Chapter 9 Estimation and Confidence Intervals
30. A random sample of 85 group leaders, supervisors, and similar personnel revealed that a person spent an average 6.5 years on the job before being promoted. The population standard deviation was 1.7 years. Using the 0.95 degree of confidence, what is the confidence interval for the population mean? A) 6.99 and 7.99 B) 4.15 and 7.15 C) 6.14 and 6.86 D) 6.49 and 7.49 Answer: C 31. The mean weight of trucks traveling on a particular section of I-475 is not known. A state highway inspector needs an estimate of the mean. He selects a random sample of 49 trucks passing the weighing station and finds the mean is 15.8 tons. The population standard deviation is 3.8 tons. What is the 95 percent interval for the population mean? A) 14.7 and 16.9 B) 13.2 and 17.6 C) 10.0 and 20.0 D) 16.1 and 18.1 Answer: A 41. Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for president. Using the 0.95 degree of confidence, what is the interval estimate for the population proportion (to the nearest tenth of a percent)? A) 78.2% to 81.8% B) 69.2% to 86.4% C) 76.5% to 83.5% D) 77.7% to 82.3% Answer: A Use the following to answer questions 81-83: A student wanted to quickly construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. 81. What is the best point estimate for the population mean? A) 2.1 years B) 1.5 years C) 19.1 years D) 9 years Answer: C 82. What is the 95% confidence interval for the population mean? A) [0.97, 3.27] B) [15.64, 22.56] C) [17.97, 20.23] D) [17.95, 20.25] Answer: D

Statistical Techniques in Business & Economics, Lind/Marchal/Wathen, 13/e

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83. What is the 99% confidence interval for the population mean? A) [17.42, 20.78] B) [17.48, 20.72] C) [14.23, 23.98] D) [0.44, 3.80] Answer: A Use the following to answer...

...
QoS Techniques: MPLS
CET 2486C – Network Technologies
Professor:
November 27, 2012
Abstract
MPLS or Multi Protocol Label Switching is a networking technology that functions between layers 2 and 3 of the OSI model. MPLS constitutes of adding a label (sometimes called “Shim” because of their placement between layer 3 and layer 2 headers.) to the data package, this label contains special addressing and sometimes prioritization information. Because the MPLS label contains all the information necessary for the router to forward the package to the next hop, the router does not have to spend time analyzing the entire package thus improving network latency or bottlenecks. Due to its multi protocol capabilities MPLS can be integrated with different networking technologies from ATM to native IP environments; in addition, this multi protocol capability also provides a way to converge different types of traffic such as data, voice and video onto one network. MPLS technology also provides some other advantageous features like Traffic Engineering (TE), VPN, Any Transport over MPLS (AToM) and Quality of Service (QoS). This paper will help provide an understanding of how MPLS works and the QoS capabilities it can provide.
History of MPLS
In 1996 a group from Ipsilon Networks introduced a “flow management protocol”, this technology only worked with ATM transmissions and did not become very popular in the market. Not long after Cisco...

...taken.
Summary of Statistics
Sample 1
Sample 2
Sample 3
Sample 4
Mean
11.96
12.10
12.14
12.15
Standard Error
0.04
0.04
0.04
0.02
Standard Deviation
0.22
0.23
0.23
0.16
Sample Variance
0.05
0.05
0.05
0.03
Sum
358.69
362.90
364.08
364.46
From the summary of statistics we can see that mean has an upward trend. Mean value differ from sample to sample. Here we can observe that mean value has a relation with the amount of sum. We can also notice that amount of sum is differing from sample to sample and follow the same trend as mean.
Introduction of the test:
By analyzing the new samples, the client would be able to know that, whether the process was operating satisfactorily or not. If the process was not operating satisfactorily, then corrective action would be taken to eliminate the problem. In this case the mean for the process is 12. The hypothesis test suggested by Quality Associates follows:
H0: μ=12
Ha: μ≠ 12
Corrective action will be taken if H0 is rejected. In this case α = .01 that means probability of making a Type I error when μ = 12 is .01.
Sample 1
Sample 2
Sample 3
Sample 4
Z statistics
-1.14
2.52
3.55
3.88
P value
0.26
0.01
0.00
0.00
Accept
Reject
Reject
Reject
Here we can see in Sample 1, P-value is higher than α. So sample 1 is operating satisfactorily, so in...

