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A. Parameter Estimation
1. Review of the Normal Distribution
A1
2. Point and Interval Estimation
3. A Confidence Interval for the Mean when the Population Variance is Known.
4. A Confidence Interval for the Mean when the Population Variance is not Known.
A2, text 8.20, 8.50 [8.21, 8.50] (8.21, 8.50) – Answers from both editions will be provided for 8.21. 8.95 on CD (8.93)
Graded Assignment 1 (Will be posted)
5. Deciding on Sample Size when working with a Mean
A3, 8.38 [8.36] (8.36)
6. A Confidence Interval for a Proportion.
Text 8.24, 8.25, 8.26, 8.58, 8.94 on CD [8.22, 8.23, 8.24, 8.58, 8.93a,c on CD] (8.22, 8.23, 8.25, 8.58, 8.91a,c)
7. A Confidence Interval for a Variance.
Text 12.1-12.2 [9.72] (9.67), A4
8. (A Confidence Interval for a Median.)
Optional - A5 -- solution is posted.
----------------------------------------------------------------------------------------------------------------------------- ---Problems 8.24 through A4 are in this document.
Problems involving a Confidence Interval for a proportion:
Exercise 8.24 [Exercise 8.22 in 9th edition] (Exercise 8.22 in 8th edition): The problem says that
n  200 and x  50 , and wants a 95% confidence interval for the fraction (proportion).
x
50
 .250 . So the confidence
Solution; Since 1    .95 ,   .05 and z  z  z.025  1.960 . p  
2
n 200
.250 .750 
pq
 .250  1.960
 .250  1.960 .0009375
n
200
 .250  1.960 .03062   .250  .060 and .190  p  .310. More formally, we can say
P.190  p  .310  .95 or make a diagram.
interval is p  p  z  2 s p  p  z  2
Exercise 8.25 (Exercise 8.23 in 8th and 9th edition): The problem says that n  400 and x  25 , and
wants a 99% confidence interval for the fraction (proportion).
x
25
 .0625 . So the
Solution: Since 1    .99 ,   .01 and z  z  z.005  2.576 . p  
2
n 400
confidence interval is p  p  z  2 s p  p  z  2
pq
n
.0625 .9375 
 .0625  2.576 .000146484  .0625  2.576 .012103   .0625  .0312 and
400
.0313  p  .0937. More formally, we can say P.0313  p  .0937   .99 or make a diagram.
 .0625  2.576
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Exercise 8.26 [Exercise 8.24 in 9th edition] (Exercise 8.25 in 8th edition): The problem says that
n  500 and x  135 , (where x represents the number of households in a sample of 500 that would buy an
additional line at a reduced installation cost) and a) wants a 99% confidence interval for the fraction
(proportion) that would buy the line. b) It then asks how a manager would use these results.
x 135
Solution: a) Since 1    .99 ,   .01 and z  z  z.005  2.576 . p  
 .270 . So the
2
n 500
confidence interval is p  p  z  2 s p  p  z  2
 .270  2.576
pq
n
.270 .730 
 .270  2.576 .0003942  .270  2.576 .01985   .270  .051 and
500
.219  p  .321.
The Minitab output follows. The ‘pone’ instruction does a confidence interval and hypothesis test for one
proportion, the word ‘pone’ is followed by the number of tries and the number of successes. The semicolon
prompts the system to request a subcommand. If there is no subcommand, Minitab will assume a
confidence level of 95%. Most Minitab commands that give a confidence interval also do a hypothesis test.
Since I did not specify a hypothesis test, Minitab assumes that I am testing to see if the population
proportion is .5. The p-value represents the probability that the tested hypothesis is correct and, since it is
zero, tells us that there is virtually no chance that the population proportion is .5.
MTB > Pone 500 135;
SUBC>
Confidence 99.0.
Test and CI for One Proportion
Test of p = 0.5 vs p not = 0.5
Exact
Sample
X
N Sample p
1
135
500 0.270000
99.0% CI
(0.220272, 0.324200)
P-Value
0.000
The Instructor’s Solution Manual provides an Excel solution of the same problem and the answer to Part b).
8.24
(a)
(b)
