Qnt 561 Week2

Topics: Normal distribution, Sample size, Standard deviation Pages: 5 (1289 words) Published: September 14, 2010
Central Limit Theorem and Confidence Intervals Problem Sets
Tiffany Blount
QNT 561
September 7, 2010
Michelle Barnet
University of Phoenix

Central Limit Theorem and Confidence Intervals Problem Sets
Chapter 8 Exercises:
21. What is sampling error? Could the value of the sampling error be zero? If it were zero, what would this mean? * Sampling error is the difference between the statistic estimated from a sample and the true population statistic. It is not impossible for the sampling error to not be zero. If the sampling error is zero then the population is uniform. For example if I were evaluating the ethnicities of a populations and everyone is the population was Black then taking any sample would give me the true proportion of 100% Black. 22. List the reasons for sampling. Give an example of each reason for sampling. * The population size is too large and costly for making the study feasible in reasonable period.

For example, if I want to know how watching the violent shows on television affects the behavior of children, it won’t be realistic to study each child in the population, so I would use sampling.

* Only estimation of particular section of population is required

For example, if I want to take an example of nation which is combined unit of states. I can choose the random samples of states which can be further divided into smaller units like cities. These cities can be clustered into smaller areas for observation. Researchers can define his pattern of selecting the sample data until data condition of observation is fully satisfied.

* It is not possible to study the entire population and accessibility of them is time consuming and difficult

For Example, if I wanted to prepare a list of all the customers from a chain of hardware stores. This would be a tedious task. But it is convenient to choose a subset of stores in stage one of cluster sampling which can be used for interviewing the customers from those stores in the second stage of cluster sampling.

34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.

A. If we select a random sample of 50 households, what is the standard error of the mean?

* Standard error of the sample mean = σ/√n == 40000/√50 = 5656.85

B. What is the expected shape of the distribution of the sample mean?

* Since sample size is greater than 50, it should be normally distributed according to the Central Limit Theorem.

C. What is the likelihood of selecting a sample with a mean of at least $112,000?

* z = (X - μ) / σx, Where X is a normal random variable, μ is the mean, and σ is the standard deviation. P(X > 112000) = P(z > (112000 –110000)/5656.85) = P(z>0.3535) = 0.5 – P (0<z<0.3535) = 0.5 – 0.1368 = 0.3632.

D. What is the likelihood of selecting a sample with a mean of more than $100,000? * P(X > 100000) = P (z > (100000 – 110000)/5656.85) = P (z > –1.7677) = 0.5+ P (–1.7677 < z < 0) = 0.5 + P (0 < z < 1.7677) =0.5 + 0.4616 = 0.9616.

E. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000.

* P (100000 < X < 112000) = P(X > 100000) – P(X >112000) = 0.9616 – 0.3632 =0.5984.

Chapter 9 Exercises
32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds.

a. What is the estimated population mean?
* 3.01.

b. Determine a 95 percent confidence interval for the population mean. * 3.01 ± 1.96*0.03/sqrt(36)= 3.0002 , 3.0198

34. A recent survey of...
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