...i Chapter9 Notes
What is Audit Sampling?
* Audit Sampling – applying a procedure to less than 100% of a population to estimate some characteristic of that population
* Sampling Risk – risk that a sample may not be representative of the population
* Risk that the auditor’s conclusion based on the sample may be different from the conclusion they would reach if they examined every item in the population
* Non-sampling Risk – risk pertaining to non-sampling errors (due to human error)
* The sample is good but the auditor simply misses a deviation from a control, or misunderstands the procedure
* Can be reduced to low levels through effective planning and supervisions of audit engagements
SLIDE 9-3
Statistical Sampling
* Relies on the laws of probability, but does not eliminate judgment
* Allows auditors to measure risk and control sampling risk, which helps:
* Designs efficient samples – provides the smallest sample size for a given risk/confidence level
* Measures sufficiency of evidence – adds an allowance for sampling risk depending on the confidence or risk level desired
* Objectively evaluates sample results – indicates whether you’ve exceeded the tolerable deviation rate or tolerable misstatement
Non-statistical Sampling
* Auditor uses judgment, rather than statistical techniques
* This provides no means of quantifying sampling...

... Statistics – Case Chapter9
1. It is not proper to multiply the average order size by the number of addresses (1.3 million people = population) in the target mailing because the sample is representative of the 600,000 people in the database, not the 1.3 million target population, thus you cannot use the average of the sample as an estimate for the population. Also, multiplying by 1.3 million would suggest that the entire 1.3 million people would be purchasing.
2. It is better to multiply the endpoints of the confidence interval by the population because now there will be a range of values, but it still will not be accurate because the confidence interval is still using the mean from the sample, which isn’t representative of the population, thus it cannot be used as an estimate for the population. In this situation, there will be far less than 1.3 million people purchasing from the catalog thus using the lower limit of the confidence interval will still be far higher than the actual amount.
3. It is also better to use the size of the frame (600,000 people from the database) since our sample is representative of the 600,000 people in the database, thus we can use the mean of the sample as an estimator for the 600,000 people in the database. However, only 13 people of the 600 sample responded with a yes and a dollar amount, thus multiplying the mean of the 13 people who responded with an intent to purchase by the 600.000 people in the database...

...CHAPTER EIGHT
INTERVAL ESTIMATION
MULTIPLE CHOICE QUESTIONS
In the following multiple choice questions, circle the correct answer.
1. When s is used to estimate (, the margin of error is computed by using
a. normal distribution
b. t distribution
c. the mean of the sample
d. the mean of the population
Answer: b
2. As the number of degrees of freedom for a t distribution increases, the difference between the t distribution and the standard normal distribution
a. becomes larger
b. becomes smaller
c. stays the same
d. None of these alternatives is correct.
Answer: b
3. For the interval estimation of ( when ( is known and the sample is large, the proper distribution to use is
a. the normal distribution
b. the t distribution with n degrees of freedom
c. the t distribution with n + 1 degrees of freedom
d. the t distribution with n + 2 degrees of freedom
Answer: a
4. An estimate of a population parameter that provides an interval of values believed to contain the value of the parameter is known as the
a. confidence level
b. interval estimate
c. parameter value
d. population estimate
Answer: b
5. The value added and subtracted from a point estimate in order to develop an interval estimate of the population parameter is known as the
a. confidence level
b. margin of error
c. parameter estimate
d. interval estimate
Answer: b
6. If an interval estimate is...

