Example 7.8 Application of Concepts
The problem gives us the following information:
represents a sample of n=25 concrete block sections on the western end of the runway.
µ= 60 Mean LCN of the original pavement quality concrete at Western end of Runway
σ=10
Standard Deviation for above µ
a.
If mean LCN of the new concrete block surface = mean LCN of old surface, describe the sampling
distribution of .
We can apply the Central Limit Theorem.
Based on this assumption, the mean LCN of the sample is normally distributed. Further
= µ=60,
meaning that the mean of the population is the mean from which the sample was taken and the
standard deviation
b.
=
. Therefore,
=
= 2.
If mean of new concrete block surface is no different than that of the original surface, find probability
that result ≥ 65.
Since we already established that mean LCN is normally distributed, we can use
. We
have ,
. Then z =2.5. Using the standard normal tables to find the area above z=2.5, we find
subtract the value from the table from .5. In this case we have .5-.4938=0062.
c.
If
= 73, what can we infer? Use
with =73. The resulting z is 4.0. Notice that at
=65
the resulting probability is.62%, so an of 73 is much higher from what we would expect. I found a
normal distribution calculator online and plugged a z value of 4 into it and the result was .000032 or
.0032%. If we can use the central limit theorem, then the individual
results we would generate
from various samples should not be that high. Either our sample was tainted or our µ is wrong.
Example 8.4 Theoretical Interpretation of Confidence Interval
We are given data from 40 random samples of 100 sale prices along with upper and lower limits or their
95%confidence intervals. From a previous problem, we have a µ=$106,405. This µ should be within the ranges of
each upper and lower limits or each confidence interval listed. There are two instances where this is not the case.
Example 8.5 Finding z for a 90% Confidence Interval
We recall that the total probability is 1.0. In looking for a 90% confidence interval we subtract .9 from 1.0 and get
.10. Also recall that for a standard normal distribution is symmetrical with the mean at zero. Taking the .1 and
dividing by2 we get .05. We are looking for a z.05. For a half of the normal table we subtract.5-.05 = .45. Looking in
the standard normal table for the value that corresponds to .45 we see that we have 1.64 corresponding to .4495
and 1.65 corresponding to .4505. Interpolating between the values and the corresponding Z =1.645.
The confidence interval is therefore
=1.645±
.
Example 8.6 99% Confidence Interval
Given the random sample of 50 ratios of sale price to appraised values, I calculated the mean and standard
deviation for the sample using excel.
=1.315 and s=.3658. The text gave us the confidence coefficient for 99% or
2.58. We do not know the actual population standard deviation, but can use the sample standard deviation, s,
because the sample is sufficiently large (sample size of 50 is ≥30). So far we have
=1.3158 ± 2.58(
).
Continuing
=1.315 ± 2.58(
) =1.315 ± .1335. We have lower limit of 1.18 and upper limit of 1.449.
Therefore, we can state that we can be 99% confident that the true mean ratio of sale price to appraised value is
contained within the interval from 1.18 to 1.45.
Example 8.7 Effect of (1-α) on the width of the Confidence Interval
Using data from example 8.6, construct a 95% confidence interval. Using confidence coefficient given in the text of
1.96 we have
=1.315 ± 1.96(
) = 1.315±.101. We have lower limit of 1.214 and upper limit of 1.416. The
confidence interval has widened or increased with the decrease in confidence from 99% to 95%.
a.
Since the confidence interval widens with a decrease in confidence, with the same mean, standard
deviation and sample size, the only thing that can differ is the confidence interval coefficient.
=1.315 ± 1.96(
) 95% Confidence
=1.315 ± 2.58(
) 99% Confidence
With a decrease in confidence, the confidence interval coefficient must also decrease.
Example 8.8 Effect of n on the width of the Confidence Interval
We are to determine the effect of a change in sample size on the confidence interval. Using the previous data, we
want to see what happens to the confidence when sample size changes from n=50 to n=100.
=1.315 ± 2.58(
) 99% Confidence Sample size n=50
=1.315 ± .1335 (lower limit of 1.18 and upper limit of 1.449)
=1.315 ± 2.58(
) 99% Confidence Sample size n=100
=1.315 ± .0944 (lower limit of 1.221 and upper limit of 1.409)
We see that the width of the confidence interval increases as sample size decreases all other things being
unchanged.
Example 8.12 Computer Analysis
The confidence interval from the computer program was lower limit of 1.1766 and upper limit of 1.4538. The
confidence interval from my calculation was lower limit of 1.18 and, upper limit of 1.45. (My actual data was
1.182258 and 1.449343 but I rounded). If I round the computer results then there is no difference. If we
standardize how far we carry the decimal point then the results should be relative the same. The computer
program apparently uses the t statistic to calculate the confidence interval.
