Final Practice 3

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McGill University *

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271

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Jun 12, 2024

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MGCR-271-PRACTICE-RGG01 1 MGCR-271 Business Statistics PRACTICE Final Exam This exam consists of 30 equally weighted multiple-choice questions divided across four sections as shown below. You should attempt all the questions in this exam paper. Statistical tables are provided at the back of this booklet. You may use a permitted calculator. You may bring one handwritten or electronically handwritten crib sheet on letter sized paper, double sided. You may not take it out of the exam afterwards. Section Questions 1: Probability 1 – 10 2: Inference 11 - 15 3: Regression 16 - 20 4: Long Questions 21 - 30
MGCR-271-PRACTICE-RGG01 2 Section 1: Probability and Data 1. In the Montreal Botanical Garden, there are two types of rare flowers: the Red Velvet Rose (RVR) and the Blue Moon Orchid (BMO). The marketing department decides to conduct a customer survey on a sample of customers to determine their preferences for the two rare flowers. After this survey, it was found that 60% of visitors like the scent of the RVR and 45% like the scent of the BMO. The preference for the scent of the RVR is independent of the preference for the scent of the BMO. What is the probability that a random visitor likes the scent of neither of the two flowers? Answer: 22% P(likes RVR) = 0.6, P(likes BMO) = 0.45 [independently] P(dislikes RVR) = 1 – 0.6 = 0.4, P(dislikes BMO) = 1 – 0.45 = 0.55 P(dislikes both) = 0.4 * 0.55 = 22%
MGCR-271-PRACTICE-RGG01 3 2. Brewster's Ale Co. is crafting a large batch of their signature lager using 200 barrels. The entire batch will be deemed unsellable if 10 or more barrels get contaminated during the brewing process. From historical records, each barrel has a 6% chance of getting contaminated during the process, and the contamination status of each barrel is independent of the others. Assuming all barrels are uncontaminated at the start, what is the (approximate) probability that the entire batch will be unsellable using the normal approximation? Answer: 77.2% This question requires using the Normal approximation to the binomial (we can check that np = 12 > 10). For the continuity correction, we will look for P(X >= 9.5). The Z-score is given by: 9.5 − 12 3.36 = −0.744 From the Normal distribution tables, we find a probability of 77.2%
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MGCR-271-PRACTICE-RGG01 4 3. During the holiday season, Luigi's Ristorante experiences a surge in customers coming in for dinner. On any given night, the number of customers dining at the restaurant follows a normal distribution with a mean of 120 customers and a standard deviation of 20 customers. Luigi wants to ensure that they have enough seating available such that they must turn customers away on no more than 15% of the nights. How many seats should the restaurant have available as a minimum to ensure this accommodation? Answer: 141 To determine the number of seats required, we need to find the number of customers represented by the 85th percentile (since we want to accommodate at least 85% of the nights). We have a mean of 120 and a standard deviation of 20, so we find the z-score for the 85 th percentile and find the number of seats from the z-score formula.
MGCR-271-PRACTICE-RGG01 5 4. At Falcon Financials, the daily number of stock trade requests follows a normal distribution. The mean number of requests is 800, and the standard deviation is 100. The company's trading platform has the capacity to handle 950 trade requests per day. What is the probability that the trading platform will be overwhelmed (i.e., receive more than 950 trade requests) on any given day? Answer: 6.68% To determine the probability that the trading platform will be overwhelmed, we need to find the probability associated with more than 950 trade requests. We can use the z-score formula and we know that μ=800 and σ=100, and with x=950, we can calculate the z-score for 950 trade requests and subsequently find the probability associated with it.
MGCR-271-PRACTICE-RGG01 6 5. Sophia is mixing paint for a DIY home project. She starts with a certain number of cans of paint, each with an average volume of 3 liters. She then adds a larger can of paint with a volume of 10 liters, and the average volume of all the paint cans, including the new one, becomes 3.5 liters. How many cans of paint did Sophia start with? Answer: 13 Let's denote the number of original cans as n. The total volume of the original cans would then be 3n liters. When Sophia adds the 10-liter can, the total volume becomes 3n+10 liters, and the number of cans becomes n+1. Given that the new average volume is 3.5 liters, we can set up the following equation: (3n + 10)/(n+1) = 3.5 and solve for n.
