6.1 Point Estimation
The following
table contains some of the well known population parameters and their point
estimates based on a random sample.
Table 1
Population parameters and their corresponding sample estimates

Population 
Sample 
Mean 
_{} 
_{} 
Variance 
_{} 
_{} 
Proportion 
_{} 
_{} 
6.2 Confidence Interval Estimation for
the Population Mean
Point estimates
may be far away from the true parameter if the estimators have large variances.
So we want to estimate parameters by confidence intervals that consider the
variability and the sampling distribution.
Confidence Interval Estimation on
the Mean of a
A _{} confidence interval
for mean _{}is given by:
_{}
or
_{}
where is the _{} is the _{} percentile of the standard
normal distribution.
Example 6.1 The average zinc
concentration recovered from a sample of zinc measurements in 36 different
locations is found to be 2.6 grams per millimeter. Find a 95% confidence
interval for the mean zinc concentration in the river. Assume that the
population standard deviation is 0.3 (Walpole, Myers, Myers and Ye, 2003, pp.
236237).
Solution Since the population
standard deviation _{}, we assume that zinc measurements follow a
normal distribution _{}. With _{}, _{} so
that a 95% CI for _{} is given by
_{}, i.e. _{}
Large
Sample Confidence Interval for the Population Mean
A _{}% confidence interval (CI) for the population mean _{}is given by
_{}
Example 6.2 A meat inspector has
randomly measured 30 packs of acclaimed 95% lean beef. The sample resulted in
the mean 96.2% with the sample standard deviation of 0.8%. Find a 98%
confidence interval for the mean of similar packs (
Solution A 98% CI for _{} is given by
_{}, i.e.
_{}
An
Experiment for Large Sample Confidence Interval
One hundred samples each of size 30 have been drawn from an exponential
distribution with mean 2, and 95% confidence interval have been calculated for
each sample using Statistica. The sample
mean, LCL (Lower Confidence Limit) and
UCL (Upper Confidence Limit) are given
in Table 6.2. The interval that contains the true mean 2 is followed by a Y,
otherwise by N.
Table 6.2 Large Sample Confidence Intervals by Statistica
Sample 
Mean 
LCL 
UCL 
Y/ N 
NewVar1 
2.1790 
1.5788 
2.7792 
Y 
NewVar2 
1.9778 
1.1719 
2.7836 
Y 
NewVar3 
2.5806 
1.6660 
3.4951 
Y 
NewVar4 
2.1161 
1.2453 
2.9868 
Y 
NewVar5 
1.9929 
1.2145 
2.7714 
Y 
NewVar6 
2.3214 
1.4795 
3.1633 
Y 
NewVar7 
2.0379 
1.3489 
2.7271 
Y 
NewVar8 
2.4378 
1.3352 
3.5405 
Y 
NewVar9 
2.1490 
1.3694 
2.9287 
Y 
NewVar10 
1.7582 
1.2117 
2.3047 
Y 
NewVar11 
1.7566 
1.1819 
2.3312 
Y 
NewVar12 
2.0279 
1.1847 
2.8710 
Y 
NewVar13 
2.0105 
1.2118 
2.8094 
Y 
NewVar14 
1.8774 
1.1075 
2.6473 
Y 
NewVar15 
1.7119 
1.1854 
2.2386 
Y 
NewVar16 
1.8232 
1.1598 
2.4867 
Y 
NewVar17 
2.7502 
1.8101 
3.6902 
Y 
NewVar18 
2.1466 
1.2467 
3.0465 
Y 
NewVar19 
