Homework 6(Dennis)
1. When a pizza restaurant’s delivery process is operation effectively, pizzas are
delivered in an average of 45 minutes with a standard deviation of 6 minutes. To
monitor its delivery process, the restaurant randomly selects five pizzas each night
and records their delivery times.
For the sake of argument, assume that the population of all delivery times on a
given evening is normally distributed with a mean of
 =45
minutes and a
standard deviation of  =6 minutes.
a. Describe the shape of the population of all possible sample means. How do
you know what the shape is?
b. Find the mean and the standard deviation of the population of all possible
sample means.
c. Calculate an interval containing 99.73 percent of all possible sample means.
Solution
a. Normally distributed, because population is normal.

6
x 

 2 .6 8 3 3
b.  x    45
n
5
c. 45 ± 3(2.6833) = [45 ± 8.0499] = [36.9501, 53.0499]
2. Quality Progress, February 2005, reports on improvements in customer satisfaction
and loyalty made by Bank of America. A key measure of customer satisfaction is the
response (on a scale from 1 to 10) to the question: “Considering all the business you
do with Bank of America, what is your overall satisfaction with Bank of America?”
Here, a response of 9 or 10 represents “customer delight”. Consider the survey of 350
Bank of America customers, and assume that 48% of Bank of America customers
would currently express customer delight. That is, assume p=0.48. Find the
probability that the sample proportion obtained from the sample of 350 Bank of
America customers would be within three percentage points of the population
proportion. That is, find P(0.45  pˆ  0.51) .
Solution
If p=0.48, np=350(0.48)=168  5 , n (1-p)=350(1-0.48)=182  5 , so
p̂
is
approximately normal with
 pˆ  p  0.48
 pˆ 
p(1  p)
0.48(1  0.48)

 0.0267
n
350
.51 .48 
 .45 .48
P(.45  pˆ  .51)  P
z
  P(1.12  z  1.12)  0.7372
.
0267
.0267 

3. The mean of the sample of 65 customer satisfaction ratings is x =42.95. If we let
 denote the mean of all possible customer satisfaction ratings for the XYZ Box
video game system, and assume that  equals 2.64:
a. Calculate 95 percent and 99 percent confidence intervals for  .
b. Using the 95 percent confidence interval, can we be 95 percent confident that 
is at least 42? Explain.
c. Using the 99 percent confidence interval, can we be 99 percent confident that 
is at least 42? Explain.
d. Based on your answers to parts b and c, how convinced are you that the mean
satisfaction rating is at least 42?
Solution

 2.64 
  [42.308,43.592 ]
 65 
a. 42.95  1.96


 2.64 
  [42.107 ,43.793]
42.95  2.575
 65 

b.
Yes, 95% interval is above 42.
c.
Yes, 99% interval is above 42.
d.
Very confident based on the 99% confidence interval being above 42.
4. Whole Foods is an all-natural grocery chain that has 50,000 square foot stores, up
from the industry average of 34,000 square feet. Sales per square foot of supermarkets
average just under $400 per square foot, as reported by USA Today in an article on “A
whole new ballgame in grocery shopping.” Suppose that sales per square foot in the
most recent fiscal year are recorded for a random sample of 10 Whole Foods
supermarkets. The data (sales dollars per square foot) are as follows:
854,858,801,892,849,807,894,863,829,815.
a. Find a 95 percent confidence interval for the true mean sales dollars per square foot
for all Whole Foods supermarkets during the most recent fiscal year.
b. Are we 95 percent confident that this mean is greater than $800, the historical
average for Whole Foods?
Solution
From the data given, we can obtain
x =846.2 and s=32.866

 32.866 
  [822 .69,869 .71]
 10 
a. 846 .2  2.262

b.We are at least 95% confident the mean is greater than 800 because the interval is
above 800.
5. Referring to the above question, Q5, regard the sample of 10 sales figures for which
s=32.886 as a preliminary sample. How large a sample of sales figures is needed to
make us 95 percent confident that x , the sample mean sales dollars per square foot,
is within a margin of error of $10 of  , the true mean sales dollars per square foot
for all Whole Foods supermarkets.
Solution
2
2
t s
 2.262(32.866) 
n   0.025   
  56 sales figures
10


 E 
Where t0.025  2.262 which is based on n-1=9 degrees of freedom..
6. Suppose we conduct a poll to estimate the proportion of voters who favor a major
presidential candidate. Assuming that 50 percent of the electorate could be in favor of
the candidate, determine the sample size needed so that we are 95 percent confident
that p̂ , the sample proportion of voters who favor the candidate, is within a margin
of error of 0.01 of p, the true proportion of all voters who are in favor of the
candidate.
Solution:
2
 1.96 
n  .5(1 – .5)
  .25(38,416 )  9604 voters
 .01 