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MGCR-271-PRACTICE-RGG01 7 6. A bookstore specializes in a particular genre of books. The monthly demand for their top- selling book follows a normal distribution with a mean of 5,000 copies and a standard deviation of 1,000 copies. On the supply side, the publisher provides monthly shipments in the following manner: 4,500 copies with a probability of 20%, 5,000 copies with a probability of 50%, or 5,500 copies with a probability of 30%. Assuming that the monthly demand and the monthly supply are independent, what is the probability that demand exceeds supply in a given month? Answer: 48.1% To solve this, we need to consider all possible supply scenarios and calculate the probability that demand exceeds each supply scenario, then sum up these probabilities, weighted by the probability of each supply scenario: 1. Probability that demand > 4,500 copies and the supply is 4,500 copies. 2. Probability that demand > 5,000 copies and the supply is 5,000 copies. 3. Probability that demand > 5,500 copies and the supply is 5,500 copies. This gives an answer of 48.1%
MGCR-271-PRACTICE-RGG01 8 The Following THREE Questions are based on this information: The weights of loaves of a particular type of bread at "Bakery Bliss" are normally distributed with a mean of 500 grams and a standard deviation of 20 grams. 7. If 5 loaves of this bread are selected at random, what is the probability that their average weight is greater than 505 grams? Answer: 28.1% HINT: This requires the use of the sample distribution for the sample mean, where we calculate the new standard error based on the sample size of n = 5. We want to find the probability that X-bar > 505 from this sampling distribution
MGCR-271-PRACTICE-RGG01 9 8. If 10 loaves are selected at random, what is the probability that at least one of them has a weight of greater than 525 grams? Answer: 67.26% HINT: The probability that all 10 selected loaves weigh 525 grams or less is: P(X ≤ 525) 10
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MGCR-271-PRACTICE-RGG01 10 9. "Bakery Bliss" has introduced a new type of muffin. The weights of these muffins are normally distributed with a mean of 150 grams and a standard deviation of 10 grams. Every morning, the bakery randomly selects 16 muffins for quality control to ensure consistent weight. What is the probability that the average weight of these 16 muffins is less than 147 grams? Answer: 11.51% HINT: This uses the sample distribution again. Compute the new standard error and we are looking or P(X-bar < 147) in a sample of 16 muffins.
MGCR-271-PRACTICE-RGG01 11 10. Malleswara is planning to travel next summer and he's considering two destinations: France and Japan. From his past travel choices: The probability that he chooses France is P(F)=0.7. The probability that he chooses Japan is P(J)=0.3. He has also been browsing travel blogs recently. Given that he chooses France, the probability that he reads blogs (B) about France is P(B ∣ F)=0.9. On the other hand, given that he chooses Japan, the probability that he reads blogs about France (maybe due to comparison or general interest) is P(B ∣ J)=0.2. Today, Rob notices that Malleswara is reading a blog about France. What is the probability that he has decided to travel to France? Answer: 91.3% HINT: This requires an application of Bayes Rule to determine P(F|B). You will need to determine P(B) from the combination of P(B|J) * P(J) and P(B|F) * P(F).
MGCR-271-PRACTICE-RGG01 12 Section 2: Inference & Sampling 11. The data in the following table represents the average monthly temperatures (in Celsius) of a town over a year. It's believed that the town has an average annual temperature of 18°C. Residents, however, feel that the average temperature has been cooler than that. Assuming that the underlying temperature distribution is Normally distributed, conduct a hypothesis test for the mean temperature of the town to be less than 18°C and determine the p-value. Month Sample Temperature January 16 February 15 March 16 April 17 May 17 June 18 July 19 August 18 September 17 October 16 What is the range of the correct p-value? Answer: 0.01 > p > 0.005 HINT: The is a t-test and you need to determine the sample statistics yourself. The null hypothesis is that the mean is 18C.
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MGCR-271-PRACTICE-RGG01 13 12. On average, Mazda hybrid cars travel 10.6 km/liter. To check if a certain Mazda hybrid car requires maintenance, a mechanic took a random sample of 45 days that showed an average of 10.1 km/liter. If the population standard deviation is known to be 1.6 km/liter, the P-value for this significance test is…? Answer: 1.8% HINT: This is a one-tailed hypothesis test with a mean of 10.6 for the null hypothesis. We know the sample size and standard deviation for the sampling distribution calculations.