...units from a normally distributed large population. If u = 15, and c2=4, what is the probability that we will obtain a sample mean of less than 14? .0062 5. The normal approximation of the binomial distribution is appropriate when. Np> 5 and n(1-p) >5 6. A newly married couple plans to have four children. Suppose that boys and girls are equal likely each time a child is born. What is the probability the couple will have no more than 2 boys? 62.5% 7. A random variable is said to be discrete if: Its outcome are countable 8. The mean life of pair of shoes is 40 months with a standard deviation of 8 months. If the life of the shoes is normally distributed, how many pairs of shoes out of one million will need replacement before 36 months? 308,500 9. If the sampled population has a mean of 48 and standard deviation 16, then the mean and the standard deviation for the sampling distribution x for n=16 48 and 4 10. The MPG (MILES PER GALLON) for a mid-size car is normally distributed with a mean of 32 and a standard deviation of .8. what is the probability that the MPG for a selected mid-size car would be less that 33.2? 93.32% 11. If the random variable X has a mean of u and a standard deviation g, then (X-u)/g) has a mean and standard deviation respectively. 0 and 1 12. For a binomial probability experiment, with n=150 and p=.2, it is appropriate to use the normal approximation to the binomial distribution. TRUE 13. A computer system uses 4 bits address size to...

...Chapter9 (Group & Teams)
How can we characterize groups?
Roles, norms, status, size, cohesiveness, diversity
To explain special types of team dynamics:
Social loafing, group link, group shift
Individual concepts – Psychological contacts, deviance
A work group is 2 more more people who: Maintain stable patterns of relationships in which they have influence over each other, they share common goals, and perceive themselves as being in a group. Members can identify other members.
How do groups and team from over time?
There are two models:
Five stage model of group development, or Group Life Cycle. Each stage involves activates directed at performing the task (task activity) and dealing with interpersonal interactions within the group ( group process)
1. Forming: Meeting with everyone on the team, and getting to know everyone.
2. Storming: May have conflict within the group, like who will be the lead, who will be doing what, etc…
3. Norming: Post the storming stage. Getting thru the logistics of how we will be working together, organizing the group activity.
4. Performing: Start to get work done. In production mode.
5. Adjourning: Applies to groups that adjourn. Break off and close the project.
Punctuated Equilibrium Model: For groups with a set deadline.
Group Properties:
1. Roles: The behavior that characterizes a person in a specific social contact and the behavior expected of that role.
a. Role perception – one’s view of how to act in a...

...Chapter9 closing case
Ashford University
BUS 650 Managerial Finance
When should Bunyan Lumber, harvest the forest?
The cash flow will grow at the inflation rate of 3.7%. Utilizing the real cash flow formula
(1+R) =v (1+R)(1+H)
1.10 = (1+R)(1.037)
R= 6.08%
The conservation funds are anticipated to grow slower than the inflation rate. The return for the conservation fund will be,
(1+R) = (1+R) (1+H)
1.10 = (1+R) (1.032)
R= 6.5%
The cash flow from the thinning process is as follow,
Cash flow from thinning = Acres thinned x cash flow per acre
Cash flow from thinning = 7,500 ($1,200)
Cash flow from thinning = $9,000,000
Thinning beyond the initial thinning is conducted on a schedule and can be included.
After tax cost of the conservation fund will be,
After tax conservation fund cost = (1”C.35) ($250,000)
After tax conservation fund cost = $162,500
For each analysis the cost and revenue are;
Revenue [ E (% of grade )(harvest per acre)(value of board game)](acres harvested) (1”C defect rate)
Tractor cost = (Cost MBF)(MBF per acre)(acres)
Road cost = (Cost MBF)(MBF per acre)(acres)
Sale preparation and administration = (Cost MBF) (MBF acre) (acres)
It is assumed that there is no depreciation as a result of the harvest. This is an indicator that operating cash flow is equal to net income. The NPV of the thinning, the NPV of all future harvests, minus the present value of the conservation fund costs.
Revenue...

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