999 Data(Y ) 93 184.795 10.03 i) If we used this data to test 𝐻0 : 𝜇𝑥 − 𝜇𝑦 = 0 against the alternative 𝐻𝑎 : 𝜇𝑥 − 𝜇𝑦 > 0 then what would the p value have been? j)If we used this data to test 𝐻0 : 𝜇𝑥 − 𝜇𝑦 = 0 against the alternative 𝐻𝑎 : 𝜇𝑥 − 𝜇𝑦 ≠ 0 then what would the p value have been? Answer: Step 1 i) Null hypothesis: 𝐻0 : 𝜇𝑥 − 𝜇𝑦 = 0 Alternative hypothesis: 𝐻𝑎 : 𝜇𝑥 − 𝜇𝑦 > 0 The test statistic is as follows: 𝑍= 𝑥1 −𝑥2 2 2 𝜎 𝜎 √ 1+ 2 𝑛1 𝑛2 = 189.030−184.795 2 √(14.04) +(10.03) 77 2 93 = 2.2192 Step 2 Thus, the test statistic value is 2.2192. Computation of P-value: The P-value can be obtained using the excel formula “ = (1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆 (2.2192,1)”. The p-value is 0.01323. Step 3 j) Null hypothesis: 𝐻0 : 𝜇𝑥 − 𝜇𝑦 = 0 Alternative hypothesis: 𝐻𝑎 : 𝜇𝑥 − 𝜇𝑦 ≠ 0 Thus, the test statistic value is 2.2192. Computation of P-value: The P-value can be obtained using the excel formula “ = 2 × (1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆 (2.2192,1) ”. The p-value is 0.02647. Question #8 Use the exponential growth model 𝐴 = 𝐴0 𝑒 𝑘𝑡 to solve: In 2000, there were 110 million cellphone subscribers in the United States. By 2010, there were 303 million subscribers a. Find the exponential function that models the data. b. According to the model, in which year were there 400 million cellphone subscribers in the United States? Answer: Step 1 Given Data: The exponential growth model is 𝐴 = 𝐴0 𝑒 𝑘𝑡 In 2000 the number of telephone users was 110 million. In 2010 the number of telephone users was 303 million. Taking 2000 as the base year (𝑡 = 0) and A as the number of cellphone users then, boundary condition can be written as, 𝑡 = 0, 𝐴 = 110 million 𝑡 = 10, 𝐴 = 303 million Step 2 Substituting the boundary condition in the model equation, 101 million= 𝐴0 𝑒 𝑘×0 𝐴0 = 110 million 303million = 𝐴0 𝑒 𝑘×10 303 million = 110 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 × 𝑒 𝑘×10 𝑒 10𝑘 = 2.75 10𝑘 = ln 2. 75 𝑘 = 0.101325 Thus, the required model equation is, 𝐴 = 110𝑒 0.101325𝑡 The time after which the telephone users will be 400 million can be determined as, 400 = 110𝑒 0.101325𝑡 𝑒 0.101325𝑡 = 3.636 0.101325𝑡 = 1.290 𝑡 = 12.74 Thus, the year in which the number of telephone users will be 400 million is 2000 + 12.74 = 2012.74 Thus, in the year 2013 the number of telephone users will be 400 million. Question #9 Step 1 From the given information, O blood group -210, A blood group -223, B blood group -51, AB blood group -16. Step 2 The frequency distribution for the given data is shown below: Blood type Frequency 0 210 A 223 B 51 AB 16 Total 500 The probability that a randomly selected person from the population has the O blood type is obtained as follows: 𝑃(𝑂𝑏𝑙𝑜𝑜𝑑𝑡𝑦𝑝𝑒) = = Number of favorable cases for blood type O Total number of cases 210 500 = 0.42 Answer: Step 1 From the given information, O blood group -210, A blood group -223, B blood group -51, AB blood group -16. Step 2 The frequency distribution for the given data is shown below: Blood type Frequency 0 210 A 223 B 51 AB 16 Total 500 The probability that a randomly selected person from the population has the O blood type is obtained as follows: 𝑃(𝑂𝑏𝑙𝑜𝑜𝑑𝑡𝑦𝑝𝑒) = = Number of favorable cases for blood type O 210 500 = 0.42 Question #10 Total number of cases Use linear regression to find the equation for the linear function that best fits this data. Click Read Button For online regression calculator. Round both numbers to two