ITIS 1P97 WINTER 2016 ASSIGNMENT#1 Due Date: February 12, 2016 @12:00AM via Sakai; only one submission per group. INSTRUCTIONS: There are four problems in this assignment. You may use 'Excel' spreadsheet applications to solve the problems. If the answers to the questions need to be supported with written arguments or explanations you must do it in the file itself for all questions. The group must have BETWEEN 4 AND 6 GROUP MEMBERS within the same section. Ensure that you have included everyone's first and last names, student numbers, and the course and section number on a cover sheet. Individual assignments are not acceptable, nor will late assignments be accepted. Assignments not having a coversheet will be deducted 20% of the assignment grade. Remember to include a letter stating the contribution of all students. THE TOTAL MARKS FOR THIS ASSIGNMENT ARE 100. PROBLEM 1: The Manager of a local StarBuckes Coffee Shop is interested in improving the service provided to customers at the coffee shop. As a first step in the process, the manger asks his assistant to record the number of arrivals between 9:00 A.M and 10:00 A.M., since this is the busiest time of the day. The number of arrivals during this time interval, for a number of days, are included in the excel file titled \"StarbuckesArrivals_140922.xls\". This file is posted on Sakai in the Data Folder. Given the background and the data, answer the following questions: TOTAL 40 MARKS a) Use the \"Excel\" spreadsheet application to calculate the average, standard deviation, variance, maximum, minimum, and range for the observations entered in the attached excel file. Interpret these values from a business point of view. In other words instead of saying 'the variance of the distribution is \"x.xx\"', provide a one sentence explanation of the value in non-technical terms, so that someone not familiar with statistical jargon can understand the meaning of the results you have obtained. (10 MARKS) b) Develop a frequency chart for the data. Without performing any technical analysis, interpret if the frequency charts resembles any of the probability distributions you learned in this course. Use the Bin Values of 0, 2, 4, 6, 8, and 10. (10 MARKS) c) Do a Chi-Square goodness of fit test to determine if the data is from a Poisson distribution. Use for your test. (15 MARKS) = 0.05 d) Assuming that the data does conform to Poisson distribution, what is the probability that there are exactly 3 arrivals between 9:00 A.M and 10:00 A.M.? What is the probability that there are between 2 and 6 arrivals, inclusive, between 9:00 A.M and 10:00 A.M? What is the probability that there are more than 7 arrivals between 9:00 A.M and 10:00 A.M? To solve for the probabilities, use two approaches. First calculate the probabilities using POISSON.DIST excel function, using the sample mean as your estimate for Lambda. Then, solve the probabilities from the data. For the second method, use the observations, and the relative frequencies, which you can find using the Frequency Function with Bin Values of 0, 1, 2,...,7. (5 MARKS) ITIS 1P97 Assignment # 1 Page 1 of 2 Pages ITIS 1P97 WINTER 2016 PROBLEM 2: The following question needs you to develop a histogram for a Binomial distribution. TOTAL 20 MARKS a) Use \"Excel\" to calculate the probabilities of getting 0 to 30 heads in 40 tosses of a coin. For the coin used in this experiment the probability of getting a head is 0.60. Label the one column as Number of heads, and the second column as Probability, and the third column as Cumulative Probability. (10 MARKS) b) Develop a probability distribution chart for the probabilities, not the Cumulative Probabilities though, that you obtained in the previous part of this question, see page 39 of the text (Figure 6) for an example chart. Your chart must conform to the following guidelines (10 MARKS): a. The chart title must read \"Binomial Distribution for the number of heads appearing in 40 tosses of a coin.