1.9332 
1.2452 
2.6212 
Y 
NewVar20 
2.0821 
1.3345 
2.8297 
Y 
NewVar21 
1.4457 
0.9776 
1.9138 
N 
NewVar22 
1.7582 
1.2355 
2.2809 
Y 
NewVar23 
1.9975 
1.2987 
2.6962 
Y 
NewVar24 
1.7588 
0.8082 
2.7094 
Y 
NewVar25 
1.8509 
1.1329 
2.5689 
Y 
NewVar26 
1.3934 
0.8597 
1.9271 
N 
NewVar27 
1.8494 
1.2896 
2.4092 
Y 
NewVar28 
1.8064 
1.1125 
2.5002 
Y 
NewVar29 
2.3428 
1.5917 
3.0939 
Y 
NewVar30 
1.8725 
1.2316 
2.5134 
Y 
NewVar31 
1.9489 
1.2072 
2.6906 
Y 
NewVar32 
1.8292 
0.9996 
2.6588 
Y 
NewVar33 
1.4579 
1.0048 
1.9111 
N 
NewVar34 
2.1429 
1.2687 
3.0173 
Y 
NewVar35 
2.0964 
1.0399 
3.1528 
Y 
NewVar36 
1.9006 
1.2708 
2.5310 
Y 
NewVar37 
2.0545 
1.2358 
2.8732 
Y 
NewVar38 
2.0777 
1.1938 
2.9616 
Y 
NewVar39 
2.1266 
1.2661 
2.9872 
Y 
NewVar40 
2.0915 
1.3087 
2.8744 
Y 
NewVar41 
2.3041 
1.3396 
3.2686 
Y 
NewVar42 
1.5656 
0.9763 
2.1549 
Y 
NewVar43 
2.7001 
1.6032 
3.7970 
Y 
NewVar44 
2.3216 
1.1147 
3.5285 
Y 
NewVar45 
2.2822 
1.4709 
3.0935 
Y 
NewVar46 
1.5049 
0.9677 
2.0423 
Y 
NewVar47 
2.3985 
1.5324 
3.2645 
Y 
NewVar48 
2.2272 
1.1277 
3.3267 
Y 
NewVar49 
2.3145 
1.5413 
3.0877 
Y 
NewVar50 
1.8669 
1.0764 
2.6576 
Y 
NewVar51 
1.6973 
0.9257 
2.4689 
Y 
NewVar52 
1.4834 
0.9449 
2.0219 
Y 
NewVar53 
2.1219 
1.4426 
2.8013 
Y 
NewVar54 
2.0054 
1.3186 
2.6923 
Y 
NewVar55 
2.2493 
1.3744 
3.1243 
Y 
NewVar56 
1.7336 
1.1152 
2.3519 
Y 
NewVar57 
1.7186 
1.0996 
2.3376 
Y 
NewVar58 
1.2960 
0.7826 
1.8095 
N 
NewVar59 
2.3322 
1.4760 
3.1884 
Y 
NewVar60 
1.9447 
1.2595 
2.6299 
Y 
NewVar61 
2.3604 
1.5601 
3.1608 
Y 
NewVar62 
2.8159 
1.9349 
3.6969 
Y 
NewVar63 
2.4363 
1.5238 
3.3489 
Y 
NewVar64 
2.1681 
1.3227 
3.0135 
Y 
NewVar65 
1.6833 
1.1838 
2.1828 
Y 
NewVar66 
2.2904 
1.3307 
3.2501 
Y 
NewVar67 
1.6014 
0.8826 
2.3203 
Y 
NewVar68 
1.6944 
1.19022 
2.1986 
Y 
NewVar69 
2.2796 
1.5889 
2.9702 
Y 
NewVar70 
1.7836 
1.2667 
2.3005 
Y 
NewVar71 
1.5878 
1.0871 
2.0885 
Y 
NewVar72 
2.6200 
1.6938 
3.5463 
Y 
NewVar73 
1.6798 
0.8418 
2.5177 
Y 
NewVar74 
1.2743 
0.9735 
1.5749 
N 
NewVar75 
2.1606 
1.3443 
2.9769 
Y 
NewVar76 
1.3105 
0.7919 
1.8289 
N 
NewVar77 
2.1495 
1.3038 
2.9952 
Y 
NewVar78 
2.2101 
1.4763 
2.9438 
Y 
NewVar79 
2.4033 
1.4772 
3.3293 
Y 
NewVar80 
1.8017 
1.3024 
2.3009 
Y 
NewVar81 
1.7591 
1.1780 
2.3401 
Y 
NewVar82 
2.0183 
1.3443 
2.6923 
Y 
NewVar83 
1.5494 
0.9228 
2.1761 
Y 
NewVar84 
2.4460 
1.5185 
3.3736 
Y 
NewVar85 
1.7547 
1.2521 
2.2572 
Y 
NewVar86 
1.9683 
1.1867 
2.7499 
Y 
NewVar87 
1.7297 
1.2564 
2.2032 
Y 
NewVar88 
1.9599 
1.3147 
2.6051 
Y 
NewVar89 
2.0622 
1.4277 
2.6968 
Y 
NewVar90 
1.8086 
1.2125 
2.4047 
Y 
NewVar91 
1.3786 
0.7675 
1.9897 
N 
NewVar92 
1.9505 
1.2360 
2.6649 
Y 
NewVar93 
1.8886 
1.1770 
2.6001 
Y 
NewVar94 
2.0094 
1.3489 
2.6699 
Y 
NewVar95 
2.1840 
1.4007 
2.9673 
Y 
NewVar96 
1.8148 