decimal places. Write your final answer in a form of an equation 𝑦 = 𝑚𝑥 + 𝑏𝑦 = 𝑚𝑥 + 𝑏 x 1 2 3 4 5 6 y 74 90 109 123 143 155 Answer: Step 1 Given data, X Y X Y 1 74 74 2 90 180 3 109 327 4 123 492 5 143 715 6 155 930 X X 1 4 9 16 25 36 Step 2 Regression equation is 𝑌 = 𝑎 + 𝑏𝑥 where ∑ 𝑋 = 21, ∑ 𝑌 = 694, ∑ 𝑋 ⋅ 𝑌 = 2718, ∑ 𝑋 2 = 91 Use the following equations to find a and b: 𝑎= 𝑏= ∑ 𝑌⋅∑ 𝑋 2 −∑ 𝑋⋅∑ 𝑋𝑌 = 694⋅91−21⋅2718 𝑛⋅∑ 𝑋 2 −(∑ 𝑋)2 𝑛⋅∑ 𝑋𝑌−∑ 𝑋⋅∑ 𝑌 6⋅91−212 6⋅2718−21⋅694 𝑛⋅∑ 𝑋 2 −(∑ 𝑋)2 6⋅91−(21)2 = ≈ 57.87 ≈ 16.51 Substitute a and b in regression equation formula 𝑦 =𝑎+𝑏 ⋅𝑥 𝑦 = 57.87 + 16.51 ⋅ 𝑥 𝑦 = 57.87 + 16.51𝑥 Question #11 Five hundred people were chosen at random from a large population and each was asked whether he or she favoured capital punishment for murderers of police officers. Suppose that 60% of those interviewed said that they were in favour. Let 𝜃 denote the proportion of the population who are in favour. Construct a 95% confidence interval for \theta from the observed data. Answer: Step 1 Given that, Sample size, 𝑛 = 500 Proportion, 𝜃 = 60% = 0.60 Then 𝑋 ∼ 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙(500,0.60) Confidence interval for 𝜃 is given by, (𝜃 ± 𝑍 ⋅ √𝜃 ⋅ 1 − 𝜃/𝑛) Z at 95% level of significance is 1.96 Step 2: ⇒ (0.60 ± 1.96 ⋅ √0.60 ⋅ 0.40/500) = (0.60 − 0.043,0.60 + 0.043) = (0.557,0.643) Question #12 Suppose you are a fan of baseball, which has two different leagues in the US, the National League and the American League. The winners of both the leagues play in the World Series Championship each year. You want to test whether both the leagues are equally strong, or whether one of the leagues is stronger than the other. Assume that each year the AL champion wins the world series with probability p, and the NL champions win with probability 1 − 𝑝 , where p is an unknown number in (0, 1). Starting from 1903, the world series championship has been played 115 times, and the AL champions won it 66 times, and the NL champions won it 49 times. Based on this data, carry out the following tests of hypothesis: (a) 𝐻0 : 𝑝 = 1/2 𝑣𝑒𝑟𝑠𝑢𝑠 𝐻1 : 𝑝 > 1/2 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝛼 = 0.1 . (b) 𝐻0 : 𝑝 = 1/2 𝑣𝑒𝑟𝑠𝑢𝑠 𝐻1 : 𝑝 ≠ 1/2𝑎𝑡 𝑙𝑒𝑣𝑒l 𝛼 = 0.1 . Answer: Step 1 a) The following information is provided: The sample size is 𝑁 = 115 , the number of favorable cases is 𝑋 = 66 , and the sample proportion is 𝑝 = 𝑋 𝑁 = 66 115 = 0.5739 , and the significance level is 𝛼 = 0.1 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: 𝐻𝑜 : 𝑝 = 0.5 𝐻𝑎 : 𝑝 > 0.5 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used. Rejection Region Based on the information provided, the significance level is 𝛼 = 0.1 , and the critical value for a righttailed test is 𝑧𝑐 = 1.28 The rejection region for this right-tailed test is 𝑅 = {𝑧: 𝑧 > 1.28} Step 2 Test Statistics The z-statistic is computed as follows: 𝑧= 𝑝−𝑝0 √𝑝0 (1−𝑝0 )/𝑛 = 0.5739−0.5 √0.5(1−0.5)/115 = 1.585 Decision about the null hypothesis Since it is observed that 𝑧 = 1.585 > 𝑧𝑐 = 1.28 , it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is 𝑝 = 0.0565 , and since 𝑝 = 0.0565 < 0.1 , it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than 𝑝0 , at the 𝛼 = 0.1 significance level. b) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: 𝐻𝑜 : 𝑝 = 0.5 𝐻𝑎 : 𝑝 ≠ 0.5 This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used. Rejection Region Based on the information provided, the significance level is 𝛼 = 0.1 , and the critical value