## (Solved) Problem Set 3 Due on Friday, March 10, 2017, 12PM 1. Quality Progress reports on improvements in customer satisfaction and loyalty made by OCBC. A

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Problem Set 3
Due on Friday, March 10, 2017, 12PM 1. Quality Progress reports on improvements in customer satisfaction and loyalty made by OCBC. A key measure of customer satisfaction is the
response (on a scale from 1 to 10, 10 being most satised) to the question:
Considering all the the business you do with OCBC, what is your overall
satisfaction with OCBC?
(a) Historically the % of OCBC customers choosing a score of 9 or more
(extremely satised) is 48%. Suppose we wish to use a survey of 350 OCBC customers to justify the claim that more than 48% of all
current OCBC customers are extremely satised. The survey nds
that 189 of 350 customers rate 9 or more. Compute the probability
of observing a sample proportion greater than or equal to .54.
(b) Based on the probability you computed in part (a) would you conclude that more than 48% of current OCBC customers are extremely
satised? Explain.
(c) Find the probability that the sample proportion from the sample of
350 OCBC customers would be within
i. 3 percentage points of the population proportion.
ii. 6 percentage points of the population proportion.
(d) In 418 telephone interviews conducted, 67% of the respondents gave
a high rating for overall satisfaction.
i. Calculate a 99% condence interval for the proportion of all customers that are extremely satised.
ii. Can we be 99% condent that more than 60% of all customers
are extremely satised.
2. A computer supply house receives a large shipment of ash drives each
week. Past experience shows that the number of aws per drive is either
0, 1, 2 or 3 with probabilities .65, .2, .1 and .05 respectively.
(a) Calculate the mean and the standard deviation of the number of
aws.
(b) Suppose we select a sample of 100 ash drives. What is the distribution of the mean number of aws. Explain. 1 (c) A shipment is rejected if the mean number of aws per drive for a
sample of 100 is greater than .75. Suppose the actual number of aws
per drive in this weeks shipment is .55. What is the probability that
this shipment will be rejected.
3. Each day a manufacturing plant receives a large shipment of drums of
Chemical ZX-900. These drums are suppose to have a mean ll of 50 gallons and a s.d. of .6 gallon.
(a) If we draw a sample of 100 drums from the shipment, what is the
probability that the average ll for the 100 drums is between 49.88
gallons and 50.12 gallons.
(b) The plant manager will reject a shipment if the average ll of 100
drums is less than 49.85 gallons. Suppose a shipment which has a mean ll of 50 gallons is received. What is the probability that this
shipment is rejected?
4. A company is considering a new bottle design for a popular soft drink
and takes a sample of 60 consumer ratings of this new bottle design. The
mean and the s.d of the 60 bottle design ratings are 30.35 and 3.1073
respectively. Compute a 95% condence interval for the mean. Interpret
this interval.
5. An ad agency output is described by nding the shares of dollar billing
volume coming from various media categories such as network TV, spot
TV, newspapers, radio and so forth.
(a) Suppose that a random sample of 400 advertising agencies gives an
average percentage share of billing volume from network TV equal
to 7.46% and assume that the population s.d. equals to 1.42%. Compute a 95% condence interval for the mean percentage share
of billing volume from network TV.
(b) Suppose that a random sample of 400 advertising agencies gives an
average percentage share of billing volume from spot TV equal to
12.44% and assume that the population s.d. equals to 1.55%. Compute a 95% condence interval for the mean percentage share of
billing volume from spot TV.
(c) Compare the intervals in (a) and (b). Does it appear that the mean
percentage share of billing volume from spot TV is greater than the
mean percentage share of billing volume from network TV? Explain.
6. Suppose we conduct a poll to estimate the proportion of voters who favour
a major presidential candidate. Assuming that 50% of the electorate could
be in favour of the candidate, determine the sample size needed so that we
are 95% condent that pË†, the sample proportion of all voters who favour athe candidate, is within a margin of error of .01 of 2 p.

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