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
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
=
.
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.
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%.
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 (
versus coordinative (
.
Data is given for a sample for each with n=8.
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:
The population of each have approximate normal distributions
The variances of the two are equal
The samples were random and unbiased.
I performed the calculations using Excel and the results were as follows:
N |
Competitive |
Coordinative |
Excel Function |
1 |
1857 |
1544 |
|
2 |
1700 |
2640 |
|
3 |
1829 |
1645 |
|
4 |
2644 |
2275 |
|
5 |
1566 |
2137 |
|
6 |
663 |
2327 |
|
7 |
1712 |
2152 |
|
8 |
1679 |
2130 |
|
Sum |
13650 |
16850 |
|
|
1706.25 |
2106.25 |
Average |
s |
538 |
357 |
STDEV |
|
289,432 |
127,797 |
VAR |
|
208,614 |
Average 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 . Substituting our values we get (-400±490) or (-890, 90).
Therefore, we estimate, with 95% confidence, that the difference ( ) falls within the interval -890 to 90.
B. 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 is within the range (-890, 90) of . This range includes zero so we can’t say that is less than .
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:
= 11.06 mg and s=4.93 mg for 500 cigarette brands. We are to
construct a 95% confidence interval for
.
Here we are using
.
The confidence interval is 95% which corresponds to an
of .025. We have a formulas to use which requires the use of
and
.
The values help to determine the upper and lower limits for the
.
Using the table we find
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.
1 5 EXAMPLES OF AUTHORISED AND UNAUTHORISED USE OF
1 MAIN ACTIVITY PACKAGE COMEXAMPLEJMAJAKTRAVEL IMPORT ANDROIDCONTENTINTENT IMPORT ANDROIDSUPPORTV7APPACTIONBARACTIVITY
1) PRESIDENT BARACK OBAMA IS A GOOD EXAMPLE OF
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