Put all answers (including SPSS output) in a single Word or Excel file and submit via Canvas. Clearly label all questions. Keep all parts of a question together (e.g., do not answers part a of...

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Put all answers (including SPSS output) in a single Word or Excel file and submit via Canvas. Clearly label all questions. Keep all parts of a question together (e.g., do not answers part a of question 1, then question 3, then part b of question 1). Work parts of a question in order. Do not lock the file. Do not submit SPSS files. ? You may use a calculator, Excel, SPSS, MS Word and your notes. ? Be concise and to the point. Good answers prioritize what is most important. Watch your rounding on the exam. I will deduct points for rounding that causes substantively different answers. Show work to receive partial credit. ? You may never share this exam with another student in the IMC program.


IMC 451: Marketing Research & Statistics Final Examination 
 WINTER 2015 1 IMC 451: Marketing Research & Statistics Professor Wheeler Final Examination 
 WINTER 2015 Read all of the following instructions before beginning the exam:  If you are reading this, you now have three hours to complete this exam.  Put all answers (including SPSS output) in a single Word or Excel file and submit via Canvas. Clearly label all questions. Keep all parts of a question together (e.g., do not answers part a of question 1, then question 3, then part b of question 1). Work parts of a question in order. Do not lock the file. Do not submit SPSS files.  You may use a calculator, Excel, SPSS, MS Word and your notes.  Be concise and to the point. Good answers prioritize what is most important. Watch your rounding on the exam. I will deduct points for rounding that causes substantively different answers. Show work to receive partial credit.  You may never share this exam with another student in the IMC program. IMC 451: Marketing Research & Statistics Final Examination 
 WINTER 2015 2 1. (25 points) A recent NY Times article reports: “More people in New York are looking for caterers to prepare the holiday meal than anywhere else in the country. Live in the Southeast? Brace yourself for a big scoop of broccoli casserole.” Moreover, “sweet potato casserole” is much more common in the south than the rest of the country. These results are derived from searches at allrecipes.com. Sweet potato casserole is the most searched term, with 310 searches per 10,000. In Tennessee it accounts for 550 searches per 10,000, which is a (550-310)/310 = 77% increase over the national average. (a) (5 points) Identify the implied population. (b) (5 points) Identify the sampling frame / sampled group. (c) (5 points) What type of sampling plan is this (e.g., simple random, convenience, judgment, quota, stratified, cluster, systematic, area, snowball)? Briefly explain. (d) (5 points) Briefly discuss possible sampling biases. (e) (5 points) The New York Times does not report a margin of error. Should it have? 2. (12 points) Your company maintains a database with information on your customers, and you are interested in analyzing patterns observed over the past quarter. In particular, 30% of customers in the database planed new orders within this period. However, for those customers who had a salesperson assigned to them, the new order rate was 60%. Overall, 10% of customers within the database had salespeople assigned to them. (a) (3 points) What percentage of customers in the database placed a new order but did not have a salesperson assigned to them? (b) (3 points) Given that a customer did not place a new order, what is the probability that they had a salesperson assigned to them? (c) (3 points) If a customer did not have a salesperson assigned to them, what is the probability that they placed a new order? (d) (3 points) What is the probability that a customer placed a new order or has a salesperson assigned to them? IMC 451: Marketing Research & Statistics Final Examination 
 WINTER 2015 3 3. (16 points) Crates to be delivered by a freight company have a mean weight of 250 pounds and a standard deviation of 66 pounds. Suppose that 36 crates are taken at random and loaded onto a truck. (a) (4 points) Can you find the probably that a single crate selected at random weighs more than 275 pounds? Explain briefly. (b) (4 points) Find the mean, standard deviation, and distribution shape of the random variable for the total weight of all 36 crates. How do you know the shape? (c) (4 points) What is the approximate probability that the total weight of 36 crates taken at random and loaded onto a truck will exceed the specified weight capacity of the truck, which is 10,000 pounds? (d) (4 points) What is the 95th percentile of the total weight of 36 crates? 4. (35 Points) A company that sells memberships for its services wants to study the long- term value of its customers. Customers pay a monthly fee for the service and may cancel at any time. They draw a sample of customers who were acquired during October, 2013. They compute the number of months that a customer was a member (“active”) for the next 12 months (November, 2013 through October, 2014). A frequency distribution is given below (e.g.,10,460 canceled the membership immediately after signing up and were “active” 0 months over the next year). Answer the questions below. (a) (5 points) What is the sample size? (b) (5 points) What percentage of customers were active 6 months or more? (c) (5 points) True or false: the distribution is approximately normal. Explain. (d) (5 points) What is the median? (e) (5 points) Find the first and third quartiles. (f) (5 points) What is the range? (g) (5 points) What is the interquartile range? IMC 451: Marketing Research & Statistics Final Examination 