for a righttailed test is Critical value 𝑍𝑐 : ±1.6449 Critical region: (−∞, −1.6449] ∪ [1.6449, ∞) Test Statistics The z-statistic is computed as follows: 𝑧= 𝑝−𝑝0 √𝑝0 (1−𝑝0 )/𝑛 = 0.5739−0.5 √0.5(1−0.5)/115 = 1.585 Decision about the null hypothesis Since it is observed that 𝑧 = 1.585 does not fall in critical region, it is then concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is not enough evidence to claim that the population proportion p is not equal to 𝑝0 , at the 𝛼 = 0.1 significance level. Question #13 A data set includes data from 500 random tornadoes. The display from technology available below results from using the tornado lengths (miles) to test the claim that the mean tornado length is greater than 2.9 miles. Use a 0.05 significance level. 1. Identify the null and alternative hypotheses, 2. Identify the test statistic, 3. Identify the P-value, 4. State the final conclusion that addresses the original claim. Hypothesis test results: 𝜇 : Mean of variable 𝐻0 : 𝜇 = 2.9 𝐻𝐴 : 𝜇 > 2.9 Variable Sample Mean Std .Err. DF T − Stat P − value Length 3.12539 0.280011 499 0.804933 0.2106 Answer: Step 1 Solution: From the given information, the output hypothesis test result for the population mean is shown. Step 2 1. State the hypotheses. Null hypothesis: 𝐻0 : 𝜇 = 2.9 Alternative hypothesis: 𝐻𝐴 : 𝜇 > 2.9 Step 3 2. From the given output, the T-stat value is 0.804933. Thus, the test statistic is 0.804933. Step 4 3. From the given output, the P-value is 0.2106. Thus, the P-value is 0.2106. Step 5 4. Rejection rule: If the P-value is less than or equal to 0.05, then reject the null hypothesis. Conclusion: Here, the P-value is 0.2106. This is greater than 0.05. By the rejection rule, do not reject the null hypothesis. Thus, the mean tornado length is not greater than 2.9 miles. Question #14 Which of the following statements are correct? A. In general, a stem-and-leaf display based on between 15 and 25 stems is recommended. B. The observation 𝑥1 in any numerical data set 𝑥1 , 𝑥2 , … , 𝑥𝑛 is the smallest observation in the set, whereas the observation 𝑥𝑛 is the largest. C. A quick way to obtain an informative visual representation of a numerical data set is to construct a stem-and-leaf display. D. Astem-and-leaf display shows the order in which observations were obtained. Answer: Step 1: explanation (A). There is no boundary to take any specific numbers of Observations. (B). As 𝑥1 , 𝑥2 … 𝑥𝑛 is a data set where the Observations are taken in ascending order. (C). There are so many another ways according to which we can represent the data and get the information but stem and leaf is the easiest and less time consuming. (D). We always construct a stem and leaf plot in ascending order of the Observations. Step 2: So option (a) is wrong. Statement (b) is correct. (C) is not correct. (D) is correct. Question #15 Consider the following two regression models (at the population level): 𝑀1 : 𝑦 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + , 𝑀2 : 𝑦 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + 𝛽3 𝑥12 + 𝛽4 𝑥22 + 𝛽4 𝑥22 + 𝛽5 𝑥1 𝑥2 + 𝜉. Assume that we fit the two models to the same data. Can we compare the 𝑅2 of the two models without looking at the results of the fittings? Justify your answer. Answer: Step 1 The given two regression models are shown below 𝑀1 : 𝑦 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + 𝑀1 : 𝑦 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + 𝛽3 𝑥3 + 𝛽4 𝑥4 + 𝛽5 𝑥5 + 