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
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The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.
z = 1.96 (from a table)
N = 50 sd = 6.2 mean = 26
* 26 - 1.96*6.2/sqrt(50) to 26 + 1.96*6.2/sqrt(50)=24.281 to 27.719; The value of 28 weeks in not inside that confidence interval, so it is not reasonable that the population mean is 28 weeks.
46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?
1st:
z = 2.5758 p = 14/220
p - z*sqrt(p*(1-p)/N) to p + z*sqrt(p*(1-p)/N)
14/220 - 2.5758*sqrt(14/220*(1-14/220)/220) to 14/220 + 2.5758*sqrt(14/220*(1-14/220)/220)
* 0.0212 to 0.1060; 10% is within that interval, so yes, it is reasonable
2nd:
z = 2.5758 p = 14/400
p - z*sqrt(p*(1-p)/N) to p + z*sqrt(p*(1-p)/N)
14/400 -
2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?
(21) You took a sample of size 21 from a normal distribution with a known standard deviation, . In order to find a 90% confidence interval for the mean, You need to find.
A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61,400 with a standard deviation of $2,200. Find a 95% confidence interval for the true mean annual income of the business’ customers.
b. Construct a 91.5% confidence interval for the mean time it take for all workers who are employed in downtown Toronto
9. Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain
Topics Distribution of the sample mean. Central Limit Theorem. Confidence intervals for a population mean. Confidence intervals for a population proportion. Sample size for a given confidence level and margin of error (proportions). Poll articles. Hypotheses tests for a mean, and differences in means (independent and paired samples). Sample size and power of a test. Type I and Type II errors. You will be given a table of normal probabilities. You may wish to be familiar with the follow formulae and their application.
c. Based on this asymmetric Confidence Interval in 'b' above, state how much minimum mean weight could be lost, after completing the dieting programme.
Explain the importance of random sampling. What problems/limitations could prevent a truly random sampling and how can they be prevented?
b) What is the difference between the mean of the two groups? What is the difference in the standard deviation?
The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.
Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of
When determining a confidence interval for this sample/population, it is necessary to remember that the interval must be one sided. Everyone to the right of a given vertical line on the distribution will be included in the result, while everyone to the left would be excluded. The vertical line represents a selected price for the
The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.
a) Develop a 99 percent confidence interval for the mean selling price of the home.