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Estimating the Value of a Parameter Using Confidence Intervals Chapter 9

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9.1: Confidence Interval for a Population Mean – When population standard deviation is known Confidence Interval: Interval of numbers for an unknown parameter. Level of Confidence: The expected proportion of intervals that will contain the parameter. Denoted as a percent

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Example 95% level of confidence ▫Implies that if 100 different intervals are constructed based on obtained samples, we will expect 95 of the intervals to contain the true parameter. ▫i.e.: we are 95% confident that the obtained interval contains the true population parameter. ▫For example, if we constructed a 99% confidence interval with a lower bound of 52 and an upper bound of 71, we would interpret the interval as follows: “We are 99% confident that the population mean, μ, is between 52 and 71.

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Requirements Population from which the sample was drawn must be normally distributed or the sample size must be greater than or equal to 30

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Margin of Error Confidence interval estimates for the population mean are in the form of ▫Point estimate + margin of error The Margin of error is a measure of how accurate the point estimate is. For the previous example, if we constructed a 99% confidence interval with a lower bound of 52 and an upper bound of 71 then the point estimate would be in the middle [61.5], and the margin of error would be 9.5

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3 Factors on Margin of Error Level of Confidence: as the level of confidence increases, the margin of error also increases. Sample Sizes: As the size of the random sample increases, the margin of error decreases. Standard deviation of the Population: The more spread there is in the population, the wider our interval will be for a given level of confidence.

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The Role of Sample Szie Note: The larger sample size reduces the margin of error

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The Role of Level of Confidence Note: As we increase the level of confidence, the margin of error increased

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Summaries If we want to be ‘more confident’ we need to increase the sample size If we want to reduce the margin of error, we would need to decrease our level of confidence or increase our sample size

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The Z-interval If we know the population standard deviation, then the confidence interval is found using a formula called the Z-interval Lower:Upper: Use Calculator: ▫Menu: 6:statistics 6: Confidence Intervals 1: Z-interval

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#23: A simple random sample of size n is drawn from a population that is normally distributed with population standard deviation known to be 13. The sample mean is found to be 108 a)Compute the 96% confidence interval for μ if the sample size n is 25. a)Compute the 88% confidence interval for μ if the sample size is 10. How does decreasing the sample size affect the margin of error?

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#23: A simple random sample of size n is drawn from a population that is normally distributed with population standard deviation known to be 13. The sample mean is found to be 108 c)Compute the 88% confidence interval if the sample size is 25. compare to part a, how does decreasing the confidence interval affect the margin of error? d)Could we have computed confidence intervals in parts a-c if the population had not been normally distributed? Why? a)If an analysis revealed three outliers greater than the mean, how would this affect the confidence interval?

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9.2: Confidence Interval for a Population Mean – When population standard deviation is unknown Since population standard deviation is unknown, we need to use sample standard deviation: t-interval

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Determine a t-value Find the t-value such that the area under the t- distribution to the right of the t-value is 0.2 assuming 10 degrees of freedom. That is, find t 0.20 with 10 degrees of freedom. The unknown value of t is labeled, and the area under the curve to the right of t is shaded. The value of t 0.20 with 10 degrees of freedom is invt(.80,10) = 0.8791. **Remember Inverse only does area to the LEFT!

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Example: The pasteurization process reduces the amount of bacteria found in dairy products, such as milk. The following data represent the counts of bacteria in pasteurized milk (in CFU/mL) for a random sample of 12 pasteurized glasses of milk. Data courtesy of Dr. Michael Lee, Professor, Joliet Junior College. Construct a 95% confidence interval for the bacteria count.

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A Gallup poll conducted May 20-22, 2005 asked 1006 Americans “During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?” Results of the survey indicated that the mean was 13.4 books and s = 16.6 books. Construct a 99% confidence interval for the mean number of books that Americans read either all or part of during the preceding year. Interpret the interval.

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9.3: Confidence Intervals for a Population Proportion A point estimate is an unbiased estimator of the parameter. The point estimate for the population proportion is where x is the number of individuals in the sample with the specified characteristic and n is the sample size.

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Example In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder. 1123 were in favor. Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

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Confidence Interval 1-Porportion Z-interval ▫Menu: 6: Statistics, 6: Confidence Intervals, 5: 1-prop z Interval Required: ▫np(1-p) > 10 ▫n < 0.05N

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Example In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder. 1123 were in favor. Obtain a 90% confidence interval for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder P=0.63 np(1-p) = For Calculator: x = 1123, n = 1783, CI = 0.90

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a)Obtain a point estimate for the proportion of adult Americans who believe that traditional journalism is out of touch. b)Verify that the requirements for constructing a confidence interval for p are satisfied. #21: A Zogby Interactive survey conducted Feb. 20-21, 2008 found that 1,322 of 1,979 randomly selected adult Americans believe that traditional journalism is out of touch with Americans want from their news.

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c)Construct and interpret a 96% Confidence interval d)Is possible that the proportion of adult Americans is below 60? Is this likely? #21: A Zogby Interactive survey conducted Feb. 20-21, 2008 found that 1,322 of 1,979 randomly selected adult Americans believe that traditional journalism is out of touch with Americans want from their news.

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© 2010 Pearson Prentice Hall. All rights reserved 8-1 Objectives 1.Describe the distribution of the sample mean: samples from normal populations 2.Describe.

© 2010 Pearson Prentice Hall. All rights reserved 8-1 Objectives 1.Describe the distribution of the sample mean: samples from normal populations 2.Describe.

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