1.2167 
2.4129 
Y 
NewVar97 
1.8457 
1.2152 
2.4762 
Y 
NewVar98 
2.2048 
1.1911 
3.2184 
Y 
NewVar99 
2.5565 
1.6199 
3.4932 
Y 
NewVar100 
1.9319 
1.1761 
2.6877 
Y 
We observe that 93% of the interval envelope
or trap the true mean _{}, whereas the theory says that 95 out of 100 should include _{}. This however, is a fairly good agreement between theory and
application. You may draw 100 samples each of
size 50, calculate 100 confidence intervals and see the difference!
Confidence Interval on the Mean of a
A _{}% confidence interval (CI) for the population mean _{} is given by
_{}
where _{} is the _{} percentile of student _{} distribution with _{}degrees of freedom.
Example 6.3 The contents of 7 similar
containers of sulfuric acid are 9.8, 10.2 10.4, 9.8, 10.0, 10.2 and 9.6 liters.
Find a 99% confidence interval for the mean of all such containers, assuming an
approximate normal distribution (Walpole, R. E. et. al, 2002, pp. 236237)
Solution For 6 degrees of freedom,
_{}using Appendix A3. A 99% confidence interval for the mean _{} is given by
_{}, i.e. _{}
6.3 Computing
Confidence Intervals Using Statistica
Normal Confidence Interval
To compute a zinterval, we compute _{} from the probability
calculator, and find _{} and _{} (if necessary)
by Descriptive Statistics and
then compute the zinterval manually using the appropriate formula. Alternatively, macros can be written to do
the task, but that will not be considered here. On the other hand, if we wish to compute a _{} normal confidence
interval, we need to find _{}by the use of Statistica as shown in Chapter 4. Next, we find
the value of _{}and _{} as we did in Chapter
2. Then we use the formula given earlier
in this chapter to find the confidence
interval, see Examples 6.1 and 6.2.
Student t Confidence Interval
The calculation of confidence intervals in
the package is based on the assumption that the variable is normally
distributed. No matter what the sample size is, Statistica always
gives a tinterval. Suppose that we need to find 99% tinterval
for the population mean in Example 6.3.
Follow the steps:
1. Enter the
sample values in an empty column, say VAR1
2. Statistics /
Basic Statistics / Tables
3. Descriptive
Statistics / OK
4. Variables / VAR1
/ OK
5. Advanced
(under Variation, moments check Conf. Limits for means)
6. Enter 99 for
Interval
7. Summary
The following
scroll sheet of results will be displayed by Statistica:
Confidence – 99.000% 
Confidence + 99.000% 
9.6037 
10.3963 
The interval
[9.6037, 10.3963] is the 99% confidence interval for mean content of sulfuric
acid in a container (Confer with the calculation by pocket calculator as done
in Example 6.3).
6.4 Confidence
Interval Estimation of the Difference Between Two
Population Means
Confidence Interval for the Difference
between the Means of Two Independent
A _{} CI for _{} is given by
_{}.