The manager in charge of promotional programs concerning residential customers can infer that
the proportion of households that would purchase an additional telephone line if it were
made available at a substantially reduced installation cost is somewhere between 0.22 and
0.32 with a 99% level of confidence.
Wouldn’t the next step be to check how much it would cost to lower the installation cost and to compare
the cost with the additional profits that would be made for various success rates between 22% and 32%?
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Exercise 8.58 (Exercise 8.58 in 8th and 9th edition): The problem says that n  300 and x  11 , (where x
represents the number of invoices in a sample of 300 from a population of 10000 that
violate an internal control) and a) wants a 95% one-sided upper confidence interval for
the fraction (proportion) that violate the internal control. It asks b) what to do if the
proportion that violates the control is above a tolerable exception rate of 4%.
pq
Solution: a) The formula for a top-side confidence interval is p  p  z s p  p  z 
. Since
n
x
11
1    .95 ,   .05 and z  z  z.05  1.645 . p  
 .03667 . In this section the authors assume
n 300
that the sample size is more than 5% of the population size, so that the formula becomes
N  n pq
N  n pq
. If we use this apparently unnecessary formula we get p  p  z 
p  p  z
N 1 n
N 1 n
 .0367  1.645
10000  300
10000  1
.0367 .9633 
, which the Instructor’s Solution Manual says is equal to
300
.0542. If we omit the finite population correction, we get p  p  z 
pq
.0367 9633 
 .0367  1.645
n
300
 .0367  1.645 .00011784  .0367  1.645 0.01856   .0546 . In any case, the conclusion is that this upper
bound is above 4%.
b) Since the upper bound is higher than the tolerable exception rate of 0.04, the auditor should request a
larger sample.
What we actually have here is a hypothesis test of the statement ‘ p  .04 .’ More on this later.
Exercise 8.94 [Exercise 8.93a,c on CD in 9th edition] (Exercise 8.91a,c in 8th edition): An automobile
dealer wants to estimate the proportion of customers who still own cars that they purchased 5 years earlier.
Sales records indicate that the population of owners is 4000.
a) Set up a 95% confidence interval estimate of the population proportion of all customers who still own
their cars 5 years after they were purchased if a random sample of 200 customers selected without
replacement from the automobile dealer’s records indicate that 82 still own cars that were purchased 5
years earlier.
b) (This was not assigned, but is worth learning about.) What sample size is necessary to estimate the true
proportion to within .025 with 95% confidence?
c) What are your answers to a) and b) if the population consists of 6000 owners?
Solution: a) The problem says that n  200 and x  82 , (where x represents the number of customers in
the sample of 200 from a population of N  4000 that still own their cars) and wants a 95% two-sided
confidence interval for the fraction (proportion) that still own their cars. Since 200 is exactly 5% of the
population, caution might lead us to use a finite population correction. Since 1    .95 ,   .05 and
x
82
 .410 . So the confidence interval is
z  z 2  z.025  1.960 . p  
n 200
p  p  z 2 s p  p  z 2
 .410  1.960 0.9502375
N n
N 1
pq
4000  200
 .410  1.960
n
4000  1
.410 .590 
200
.0012095  .410  1.960 .9748012 .0347778   .410  .066 or we could say
P.344  p  .476   .95 . Check this – there is no answer given by the author!
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b) Our confidence level is 1    .95 ,   .05 so z  z  z.025  1.960 . The problem asks for a
2
confidence interval with an error of e  .025 . First, let us use the formula given in class. If we did not
have a value for p , we would use p  .5 and q  1  p  .5 in the formula
n0 
pqz 2