Example 8.13 Selecting a Point Estimate
Sample size is n= 576 respondents
63 quit their jobs within 1 year after winning $50,000
If we let p be the proportion of the sample, x be the number of respondents that quit their job and n be the sample
population, we can divide that part of the population over the total sample population and we get 63/576 or .11.
This means that 11% of the lottery winners in the sample quit their job during the first year after winning.
Example 8.14 95% Confidence Interval for π
We are to construct a 95% confidence interval using the formula, p±
. We just calculated p=.11. For a
confidence interval of 95%, we were given the coefficient of 1.96. q=1-p and n=576. Therefore
.11±
= .11±.03. This gives us an interval of (.08, .14).
Example 8.15 90% Confidence Interval for π
Random sample of 165 families. 101 responded positively to the question asked. p= = 101/165 = .612.
Once again using p±
= .612±1.645
= .612±.062. This gives us a 90% confidence interval of
(.55, .674). Therefore, we are 90% confident that the true proportion of TV families that watched the show is
between .550 and .674.
Example 8.19 Small Sample 95% Confidence Interval for (
Given: Two type of bargaining strategies - competitive (
sample for each with n=8.
A.
)
versus coordinative (
. Data is given for a
We are asked to construct a 95% confidence interval for the difference between the buyer
savings of the two strategies.
We are told nothing of the data so we must make certain assumptions as follows:
1.
The population of each have approximate normal distributions
2.
The variances of the two are equal
3.
The samples were random and unbiased.
I performed the calculations using Excel and the results were as follows:
N
1
2
3
4
5
6
7
8
Sum
s
Competitive
1857
1700
1829
2644
1566
663
1712
1679
13650
1706.25
538
289,432
208,614
Coordinative
1544
2640
1645
2275
2137
2327
2152
2130
16850
2106.25
357
127,797
Average Var
Excel Function
Average
STDEV
VAR
To address assumption 3, I calculated the variance to be the average of each individual
variance i.e. adding the two together and dividing by 2. This differed from the common
variance calculated in the problem solution, so I redid this calculation using the formula
for small sample with equal variances or
. Using this
formula, the variance for the two samples was the same 208,614 which still differed
from the problem solution of 208,446. To get the 95% confidence interval, therefore we
use
890, 90).
. Substituting our values we get (-400±490) or (-
Therefore, we estimate, with 95% confidence, that the difference (
the interval -890 to 90.
B.
) falls within
We are told that I theory the mean buyer savings for the competitive strategy will be
less than the corresponding mean for the coordinative strategy and to make inference
about that.
From Part A we found that
zero so we can’t say that
is within the range (-890, 90) of
is less than
. This range includes
.
Example 8.21 95% Confidence Interval for
Given that data and information for the 8.20, we are to determine the 95% confidence interval for the difference
in assertiveness for the two different training methods. I used an Excel SS to calculate the average of 11.0, and s of
6.53. Adding all values to the equation we get 11.0 ±4.7 or (6.3, 15.7). We estimate with 95% confidence that the
mean managerial assertiveness test scores for methods 1 and 2 fall within the interval 6.3 to 15.7. Both interval
limits are positive so we can say that method 1 seems to produce a test score that is larger, statistically, than
method 2. If we look at the data, which is probably not possible with large samples, we can see that except in one
instance, the test scores for method 1 are higher than method 2.
Example 8.30 95% Confidence Interval for
We are given data from a previous example:
construct a 95% confidence interval for
corresponds to an
= 11.06 mg and s=4.93 mg for 500 cigarette brands. We are to
. Here we are using
. The confidence interval is 95% which
of .025. We have a formulas to use which requires the use of
values help to determine the upper and lower limits for the
. Using the table we find
and
. The
and
with
n=500 = degrees of freedom. The resulting values are 21.51 ≤
≤ 27.57. Therefore we can say with 95%
confidence that the true variance of tar in domestic cigarettes falls between 21.51 and 27.57.
Computer Application
For example 8.19, I used Excel to calculate the mean, deviation and standard deviation for the problem. There
were no surprises in the simple calculations. The average variance I calculated with Excel did differ from the output
from the book. First I simply took the average of the 2 variance figures and did not use the formula. However, the
difference was small and virtually disappeared when taking the square root of this figure to get the standard
deviation. When finally rounding to an appropriate number of decimals there was no difference.