Example 6.4: An experiment was
conducted with two types of engines, _{}. Gas mileages in miles per gallon were measured. Fifty
experiments were conducted using engine Type A and 75 experiments were done for
Engine Type B. The gasoline used and other conditions were held constant. The
average gas mileage for Engine _{}was 42 miles per gallon and that for Engine _{}was 36 miles per gallon. Find a 96% confidence interval for
the difference between the gas mileages for the two types of engines. Assume that population standard deviations of
gas mileages are 6 and 8 for engines _{} (cf.
Solution With _{}, a 96% CI for _{} is given by
_{}, i.e. _{}.
Large Sample Confidence Interval for the
Difference Between the Means of Two Independent Populations, Variances Unknown
A _{} confidence interval
(CI) for the difference between two population means _{}is given by
_{}.
Example 6.5 Consider a tire
manufacturer who wishes to estimate the difference between the mean lives of
two types of tires, Type A and Type B, as a prelude to a major advertising
campaign. A sample of 100 tires is taken from each production process. The
sample mean lifetimes are 30100 and 25200 miles, respectively whereas the
sample variances are 1500000 and 2400000 miles squared, respectively. Find a 99% confidence interval of the
difference between the mean lives of the two types of tires.
Solution A 99% confidence interval
for _{}is given by
_{}, i.e. _{}
Confidence
Interval for the Difference Between the Means of Two Independent
A _{}% confidence interval (CI) for the difference in population
means _{}is given by
_{}
where _{} is the _{} percentile of student
t distribution with _{}degrees of freedom , and
_{}.
Example 6.6 A random sample of 15
bulbs produced by an old machine was tested and found to have a mean life span
of 40 hours with standard deviation 5 hours. Also, a random sample of 10 bulbs
produced by a new machine was found to have a life span of 45 hours with
standard deviation _{} hours. Assuming that
the life span of a bulb has a normal distribution for both machines, and true
variances are the same, construct a 95% confidence interval for the difference
between the mean lives of the bulbs produced by two machines.
Solution We have _{}. The combined (pooled) estimate for the common variance is
given by
_{}.
With _{} degrees of freedom, _{}, _{}, a 95% confidence interval is given by
_{}, i.e. _{}.
Confidence
Interval for the Difference Between the Means of Two Independent
If _{}, a _{}CI for _{} is given by the
following approximate tinterval:
_{}
where _{} has 2 (n – 1) degrees of freedom.
Example 6.7 A random sample of 16 bulbs
produced by an old machine was tested and found to have a mean life span of 40
hours with standard deviation 5 hours. Also, a random sample of 16 bulbs
produced by a new machine was found to have a mean life span of 45 hours with
standard deviation _{} hours. Assume that the
life span of a bulb has a normal distribution for both machines, construct a
99% confidence interval for the difference between the mean lives of the bulbs
produced by two machines.
Solution We have _{}. A 99% CI for _{}is given by
_{}, i.e. _{}
Confidence
Interval for the Difference Between the Means of Two Independent
A_{}CI for _{} is given by the
following approximate tinterval:
_{}
where _{} has the following
degrees of freedom
_{}
Example 6.8 A random sample of 12
bulbs produced by an old machine was tested and found to have a mean life span
of 40 hours with variance 24 hours squared. Also, a random sample of 10 bulbs
produced by a new machine was found to have a mean life span of 45 hours with
variance 30 hours squared. Assume that the life span of a bulb has a normal
distribution for both machines; construct a 95% confidence interval for the
difference between the mean lives of the bulbs produced by the two machines.
Solution For old machine, we have _{} and for new machine,
we have _{}. The degrees of freedom is given by
_{}
Since the degrees
of freedom of _{}here is 18, _{} and the required CI
is
_{}, i.e. _{}
Letting i denote
the _{}^{th} pair, with _{} and _{}representing sample observations from the respective groups,
we express the difference as _{}and the mean of the differences by _{}. The differences _{}’s are assumed to be independently and normally distributed.
A _{} CI for _{}is then given by
_{}
where _{} is the _{} percentile of student
t distribution with _{} degree of freedom .