\" b. X axis label (in black): Number of Heads c. Y axis label (in black): Probability d. Chart Background: None. PROBLEM 3: The following table shows the length of stay distribution for guests staying at a beach resort, in days. TOTAL 20 MARKS Days Probability % 1 3 2 5 3 10 4 10 5 11 6 11 7 15 8 14 9 12 10 9 The resort management makes a net profit of $380 per day per guest during the first 2 days of the stay, and $225 per day per guest after the first 2 days. How much profit will the resort owners make in a month (assuming 30 days in a month) if there are 100 guests arriving per day? (20 MARKS) [Hints: Note that a guest who only stays for two days is billed $760; find the average profit for one guest then work out for the entire month.] PROBLEM 4: A marketing manager for a chain of hardware stores needed more information about the effectiveness of the three types of advertising that the chain used. There are direct mailing, newspaper advertising, radio advertising, and television commercials. Data in the file XR2516, posted on Isaak, contain the weekly gross sales, and advertising expenditures for the four types of advertising. All values are in thousands of dollars. TOTAL 20 MARKS a) Determine the regression equation (5 MARKS) b) What is the coefficient of determination and is the model valid? (5 MARKS) c) Should any of the independent variables be removed from the model? Explain your answer. (5 MARKS) d) What are the estimates of the variance for the residuals and of the variance for the y values? (5 MARKS) ITIS 1P97 Assignment # 1 Page 2 of 2 Pages Number of Arrivals at StarBuckes Coffee Shop 5 3 2 7 10 4 5 7 7 7 10 7 6 10 9 5 9 7 9 6 8 5 9 9 3 6 6 8 2 11 7 6 4 9 4 5 1 4 6 6 7 4 10 3 9 6 9 5 10 7 4 3 6 4 7 6 2 7 4 6 5 5 4 3 4 8 10 4 7 9 5 12 8 5 10 7 12 8 8 7 6 2 4 Arrival Time Average 5.910714 St. Dev Sa 2.468996 VAR.S 6.095942 Maximum 13 Minimum 0 Range 13 Bins # of ArrivalFrequency 0 0 2 2 1 or 2 18 4 3 or 4 68 6 5 or 6 77 8 7 or 8 70 10 9 or 10 38 More than More than 7 1a) 1. On average, there are 5.91 arrivals between 9 AM and 10 AM 2. The standard deviation, which is a measure of the dispersion in the arrivals is 2.47 arrivals 3. The variance in this set of data is 6.1. The variance is the standard deviation squared and demonstrates how far the arrivals are spread out. 4. The maximum number of customers who arrive between 9 AM and 10 AM is 13. 5. The minimum number of customers arriving between 9 AM and 10 AM is 0. 6. The range, which is the difference between the maximum and minimum number of arrivals is 13 customers. b) Without any technical analysis (ie histogram), I would predict that there would be a symmetrical shape around 5/6 arrivals. The frequency numbers after and before it are close in value which is why I predict that it would cause such a shape. This is due to the bin numbers used, with different bin numbers perhaps a different shape would occur. 