MGCR-271-PRACTICE-RGG01 14 13. A law enforcement agency is concerned about the response time to emergency calls in their jurisdiction. Historically, the average response time to emergency calls has been 7 minutes. However, due to recent changes in procedures and staffing, they suspect that the average response time may have changed. To assess this, the agency randomly selects the response times of 50 emergency calls from the past month. Here are the sample statistics: Sample Mean: =6.8 minutes Sample Standard Deviation: s=1.5 minutes Using a 5% significance level, conduct a two-tailed t-distribution hypothesis test to determine if there's evidence to suggest that the true average response time differs from 7 minutes. Determine the range of the p-value. Answer: p-value between 40% and 30% so we fail to reject the null hypothesis. HINT: This is a two-tailed t-test for a population mean from a sample of 50 (use 50 degrees of freedom as we do not have 49 in the tables). You must then double the p-values you find to correct for the two-tails of this test.
MGCR-271-PRACTICE-RGG01 15 14. A factory manufactures baked beans and packages them in standard cans. As part of the quality control process, the factory wants to ensure that the mean weight of the beans in the cans meets the labeled weight. To assess this, a random sample of 40 cans is taken from a day's production, and the weights of the beans in these cans are recorded. The sample produces the following statistics: Sample Mean: 455 grams Sample Standard Deviation: 12 grams Construct a 95% confidence interval for the true mean weight of the beans in the cans produced by the machine. Answer: Approx. [451.3, 458.7] HINT: The central limit theorem applies here due to the sample size so we can use the normal distribution. Ensure you use the sampling distribution with the correction to the standard deviation.
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MGCR-271-PRACTICE-RGG01 16 15. In Formula 1 racing, tire performance is crucial. A leading tire manufacturer, SpeedTire Inc., has developed a new tire compound and is testing its longevity in terms of the number of laps it can last before significant degradation. From an initial test on a particular circuit, they took a random sample of 30 tires and recorded the number of laps each tire lasted. The sample produced the following statistics: Sample Mean: =50 laps Sample Standard Deviation: =5 laps The chief statistician is uncertain whether to use • Construct a 95% confidence interval for the true mean number of laps the new tire compound can last using the t-distribution. • Construct a 95% confidence interval for the true mean number of laps the new tire compound can last using the Normal distribution. • Compute differences between the two intervals. Answer: Normal = [48.21, 51.79], t-distribution = [48.13, 51.87], to the total difference is 0.08 laps at either end of the interval, with the t-distribution computing a slightly wider interval (as expected).
MGCR-271-PRACTICE-RGG01 17 Section 3: Regression 16. In athletics, coaches often rely on data to optimize an athlete's performance. A coach is analyzing the relationship between the number of hours of sleep an athlete gets the night before a 100m sprint and the time taken to complete the sprint. Based on collected data, the coach has established a simple linear regression model: Sprint Time (seconds)=15−0.2×Hours of Sleep One of the athletes, named Jordan, reported 8 hours of sleep the night before a competition and completed the 100m sprint in 13.5 seconds. Calculate the residual for Jordan's observation. Answer: 0.1 seconds HINT: this is relatively simple, just determine the difference between the observation and the value for this value of sleep hours.