Example 6.9 (cf. Lapin, 1997, 329). The chief engineer in a
machine parts manufacturing company is comparing ComputerAided Design (A) to
the traditional method (B). The two procedures are compared in terms of the
mean time from start until production drawings and specifications are ready. A
random sample of 10 parts has been selected, each to be designed twice, once by
an engineer using borrowed time on the CAD (ComputerAided Design) system at a
nearby facility and again by an engineer inhouse working in the traditional
manner. The two engineers designing each sample part have been matched in terms
of the quality of their past performance. The engineers in the CAD group have
each just completed an afterhours training program and have been judged
proficient in the new system. The following completion times (days) have been
obtained:
Parts 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
CAD
(X) 
9.2 
16.4 
5.6 
6.5 
9.0 
11.6 
6.3 
6.0 
20.3 
8.7 
Traditional
(Y) 
12.5 
26.2 
5.0 
7.0 
12.5 
10.4 
9.1 
7.4 
21.1 
10.3 
Find a 95%
confidence interval of the difference in mean time by CAD and by the traditional
Method.
Solution
_{} 
–3.3 
–9.8 
–0.6 
–0.5 
–3.5 
1.2 
–2.8 
–1.4 
–0.8 
–1.6 
A 95% CI for _{}is then given by
_{}.
6.5 Large Sample Confidence Interval Estimation
of a Population Proportion
A _{}confidence interval for _{}is given by
_{}.
Example 6.10 In certain waterquality studies, it is important
to check for the presence or absence of various types of microorganisms.
Suppose 20 out of 100 randomly selected samples of a fixed volume show the
presence of a particular microorganism.
Estimate the true proportion of microorganism with a 90% confidence
interval (Scheaffer and McClave, 1995, 369).
Solution Since the sample size is large, we use a zinterval.
A 90% confidence interval for _{}is given by
_{}, i.e. _{}
6.6
Large Sample Confidence Interval Estimation
of the Difference Between Two Population Proportions
A _{}confidence interval for _{}is given by
_{}
Example 6.11 We want to compare the proportion of defective bulbs
turned out by two shifts of workers. From the large number of bulbs produced in
a given week, _{} bulbs were selected
from the output of Shift I, and _{} bulbs were selected
from the output of Shift II. The sample from Shift I revealed four to be
defective, and the sample from Shift II showed six faulty bulbs. Estimate, by a
95% confidence interval, the true difference between proportions of defective
bulbs produced.
Solution Since the sample sizes are large, we use a zinterval.
Here _{} and _{}. With _{}, a 95% CI for _{}is given by
_{}_{}, i.e. _{}.
6.7 Confidence Interval Estimation of the
Variance of a
A _{} confidence interval
for _{}is given by
_{},
where _{}has _{} degree of freedom
_{}.
i.e. _{}.
Students are encouraged to find z, t and chisquare percentiles using
Statistica.
Exercises
6.1
(cf.
Walpole, R. E, et. al, 2002, 246). The following measurements were recorded for
the drying time, in hours, of a certain brand of latex paint:
3.4 
4.8 
2.8 
3.3 
4.0 
4.4 
5.2 
5.6 
4.3 
5.6 
Assuming that that the measurements represent a random
sample from a normal population, find a 99% confidence interval of the true
mean of drying time.
(a)
Assume that population standard deviation is 1.3.
(b)
Assume that population standard deviation is unknown.
6.2
(cf.
Devore, J. L., 2000, 287). A random sample of fifteen heat pumps of a certain
type yielded the following observations on lifetime (in years):

2.0 
1.3 
6.0 
1.9 
5.1 
0.4 
1.0 
5.3 
15.7 
0.7 
4.8 
0.9 
12.2 
5.3 
0.6 
(a)
Obtain a 95% confidence interval for expected (true average) lifetime.
(b)
Obtain a 99% confidence interval for expected (true average) lifetime.
6.3
(cf.
Devore, J. L., 2000, 299). Consider the following sample of fat content (in
percentage) of ten randomly selected hot
dogs:
25.2 
21.3 
22.8 
17.0 
29.8 
21.0 
25.5 
16.0 
20.9 
19.5 
Assuming that these were selected from a normal
population distribution, find a 95% C.I. for the population mean of fat
content.
(a)
Assume that population standard deviation is 2.4.
(b)
Assume that population standard deviation is unknown.