 WINTER 2015 4 5. (32 points) The director of training for a direct marketing team is interested in determining whether different training methods have an effect on the productivity of call center employees. She decides to study all 40 people who were hired in June, most of whom are college students working over the summer. She randomly assigns the 40 recent hires to two groups of 20. The first group receives a computer-assisted, individual-based training program and the other receives a team- based training program. Upon completion, the employees are evaluated on the length (in seconds) of their calls. Data are provided in comptrain.csv. The program variable equals 1 for the team-based training and 2 for the computer-based training. (a) (5 points) Draw a diagram using X’s, O’s, and R’s for this experiment. (a) (7 points) Read the data into SPSS and give value labels to the program variable. Use Compare Means to find basic descriptives (n, mean, and standard deviation) of the time variable for each value of program. Submit the output. (b) (3 points) Create boxplots of time for the two levels of time. State whether the assumptions of a t-test are satisfied. (c) (5 points) We wish to determine if there a difference in the average time to as- semble a part between the two groups? State the null and (two-sided) alternative hypothesis. Define your symbols (i.e., the parameters). (d) (5 points) Give the P-value for testing the hypotheses in part (c). Using a 0.05 level of significance, can you reject H0? What do you conclude about which training, if any, is more effective? (e) (5 points) Find the 95% confidence interval for the mean difference. (f) (2 points) What is the standard error of the sample mean time to assemble a part after computer-based training? 4. (30 points) IBM conducted a survey of 1650 CMOs from organizations from around the world. Five segments of CMOs were identified based on their control over the marketing functions (price, product, promotion and place) and their use of data in making marketing decisions. The segment labeled “D” below has full control over all four Ps and uses data. The segment “A” has little or no control over the 4Ps and does not use data. Segment “B” has substantial control over promotion and product, and uses data, but does not have much control over price or distribution. Segment “C” has control over promotion only and segment “E” has substantial, but not full, control over all four Ps and uses data. Refer to segments using the letters A, B, C, D, and E. The crosstab below shows the relationship between segment membership and the region of the world. The purpose of this exercise is to understand the relationship between segment membership and region. The number in each cell gives the cell count. IMC 451: Marketing Research & Statistics Final Examination 
 WINTER 2015 5 (a) (6 points) Enter the data into SPSS and label the rows and columns. Submit your crosstab. (b) (3 points) State the null and alternative to do a chi-square test of independence. Give the P-value and your decision. What does it tell you? Be brief. (c) (3 points) Find the adjusted residual for segment C (promote only) and Asia Pacific. What does it tell you? Be brief. (d) (3 points) Given that a CMO is based in North America, what is the estimated probability that the CMO has full control (segment D)? (e) (3 points) Consider the cell for segment C (promote only) and Western Europe, which has an observed count of 114. If segment were independent of region, how many CMOs would you expect in this cell? (f) (3 points) Find the contribution to the chi-square statistic for the segment C, Western Europe cell (i.e., the standardized residual). (g) (3 points) Find the adjusted standardized residuals for the segment C, Western Europe cell. What does it tell you? (h) (6 points) Is the proportion of CMOs in North American who have control over promotion only (Segment C) significantly different from the proportion of CMOs in Western Europe who have control over promotion only (Segment C)? To answer this question, perform an independent sample Z test of proportions, conditioning on region. Be sure to state the null and alternative hypotheses, find the P-value, and state your conclusion. Show work for partial credit.
Answered Same DayDec 25, 2021

Answer To: Put all answers (including SPSS output) in a single Word or Excel file and submit via Canvas....

David answered on Dec 25 2021
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IMC 451: Marketing Research & Statistics Final Examination WINTER 2015
1. (25 points) A recent NY Times article reports: “More people in New York are looking for caterers to prepare the holiday meal than anywhere else in the country. Live in the Southeast? Brace yourself for a big scoop of broccoli casserole.” Moreover, “sweet potato casserole” is much more common in the south than the rest of the country. These results are derived from searches at allrecipes.com. Sweet potato
casserole is the most searched term, with 310 searches per 10,000. In Tennessee it accounts for 550 searches per 10,000, which is a (550-310)/310 = 77% increase over the national average.