7 4 5 5 5 4 5 4 6 5 4 5 5 6 5 8 5 5 7 1 6 5 6 7 1 5 5 9 8 2 4 8 7 8 9 3 7 4 8 5 2 8 7 5 8 5 4 6 7 3 3 4 6 10 4 4 5 7 5 4 7 4 5 4 5 2 1 10 9 4 4 8 5 4 5 3 9 8 4 5 9 8 2 7 9 7 6 3 4 7 5 0 5 11 4 9 5 8 3 5 8 9 6 5 5 7 5 8 6 6 4 8 4 2 4 6 5 5 5 10 7 4 5 6 3 1 10 6 4 8 3 9 9 2 5 3 6 4 3 3 4 9 8 3 5 8 7 8 9 3 7 2 5 2 12 10 3 3 4 8 8 10 4 8 6 6 4 7 4 8 7 8 7 8 8 10 7 13 7 13 1 7 6 9 7 4 7 4 0 7 9 5 3 3 4 4 9 Number of Arrivals at StarBuckes Coffee Shop 5 3 2 7 10 4 5 7 7 7 10 7 6 10 9 5 9 7 9 6 8 5 9 9 3 6 6 8 2 11 7 6 4 9 4 5 1 4 6 6 7 4 10 3 9 6 9 5 10 7 4 3 6 4 7 6 2 7 4 6 5 5 4 3 4 8 10 4 7 9 5 12 8 5 10 7 12 8 8 7 6 2 4 Arrival Time Average 5.910714 St. Dev Sa 2.468996 VAR.S 6.095942 Maximum 13 Minimum 0 Range 13 Bins # of ArrivalFrequency 0 0 2 2 1 or 2 18 4 3 or 4 68 6 5 or 6 77 8 7 or 8 70 10 9 or 10 38 More than More than 7 1a) 1. On average, there are 5.91 arrivals between 9 AM and 10 AM 2. The standard deviation, which is a measure of the dispersion in the arrivals is 2.47 arrivals 3. The variance in this set of data is 6.1. The variance is the standard deviation squared and demonstrates how far the arrivals are spread out. 4. The maximum number of customers who arrive between 9 AM and 10 AM is 13. 5. The minimum number of customers arriving between 9 AM and 10 AM is 0. 6. The range, which is the difference between the maximum and minimum number of arrivals is 13 customers. b) Without any technical analysis (ie histogram), I would predict that there would be a symmetrical shape around 5/6 arrivals. The frequency numbers after and before it are close in value which is why I predict that it would cause such a shape. This is due to the bin numbers used, with different bin numbers perhaps a different shape would occur. 7 4 5 5 5 4 5 4 6 5 4 5 5 6 5 8 5 5 7 1 6 5 6 7 1 5 5 9 8 2 4 8 7 8 9 3 7 4 8 5 2 8 7 5 8 5 4 6 7 3 3 4 6 10 4 4 5 7 5 4 7 4 5 4 5 2 1 10 9 4 4 8 5 4 5 3 9 8 4 5 9 8 2 7 9 7 6 3 4 7 5 0 5 11 4 9 5 8 3 5 8 9 6 5 5 7 5 8 6 6 4 8 4 2 4 6 5 5 5 10 7 4 5 6 3 1 10 6 4 8 3 9 9 2 5 3 6 4 3 3 4 9 8 3 5 8 7 8 9 3 7 2 5 2 12 10 3 3 4 8 8 10 4 8 6 6 4 7 4 8 7 8 7 8 8 10 7 13 7 13 1 7 6 9 7 4 7 4 0 7 9 5 3 3 4 4 9 Number of Arrivals at StarBuckes Coffee Shop 5 3 2 7 10 4 5 7 7 7 10 7 6 10 9 5 9 7 9 6 8 5 9 9 3 6 6 8 2 11 7 6 4 9 4 5 1 4 6 6 7 4 10 3 9 6 9 5 10 7 4 3 6 4 7 6 2 7 4 6 5 5 4 3 4 8 10 4 7 9 5 12 8 5 10 7 12 8 8 7 6 2 4 Arrival Time Average 5.910714 St. Dev Sa 2.468996 VAR.S 6.095942 Maximum 13 Minimum 0 Range 13 Bins # of ArrivalFrequency 0 0 2 2 1 or 2 18 4 3 or 4 68 6 5 or 6 77 8 7 or 8 70 10 9 or 10 38 More than More than 7 1a) 1. On average, there are 5.91 arrivals between 9 AM and 10 AM 2. The standard deviation, which is a measure of the dispersion in the arrivals is 2.47 arrivals 3. The variance in this set of data is 6.1. The variance is the standard deviation squared and demonstrates how far the arrivals are spread out. 4. The maximum number of customers who arrive between 9 AM and 10 AM is 13. 5. The minimum number of customers arriving between 9 AM and 10 AM is 0. 6. The range, which is the difference between the maximum and minimum number of arrivals is 13 customers. b) Without any technical analysis (ie histogram), I would predict that there would be a symmetrical shape around 5/6 arrivals. The frequency numbers after and before it are close in value which is why I predict that it would cause such a shape. This is due to the bin numbers used, with different bin numbers perhaps a different shape would occur. 7 4 5 5 5 4 5 4 6 5 4 5 5 6 5 8 5 5 7 1 6 5 6 7 1 5 5 9 8 2 4 8 7 8 9 3 7 4 8 5 2 8 7 5 8 5 4 6 7 3 3 4 6 10 4 4 5 7 5 4 7 4 5 4 5 2 1 10 9 4 4 8 5 4 5 3 9 8 4 5 9 8 2 7 9 7 6 3 4 7 5 0 5 11 4 9 5 8 3 5 8 9 6 5 5 7 5 8 6 6 4 8 4 2 4 6 5 5 5 10 7 4 5 6 3 1 10 6 4 8 3 9 9 2 5 3 6 4 3 3 4 9 8 3 5 8 7 8 9 3 7 2 5 2 12 10 3 3 4 8 8 10 4 8 6 6 4 7 4 8 7 8 7 8 8 10 7 13 7 13 1 7 6 9 7 4 7 4 0 7 9 5 3 3 4 4 9