MGCR-271-PRACTICE-RGG01 18 The following TWO questions are based on the following information. The following is (part of) the Excel summary output of a simple linear regression between a dependent variable, length of carrots grown on a farm (cm) and hours of sunlight in the growing season. 17. What are the values of A and B in the output table? Answer: A = 48 and B = 0.800
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MGCR-271-PRACTICE-RGG01 19 18. What are the values of C and D in the output table? Answer: C = 5.48, D = 1.43 HINT: use b = t * SE
MGCR-271-PRACTICE-RGG01 20 The following TWO questions are based on the following information. Extending our work on carrots, the output below shows the results of a regression where carrot consumption is used to predict eyesight in a survey of fighter jet pilots, based on test scores out of 100%. 19. Which of the following statements is TRUE? A) The dependent variable in this regression is carrot consumption and the independent is eyesight. B) A hypothesis that the intercept of the population regression intercept is different to 0.443 would fail to reject the null hypothesis at the 95% level based on this sample. C) Variation in the independent variable explains about 71.4% of the variance in the dependent variable. D) A hypothesis test that the X variable coefficient is different to 0 would reject the null hypothesis at the 1% significance level. Answer: D
MGCR-271-PRACTICE-RGG01 21 20. Construct a 99% confidence interval for the x variable regression coefficient based on the information provided above. Answer: [0.0296, 0.0658]
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MGCR-271-PRACTICE-RGG01 22 Section 4: Longer Questions The following TWO questions are based on the following information. Montreal Health did a test of body fat percentage of 16 random Montreal males, and Montreal Health would like to know whether Montreal males’ average body fat percentage (denoted by μ) is significantly di ff erent from 15% with a significance level of 5%. 21. Write out the hypotheses for this test. Answer: H0 :μ=15% vs Ha: μ≠15% HINT: Two-tailed test.
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MGCR-271-PRACTICE-RGG01 23 22. After conducting the appropriate hypothesis test, Montreal Health obtained a p-value of 4.4%. Suppose the sample standard deviation (i.e., the standard deviation calculated from the sample) is ࠵?̅ = 16.52% . What is the range of the ࠵? !"#$%& ? Answer: between 2.85 and 2.70 This is a two-tailed test for the mean, so we compute the t-statistic and find the probability t-stat will be between 2.131 and 2.249. That means we have ’̅)* +/√. = ࠵?࠵?࠵?࠵?࠵?࠵?࠵? 2.131 ࠵?࠵?࠵? 2.249 . So ࠵? will be between (’̅)*)×√. 2.454 and (’̅)*)×√. 2.267
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MGCR-271-PRACTICE-RGG01 24 The following FOUR questions are based on the following information. A tech company recently ran two different marketing campaigns, A and B, for their new software product. They're interested in determining if there's a significant difference in the success rates of the two campaigns based on the region (North and South) where they were run. The table below shows the results: Campaign A Success Campaign A Failure Campaign B Success Campaign B Failure North Region 111 139 109 141 South Region 114 136 106 144 23. State the null and alternative hypotheses for testing if there's a significant difference in the success rates of the two campaigns based on the region. Answer: H0: The success rates of the two campaigns are the same in both regions. Ha: The success rates of the two campaigns are different in at least one of the regions. This is a test of whether there is a relationship between campaign success and region.
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MGCR-271-PRACTICE-RGG01 25 24. Calculate the expected counts for each cell in the table under the assumption that the null hypothesis is true. Answer: Campaign A Success Campaign A Failure Campaign B Success Campaign B Failure North Region 112.5 137.5 107.5 142.5 South Region 112.5 137.5 107.5 142.5 HINT: Example: the total success for A are 225, we expect the successes to be evenly distributed between the North and South regions, so each will be 112.5. Continue for all others.
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MGCR-271-PRACTICE-RGG01 26 25. Compute the chi-squared test statistic based on the observed and expected counts. Answer: 0.1462
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MGCR-271-PRACTICE-RGG01 27 26. At a 1% significance level, determine whether you should reject the null hypothesis. What conclusions can you draw about the effectiveness of the two campaigns in different regions? Answer: We cannot reject the null hypothesis, there appears to be no difference between the regions. HINT: determine degrees of freedom and use the tables to check.
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MGCR-271-PRACTICE-RGG01 28 The following FOUR questions are based on the following information. Researchers in the McGill University Sports Medicine department are studying which variables contribute to fast marathon times. They collect observational data from 50 runners of the 2023 Boston marathon for four variables: marathon time, number of pancakes eaten for breakfast, price of shoes, and length of legs. They obtain the following output from their regression: 27. What is the value of A in this regression output? Answer: 1.92
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MGCR-271-PRACTICE-RGG01 29 28. What is the value of B in this regression output? Answer: 170.19
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MGCR-271-PRACTICE-RGG01 30 29. What is the value of C in this regression output? Answer: 0.104
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MGCR-271-PRACTICE-RGG01 31 30. Which of these plots shows the line fit for the variable “number of pancakes consumed”? Answer: C
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