6.4
(Devore,
J. L., 2000, 303). A study of the ability of individuals to walk in a straight
line reported the accompanying data on
cadence (stride per second) for a sample of
twenty randomly selected healthy men:
0.95 
0.85 
0.92 
0.95 
0.93 
0.86 
1.00 
0.92 
0.85 
0.81 
0.78 
0.93 
1.05 
0.93 
1.06 
1.06 
0.96 
0.81 
0.96 
0.93 
Calculate
and interpret a 95% confidence interval for population mean cadence.
6.5
(Devore,
J. L., 2000, 306). The following observations were made on fracture toughness
of a base plate of 18% nickel steel:
69.5 
71.9 
72.6 
73.1 
73.3 
73.5 
75.5 
75.7 
75.8 
76.1 
76.2 
76.2 
77.0 
77.9 
78.1 
79.6 
79.7 
79.9 
80.1 
82.2 
83.7 
93.7 
Calculate a 99% CI for
the actual mean of the fracture toughness.
6.6
(Devore,
J. L., 2000, 307). For each of 18 preserved cores from oilwet carbonate
reservoirs, the amount of residual gas saturation after a solvent injection was
measured at water floodout. Observations, in percentage of pore volume, were:
23.5 
31.5 
34.0 
46.7 
45.6 
32.5 
41.4 
37.2 
42.5 
46.9 
51.5 
36.4 
44.5 
35.7 
33.5 
39.3 
22.0 
51.2 
Calculate a 98% CI for
the true average amount of residual gas saturation.
6.7
(cf.
Vining, G. G., 1998, 176). In a study of the thickness of metal wires produced
in a chipmanufacturing process. Ideally, these wires should have a target thickness
of 8 microns. These are the sample data:
8.4 
8.0 
7.8 
8.0 
7.9 
7.7 
8.0 
7.9 
8.2 
7.9 
8.1 
7.8 
8.2 
7.9 
8.2 
7.9 
7.8 
7.9 
7.9 
8.0 
8.0 
7.6 
8.2 
8.1 
8.3 
7.8 
8.0 
8.0 
8.3 
7.8 
8.2 
8.3 
8.0 
8.0 
7.8 
8.2 
7.7 
7.8 
8.3 
7.8 
7.9 
8.4 
7.7 
8.0 
7.9 
8.0 
7.7 
7.7 
7.8 
8.3 
8.0 
7.5 
Construct a 95%
confidence interval for the true mean thickness.
6.8
(cf. Vining, G. G., 1998, 177). In a study of aluminum
contamination in recycled PET plastic from a pilot plant operation at
291 
222 
125 
79 
145 
119 
244 
118 
182 
119 
120 
30 
115 
63 
30 
140 
101 
102 
87 
183 
60 
191 
511 
172 
90 
90 
Construct a 95%
confidence interval for the true mean concentration.
6.9
(cf.
Vining, G. G., 1998, 178). Researchers discuss the production of polyol, which
is reached with isocynate in a foam molding process. Variations in the moisture
content of polyol cause problems in controlling the reaction with isocynate.
Production has set target moisture content of 2.125%. The following data
represent 27 moisture analyses over a 4month period.
2.29 
2.22 
1.94 
1.90 
2.15 
2.02 
2.15 
2.09 
2.18 
2.00 
2.06 
2.02 
2.15 
2.17 
2.17 
1.90 
1.72 
1.75 
2.12 
2.06 
2.00 
1.98 
1.98 
2.02 
2.14 
2.10 
2.05 
Construct a 99%
confidence interval for the true mean moisture content.
6.10 (cf. Vining, G.
G., 1998, 178). In a study of a galvanized coating process for large pipes.
Standards call for an average coating weight of 200 lb per pipe. These data are
the coating weights for a random sample
of 30 pipes:
216 
202 
208 
208 
212 
202 
193 
208 
206 
206 
206 
213 
204 
204 
204 
218 
204 
198 
207 
218 
204 
212 
212 
205 
203 
196 
216 
200 
215 
202 
Construct 97%, 98% and
99% confidence intervals for the true mean coating weight.
