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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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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