𝜉 𝑅2 or squared gives an idea of a fitted model appropriateness to the given data. Step 3 a) Generally, the more number of predictors increases R squared by a certain amount. That is, as the number of predictors (independent variables) increases, the R squared also increases (even by a very small amount). Here, model 1 has 2 predictors while model 2 has 5 predictors. So, 𝑅2 will be greater for Model 2 when compared with Model 1. Therefore, 𝑅2 for model 1 < 𝑅2 for model 2. Question #16 Let T be the time needed to complete a job at a certain factory. By using the historical data, we know that 1 2  t P(T  t ) = 16 1 for 0  t  4 for t  4 a. Find the probability that the job is completed in less than one hour, i.e., find 𝑃(𝑇 ≤ 1) . b. Find the probability that the job needs more than 2 hours. c. Find the probability that 1 ≤ 𝑇 ≤ 3 . Answer: Step 1 Given: Let T be the time needed to complete a job at a certain factory. We know that, 1 2  t P(T  t ) = 16 1 for 0  t  4 for t  4 We want to find, (a) 𝑃(𝑇 ≤ 1) =? (b) 𝑃(𝑇 ≥ 2) =? (c) 𝑃(1 ≤ 𝑇 ≤ 3) =? Step 2 Solution: (a) From the given information 1 1 (1)2 = 𝑃(𝑇 ≤ 1) = 16 16 𝑃(𝑇 ≤ 1) = 0.0625 (b) Let 𝑃(𝑇 ≥ 2) = 1 − 𝑃(𝑇 ≤ 2) 1 = 1 − (2)2 16 1 = 1− 16 1 (4) = 1− 4 3 = 4 𝑃(𝑇 ≥ 2) = 0.75 (c) 𝑃(1 ≤ 𝑇 ≤ 3) = 𝑃(𝑇 ≤ 3) − 𝑃(𝑇 ≤ 1) 1 1 = (3)2 − (1)2 = = = 16 1 16 1 16 1 16 (9 − 1) (8) 2 𝑃(1 ≤ 𝑇 ≤ 3) = 0.5 Question #17 A freelance computer consultant keeps a database of her clients, which contains the names 𝑆 = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐹𝑙𝑜𝑦𝑑, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} . These clients owe her money: 𝐴 = {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} . These clients have done at least $3,000 worth of business with her: 𝐵 = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙} . These clients have employed her in the last year: 𝐶 = {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} . A subset of clients is described that the consultant could find using her database. The clients who do not owe her money or have employed her in the last year. Write the subset in terms of A, B, and C. 𝐴′ ∩ 𝐶 𝐴 ∩𝐵 𝐴′ ∩ 𝐵 𝐴′ ∪ 𝐶 𝐴 ∪𝐶 List the clients in that subset. {𝐴𝑐𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐶𝑜𝑟𝑒𝑠, 𝐺𝑙𝑜𝑏𝑒} {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠} ∅ {𝐴𝑐𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐶𝑜𝑟𝑒𝑠, 𝐺𝑙𝑜𝑏𝑒, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐷𝑢𝑎𝑙, 𝐹𝑙𝑜𝑦𝑑, 𝐻𝑖𝑑𝑒} {𝐹𝑙𝑜𝑦𝑑, 𝐵𝑎𝑘𝑒𝑟𝑠} Answer: Given data, 𝑆 = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐹𝑙𝑜𝑦𝑑, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} . 𝐴 = {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} . 𝐵 = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙} . 𝐶 = {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} . Step 1 The clients who do not owe her money or have employed her in the last year. The clients who do not owe her money, = 1 − The clients owe her money = 1−𝐴 = 𝐴′ The clients have employed her in the last year is C. Hence, the clients who do not owe her money or have employed her in the last year, The clients who do not owe her money ∪ The clients have employed her in the last year = 𝐴′ ∪ 𝐶 Step 2 The elements in the set is 𝐴′ ∪ 𝐶 𝐴′ = 𝑆 − 𝐴 = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐹𝑙𝑜𝑦𝑑, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} − {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒} = {𝐵𝑎𝑘𝑒𝑟𝑠, 𝐷𝑢𝑎𝑙, 𝐹𝑙𝑜𝑦𝑑, 𝐻𝑖𝑑𝑒} 𝐴′ ∪ 𝐶 = {𝐵𝑎𝑘𝑒𝑟𝑠, 