(a) (5 points) Identify the implied population.
The implied population is people live in Tennessee.
(b) (5 points) Identify the sampling frame / sampled group.
The sampling frame is allrecipes.com.
(c) (5 points) What type of sampling plan is this (e.g., simple random, convenience, judgment, quota, stratified, cluster, systematic, area, snowball)? Briefly explain.
This type of sampling plan is convenience as selecting people by this method is easy because of their availability or easy access.
(d) (5 points) Briefly discuss possible sampling biases.
The sample is biased as it did not include the people who liked sweet potato casserole but did not search it in allrecipes.com.
(e) (5 points) The New York Times does not report a margin of error. Should it have?
The sample proportion is p = 550/10000 = 0.055 is the number in the sample with the characteristic of interest. The standard error is ({p(1-p)/n} = 0.002. If the confidence level is 95%, the z-value is 1.96.
Margin of error is 1.96 (0.002) = 0.004. According to this data, we conclude with 95% confidence that 5.5% of Tennessee people searched sweet potato casserole, plus or minus 0.4%. Therefore, it should not have error.
2. (12 points) Your company maintains a database with information on your customers, and you are interested in analyzing patterns observed over the past quarter. In particular, 30% of customers in the database planed new orders within this period. However, for those customers who had a salesperson assigned to them, the new order rate was 60%. Overall, 10% of customers within the database had salespeople assigned to them.
(a) (3 points) What percentage of customers in the database placed a new order but did not have a salesperson assigned to them?
Let
A = customers in the database planed new orders
B = customers within the database had salespeople assigned
P(A) = 0.3
P(A | B) = 0.6
P(B) = 0.1
P(ABc )
= P(A) - P(AB)
= P(A) - P(A | B)P(B)
= 0.3 - 0.6(0.1
=0.24
(b) (3 points) Given that a customer did not place a new order, what is the probability that they had a salesperson assigned to them?
P(B|Ac)
= (P(B) - P(AB))/ (1 - P(A))
= (P(B) - P(A | B)P(B))/ 0.7
= (0.1 - 0.6(0.1)/0.7
=0.057
(c) (3 points) If a customer did not have a salesperson assigned to them, what is the probability that they placed a new order?
P(A | Bc )
= (P(A) - P(AB))/ (1 - P(B))
= (P(A) - P(A | B)P(B))/ 0.9
= (0.3 - 0.6(0.1)/0.9
=0.27
(d) (3 points) What is the probability that a customer placed a new order or has a salesperson assigned to them?
P(AB) = P(A | B)P(B) = 0.6(0.1 = 0.06
3. (16 points) Crates to be delivered by a freight company have a mean weight of 250 pounds and a standard deviation of 66 pounds. Suppose that 36 crates are taken at random and loaded onto a truck.
(a) (4 points) Can you find the probably that a single crate selected at random weighs more than 275 pounds? Explain briefly.
Let X be the r.v. of create weight
X ( N(250, 662)
Z = (X - 250)/66 ( N(0, 12)
P(X > 275) = P(Z > (275 - 250)/66) = P(Z > 0.38) = 1 - ((0.38) = 1 - 0.648 = 0.352
(b) (4 points) Find the mean, standard deviation, and distribution shape of the random variable for the total weight of all 36 crates. How do you know the shape?
Mean = 250
Standard deviation = 66/(36 = 11
The distribution shape will be bell shaped as the distribution is normal.
(c) (4 points) What is the approximate probability that the total weight of 36 crates taken at random and loaded onto a truck will exceed the specified weight capacity of the truck, which is 10,000 pounds?
(Xi ( N(36(250, 36(662)
i.e. (Xi ( N(9000, 3962)
Z = ((Xi - 9000)/396 ( N(0, 12)
P((Xi >10000) = P(Z > (10000 - 9000)/396) = P(Z > 2.53) = 1 - ((2.53) = 1 - 0.9943 = 0.0057
(d) (4 points) What is the 95th percentile of the total weight of 36 crates?
P((Xi >x) = 0.95
=> P(Z >(x -10000)/396)) = 0.95
=> (x -10000)/396) = 1.645
=> x = 10651.42
4. (35 Points) A company that sells memberships for its services wants to study the long-term value of its customers. Customers pay a monthly fee for the service and may cancel at any time. They draw a sample of...
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