6.11 (cf. Vining, G. G., 1998, 179). Researchers studied a batch operation at a chemical plant where an
important quality characteristic was the product viscosity, which had a target
value of 14.90. Production personnel use a viscosity measurement for each
12hour batch to monitor this process. These are the viscosities for the past
ten batches:
13.3 
14.5 
15.3 
15.3 
14.3 
14.8 
15.2 
14.9 
14.6 
14.1 
Construct a 90%
confidence interval for the true mean viscosity.
6.12 (cf. Vining, G. G., 1998, 179). Scientists looked at the average particle size of a product with a
specification of 70 – 130 microns and a target of 100 microns. Production
personnel measure the particle size distribution using a set of screening
sieves. They test one sample a day to monitor this process. The average
particle sizes for the past 25 days are listed here:
99.6 
92.1 
103.8 
95.3 
101.6 
102.3 
93.8 
102.7 
94.9 
94.9 
102.8 
100.9 
100.5 
102.7 
96.9 
103.2 
97.5 
98.3 
105.8 
100.6 
101.5 
96.7 
96.8 
97.8 
104.7 





Construct a 95%
confidence interval for the true mean particle size assuming that true standard
deviation is 5.2.
6.13
(cf.
Vining, G. G., 1998, 184). In a study of cylinder boring process for an engine
block. Specifications require that these bores be _{}in. Management is concerned that the true proportion of
cylinder bores outside the specifications is excessive. Current practice is
willing to tolerate up to 10% outside the specifications. Out of a random
sample of 165, 36 were outside the specifications. Construct a 99% confidence
interval for the true proportion of bores outside the specifications.
6.14 (cf. Vining, G.
G., 1998, 184). Consider nonconforming brick from a brick manufacturing process. Typically,
5% of the brick produced is not suitable for all purposes. Management monitors
this process by periodically collecting random samples and classifying the
bricks as conforming or nonconforming. A recent sample of 214 bricks yielded 18
nonconforming. Construct a 98% confidence interval for the true proportion of
nonconforming bricks.
6.15 (cf. Vining, G. G., 1998,
185). In a study examining
a process for manufacturing electrical resistors that have a normal resistance
of 100 ohms with a specification of _{}ohms. Suppose management has expressed a concern that the
true proportion of resistors with resistances outside the specifications has
increased from the historical level of 10%. A random sample of 180 resistors
yielded 46 with resistances outside the specifications. Construct a 95%
confidence interval for the true proportion of resistors outside the
specification.
6.16 (Vining,
G. G., 1998, 185). An automobile manufacturer gives a 5year/60,000mile
warranty on its drive train. Historically,
7% of this manufacturer's automobiles have required service under this
warranty. Recently, a design team proposed an improvement that should extend
the drive train's life. A random sample of 200 cars underwent 60,000 miles of
road testing; the drive train failed for 12. Construct a 95% confidence interval for the true proportion of
automobiles with drive trains that fail.
6.17 (Vining,
G. G., 1998, 185). Historically, 10% of the homes in
6.18 (Devore,
J. L., 2000, 382). Two types of fish attractors, one made from vitrified clay
pipes and the other from cement blocks and brush, were used during 16 different
time periods spanning 4 years at
Pipe 
6.64 
7.89 
0.42 
0.85 
0.29 
0.57 
0.63 
1.83 
0.32 
0.37 
0.00 
0.11 
4.86 
1.80 
0.23 
0.58 

Brush 
9.73 
8.21 
2.17 
0.75 
1.61 
0.75 
0.83 
0.56 
0.76 
0.32 
0.48 
0.52 
5.38 
2.33 
0.91 
0.79 
Find a 95% confidence
interval for the difference in means.
6.19 (Walpole, R. E, et al, 2002, 256). Construct
a 95% confidence interval for the difference in the mean stem weights between
seedlings that receive no nitrogen and those that receive 368 ppm of nitrogen
by using the following sample data. Assume the populations to be normally
distributed.
