EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US

CONFIGURATION EXAMPLES FOR THE DLINK NETDEFEND FIREWALL
MULTIPLE REGRESSION EXAMPLE CALIFORNIA RAIN
EXAMPLE 1 INSTALLATION OF OFFICERS AND

TEMPLATEEXAMPLE HOSPITAL PEDIATRIC DISASTER PLAN FEBRUARY 10 2012
(FIRST EXAMPLE IN BRACKETS IS FROM ENGLISH AND THE
0150781 TOYOTA MOTOR EUROPEAN (TME) SUSTAINABLE LOGISTICS AN EXAMPLE

Example 7



Example 7.8 Application of Concepts
The problem gives us the following information:
EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US 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

  1. If mean LCN of the new concrete block surface = mean LCN of old surface, describe the sampling distribution ofEXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US .
    We can apply the Central Limit Theorem.
    Based on this assumption, the mean LCN of the sample is normally distributed. Further EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = µ=60, meaning that the mean of the population is the mean from which the sample was taken and the standard deviation EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . Therefore, EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = 2.

  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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . We have EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US , EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . 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.


  1. If EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = 73, what can we infer? Use EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US with EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =73. The resulting z is 4.0. Notice that at EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =65 the resulting probability is.62%, so an EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US 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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US 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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.645±EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US .

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. EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.3158 ± 2.58(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ). Continuing EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 2.58(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) =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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 1.96(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) = 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%.

  1. 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.

EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 1.96(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) 95% Confidence

EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 2.58(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) 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.

EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 2.58(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) 99% Confidence Sample size n=50
EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± .1335 (lower limit of 1.18 and upper limit of 1.449)


EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =1.315 ± 2.58(EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) 99% Confidence Sample size n=100
EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US =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±EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . 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±EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = .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=EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = 101/165 = .612.

Once again using p±EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = .612±1.645EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = .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 (EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US )
Given: Two type of bargaining strategies - competitive (EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US versus coordinative (EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . 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:

  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

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

 

EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US

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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . 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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . Substituting our values we get (-400±490) or (-890, 90).

Therefore, we estimate, with 95% confidence, that the difference (EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ) 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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US is within the range (-890, 90) of EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . This range includes zero so we can’t say that EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US is less than EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US .


Example 8.21 95% Confidence Interval for EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US
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 EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US
We are given data from a previous example: EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US = 11.06 mg and s=4.93 mg for 500 cigarette brands. We are to construct a 95% confidence interval for EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . Here we are using EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . The confidence interval is 95% which corresponds to an EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US of .025. We have a formulas to use which requires the use of EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US and EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . The values help to determine the upper and lower limits for the EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US . Using the table we find EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US and EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US with n=500 = degrees of freedom. The resulting values are 21.51 ≤ EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US EXAMPLE 78 APPLICATION OF CONCEPTS THE PROBLEM GIVES US ≤ 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.




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