.5.51.960 2
 1536 .64 and we would need a sample of 1537. However, the population size
e2
.025 2
is N  4000 , and 1537 is more than 5% of the population, so we ought to use the finite population
correction factor. The corrected formula given on the CD was
n0 N
1536 .64 4000 
 4000 
n

 1536 .64 
  1536 .64 .7226   1110 .36 , which implies a
n 0  N  1 1536 .64  4000  1
 5535 .64 
sample size of 1111.
But, let us assume that we believe our previous estimate is correct, that is p  .41 . Then, using our original
formula n 0 
pqz 2

.41.59 1.960 2
 1486 .85 and we would need a sample of 1487. However, the
e2
.025 2
population size is N  4000 , and 1487 is more than 5% of the population, so we ought to use the finite
population correction factor. The corrected formula given on the CD was
n0 N
1486 .854000 
 4000 
n

 1486 .85
  1486 .85.7291   1084 .13 , which implies a
n 0  N  1 1486 .85  4000  1
 5485 .85 
sample of 1085. Let’s see how this works out. Using the formula we used in part a, and assuming that we
N  n pq
still find a sample proportion of .41, p  p  z  2 s p  p  z  2
N 1 n
4000  1085 .410 .590 
 .410  1.960 0.7289322 .0002229493
4000  1
1085
 .410  1.960 .8537752 .0149314   .410  .025 .
 .410  1.960
c) In this part of the problem N  6000 , which is more than 20 times n  200 , and the finite population
correction factor is not needed. We can use the formula given in class, and say
.410 .590 
pq
p  p  z 2 s p  p  z 2
 .410  1.960
n
200
 .410  1.960 .0012095  .410  1.960 .0347778   .410  .062 or we could say P.348  p  .472   .95 .
Check this – there is no answer given by the author!
Our confidence level is 1    .95 ,   .05 so z  z  z.025  1.960 . The problem asks for a confidence
2
interval with an error of e  .025 . First, let us use the formula given in class. If we did not have a value
for p , we would use p  .5 and q  1  p  .5 in the formula n 0 
pqz 2

.5.51.960 2
 1536 .64 and
e2
.025 2
we would need a sample of 1537. This is the answer I would expect if you had not done b). However, the
population size is N  6000 , and 1537 is more than 5% of the population, so we ought to use the finite
population correction factor. The corrected formula given on the CD was
n0 N
1536 .64 6000 
 6000 
n

 1536 .64 
  1536 .64 .7962   1223 .50 , which implies a
n 0  N  1 1536 .64  6000  1
 7535 .64 
sample size of 1224.
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But, let us assume that we believe our previous estimate is correct, that is p  .41 . Then, using our original
formula n 0 
pqz 2

.41.59 1.960 2
 1486 .85 and we would need a sample of 1487. However, the
e2
.025 2
population size is N  6000 , and 1487 is more than 5% of the population, so we ought to use the finite
population correction factor. The corrected formula given on the CD was
n0 N
1486 .856000 
 6000 
n

 1486 .85
  1486 .85 .8015   1191 .72 , and implies a sample
n 0  N  1 1486 .85  6000  1
 7485 .85 
size of 1192.
Problems involving a Confidence Interval for a Standard Deviation or a Variance
Exercise 12.1-2 [Exercise 9.72 in 9th edition] (9.67 in 8th edition): From the Instructor’s Solution Manual,
2
9.67
(a)
For df = 15 and  = 0.01,  = 30.578. (On my table 30.5779)
(b)
(c)
(d)
(e)
(f)
 = 0.025,  2 = 20.483. (On my table 20.4832)
2
For df = 7 and  = 0.05,  = 14.067. (On my table 14.0671)
2
For df = 27 and  = 0.95,  = 16.151. (On my table 16.1514)
2
For df = 20 and  = 0.975,  = 9.591. (On my table 9.5908)
2
For df = 4 and  = 0.99,  = 0.297. (On my table 0.2971)
For df = 10 and
PROBLEM A4: If s = 15 find a 95% confidence interval for  if a) n = 26, b) n = 99
Solution: Use the formulas from Table 3 of the syllabus supplement or from the outline.
a) This is a small sample since n  31 , so use
freedom are
n  1s 2
 22
 2 
n  1s 2
12 2
. Since the degrees of
2
 13 .1197 , the
n  1  26  1  25 ,  22   .2025  40 .6466 and 12 2   975
25 15 2
25 15 2
or 138 .388   2  428 .745 . Since an interval for
40 .6466
13 .1197
the standard deviation was requested, take the square root of both sides. 11.76    20.71 .
b) Since the degrees of freedom are n  1  99  1  98 and are too large for the chi-square table use
interval becomes
s 2DF 
z 2  2DF 
 
 2 
s 2DF 
 z 2  2DF 
.
Since
2DF   298   196  14
and
15 14 
15 14 
 
or 13.158    17.442 .
1.960  14
 1.960  14
Note that due to the larger sample size, this interval is smaller than the one in a.
z 2  z.025  1.960 , the formula becomes