Nitrogen 
0.26 
0.43 
0.47 
0.49 
0.52 
0.75 
0.79 
0.86 
0.62 
0.46 
No nitrogen 
0.32 
0.53 
0.28 
0.37 
0.47 
0.43 
0.36 
0.42 
0.38 
0.43 
6.20 (Walpole,
R. E, et al, 2002, 254). A study published in Chemosphere reported the
levels of dioxin TCDD of 20 Massachusetts Vietnam veterans who were possibly
exposed to Agent Orange. The amount of TCDD levels in plasma and in fat tissue
is listed in the table below. Find a 95% confidence interval for the difference
in means.
TCDD levels in plasma 
2.5 
3.1 
2.1 
3.5 
3.1 
1.8 
6.0 
3.0 
36.0 
4.7 
6.9 
3.3 
4.6 
1.6 
7.2 
1.8 
20.0 
2.0 
2.5 
4.1 

TCDD levels in fat tissues 
4.9 
5.9 
4.4 
6.9 
7.0 
4.2 
10.0 
5.5 
41.0 
4.4 
7.0 
2.9 
4.6 
1.4 
7.7 
1.1 
11.0 
2.5 
2.3 
2.5 
6.21 (cf. Vining, G. G., 1998,
192). Researchers
studied the impact of viscosity on the observed coating thickness produced by a
paint operation. For simplicity, they chose to study only two viscosities:
"low" and "high." Up to a certain paint viscosity, higher
viscosities cause thicker coatings. The engineers do not know whether they have
hit that limit or not. They thus wish to test whether the higher viscosity
paint leads to thicker coatings. Here are the coating thicknesses:
Low Viscosity 
1.09 
1.12 
0.83 
0.88 
1.62 
1.49 
1.48 
1.59 
1.65 
1.71 
0.88 
1.29 
1.04 
1.31 
1.83 
1.76 





High Viscosity 
1.46 
1.51 
1.59 
1.40 
0.74 
0.98 
0.79 
0.83 
1.46 
1.42 
2.05 
2.17 
2.36 
2.12 
1.51 
1.40 




Construct
a 95% confidence interval for the true difference in the mean coating
thicknesses.
6.22 (Vining,
G. G., 1998, 194). The following data are the yields for the last 8 hours of
production from two ethanolwater distillation columns:
Column
1 
70 
74 
73 
72 
72 
73 
72 
73 
Column
2 
71 
74 
72 
71 
72 
70 
72 
72 
Construct a 97 % confidence interval for the
true difference in the mean yields.
6.23 (cf. Vining, G. G., 1998,
p.192). A manufacturer
of aircraft monitors the viscosity of primer paint. The viscosities for two
different time periods are listed here:
Time Period I 
33.8 
33.1 
34.0 
33.8 
33.5 
34.0 
33.7 
35.2 
33.8 
33.3 
33.5 
33.2 
33.6 
33.0 
33.5 
33.1 





Time Period II 
33.5 
33.3 
33.4 
33.3 
34.7 
34.8 
34.8 
33.2 
35.0 
35.0 
34.8 
34.5 
34.7 
34.3 
34.6 
34.5 




Construct a 95%
confidence interval for the true difference in the mean viscosities.
6.24 (cf. Vining, G. G., 1998,
193). An independent consumer group tested radial tires from two
major brands to determine whether there were any differences in the expected
tread life. The data (in thousands of miles) are given here:
Brand 1 
50 
54 
52 
47 
61 
56 
51 
51 
48 
56 
53 
43 
58 
52 
48 
Brand 2 
57 
61 
47 
52 
53 
57 
56 
53 
67 
58 
62 
56 
56 
62 
57 
Construct a 95%
confidence interval for the true difference in the mean tread lives.
6.25 (cf. Vining, G. G., 1998,
194). In a comparison of two brands of ultrasonic humidifiers
with respect to the rate at which they output moisture, the following data are
the maximum outputs (in fluid ounces) per hour as measured in a chamber
controlled at a temperature of _{}F and a relative humidity of 30%:
Brand
1 
14.0 
14.3 
12.2 
15.1 
Brand
2 
12.1 
13.6 
11.9 
11.2 
Construct
a 90% confidence interval for the true difference in the mean viscosities.