𝐷𝑢𝑎𝑙, 𝐹𝑙𝑜𝑦𝑑, 𝐻𝑖𝑑𝑒} ∪ {𝐴𝑐𝑒𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒}. = {𝐴𝑐𝑒𝑠, 𝐵𝑎𝑘𝑒𝑟𝑠, 𝐶𝑜𝑟𝑒𝑠, 𝐷𝑢𝑎𝑙, 𝐸𝑛𝑒𝑟𝑔𝑦, 𝐹𝑙𝑜𝑦𝑑, 𝐺𝑙𝑜𝑏𝑒, 𝐻𝑖𝑑𝑒} Question #18 An article in Fortune (September 21, 1992) claimed that nearly one-half of all engineers continue academic studies beyond the B.S. degree, ultimately receiving aither an M.S. or a Ph.D. degree. Data from an article in Engineering Horizons (Spring 1990) indicated that 117 of 484 new engineering graduates were planning graduate study. Are the data from Engineering Horizons consistent with the claim reported by Fortune? Use \𝑎𝑙𝑝ℎ𝑎 = 0.05 in reaching your conclusions. What is the value of the test statistic? Answer with two decimal places Answer: Step 1 Here, a sample of 484 is observed. If 117 engineering graduates were planning graduate study, then the proportion is 117 𝑝= 484 = 0.241735 Since, it is claimed that nearly one half of engineering graduates continue the studies, then it is to be tested that 1 𝐻0 : 𝑝 = 𝐻1 : 𝑝 ≠ 2 1 2 Thus, Step 2 The test statistic is 𝑧= = 𝑝−𝑝0 (1−𝑝0 ) √𝑝0 𝑛 0.241735−0.5 √0.5(1−0.5) 484 = −0.258265 0.02273 = −11.3623 Step 3 The critical value at 𝛼 = 0.05 in two tails is 1.96. Since, |𝑧| > 1.96 , reject 𝐻0 Therefore, the claim that nearly one half of engineering graduates continue the studies is not true. Question #19 Hamza comes to office located in Islamabad through different modes of transport. From his past data, it is known that the probability of him coming through car, bus, motobrike and rickshaw are respectively 3 1 1 2 1 1 1 , , and . The probabilities that he will arrive to his office late are , , and if he comes by car, bus 10 5 10 5 4 3 12 and motorbike. Interestingly, he will not be late if he comes by rickshaw. Suppose, he arrives late on Friday. What is the probability that he came by car? Answer: Step 1 The provided probabilities are; 3 P(coming through car) = P(coming through bus) = 10 1 5 P(coming through motorbike) = P(coming through rickshaw) = P(late by car) = P(late by bus) = 1 2 1 10 5 4 1 3 P(late by motorbike) = 1 12 P(late by rickshaw) = 0 Step 2 The probability that he came by car and was late is The probability that he came by bus and was late is 3 10 1 5 1 3 4 1 40 1 3 15 1 1 10 12 × = × = The probability that he came by motorbike and was late is The probability that he came by rickshaw and was late is 2 5 × = 1 120 ×0=0 Therefore, the total probability that he was late is, 3 1 1 3 𝑃(𝑙𝑎𝑡𝑒) = + + +0 = 40 15 120 20 If he was late, the probability that he came by car is 3 40 3 20 = 1 2 Question #20 Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with a population standard deviation, 𝜎, 2.2 years . For pickup trucks the mean was 7.1 years with population standard deviation, 𝜎, 3.0 years . Construct the 95% confidence interval for the difference in the means based on these data. Answer: Step 1 3. Computation of critical value: (1 − 𝛼) = 0.95 𝛼 = 0.05 𝛼 = 0.025 2 𝛼 1 − = 0.975 2 The critical value of z-distribution can be obtained using the excel formula “ = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(0.975)”. The critical value is 1.96. Step 2 The 95% confidence interval for the difference in the means based on these data is, 𝜎12 𝐶𝑙 = (𝑥1 − 𝑥2 ) ± 𝑧𝛼 √ 2 𝑛1 = (5.3 − 7.1) ± 1.96√ + 2.22 40 𝜎22 𝑛2 + 3.02 40 = −1.8 ± (1.96 × 0.5882176) = −1.8 ± 1.1 = (−1.8 − 1.1, −1.8 + 1.1) = (−2.9, −0.7) Thus, the 95% confidence interval for the difference in the means based on these data is (-2.9, -0.7).