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Point Estimation Large Sample Confidence Intervals for a Population Mean Margin of error Lecture 14A

2. The Basic Paradigm.. Point Estimation. A point estimate of some parameter ? is a single number, calculated from sample data, that can be regarded as an educated guess for the value of ?.Examples:32.5 mpg as a point estimate of the population mean fuel efficiency

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Point Estimation Large Sample Confidence Intervals for a Population Mean Margin of error Lecture 14A

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    1. 1 Point Estimation & Large Sample Confidence Intervals for a Population Mean Margin of error Lecture 14A William F. Hunt, Jr. Statistics 361 Section 7.1 and 7.2

    2. 2 The Basic Paradigm.

    3. Point Estimation A point estimate of some parameter ? is a single number, calculated from sample data, that can be regarded as an educated guess for the value of ?. Examples: 32.5 mpg as a point estimate of the population mean fuel efficiency for all cars of a particular type. .350 is a point estimate for the proportion p of all individuals who would try a particular product again after using a free sample.

    4. Example 7.1 (Page 290) Estimating Fish Populations Capture/Recapture Experiment Biologist wants to estimate the number of fish in a certain lake, say N A sample of 100 fish is selected & each is tagged & then returned to the lake. After some time, a second sample of 250 fish is selected & 25 are tagged, what is a sensible estimate for N? Since a total of 100 fish were initially tagged, this suggests that we use 1000 as a point estimate of N.

    5. Example 7.1 Continued. Let M denote the number of fish initially tagged. Let n = the size of the recapture sample. Let x = the no. of tagged fish in the recapture sample, then N = [Mn/x], where the [c] denotes the largest whole number that is at most c; this takes care of cases where Mn/x is not a whole number.

    6. Properties of Estmators The estimator should be unbiased.

    7. Butchers thumbButchers thumb

    8. Unbiased Estimator Denote a population parameter by the letter ? and denote any estimator of this parameter by ?. Then ? is an unbiased estimator if ? =0. Otherwise, ? is said to be biased, and the quantity ? ? is called the bias of ?. ? is theta hat!? is theta hat!

    9. Consistent Estimator If the probability that an estimator ? falls close to a population parameter ? can be made as near as 1 as desired by increasing the sample size n, then ? is said to be a consistent estimator of ?.

    10. Large Sample Confidence Intervals for a Population Mean Statistic Margin of Error This interval should contain the true population parameter. We often use the term confidence interval to talk about the statistic plus or minus the margin of error. In many research settings and publications we present the confidence interval. This interval should contain the true population parameter. We often use the term confidence interval to talk about the statistic plus or minus the margin of error. In many research settings and publications we present the confidence interval. This interval should contain the true population parameter.

    11. Recall: Distribution of the mean Recall that the distribution of the mean from a large random sample is something that closely follows a normal distribution. Just as with the proportion we can use the width of this type of distribution to provide an appropriate margin of error.Recall that the distribution of the mean from a large random sample is something that closely follows a normal distribution. Just as with the proportion we can use the width of this type of distribution to provide an appropriate margin of error.

    12. Recall: Distribution of the mean The width of this normal distribution is determined by the standard deviation of the sample mean. Our margin of error will be based on the standard deviation of the mean. The width of this normal distribution is determined by the standard deviation of the sample mean. Our margin of error will be based on the standard deviation of the mean.

    13. MOE for Mean from large sample Base MOE on normal distribution standard error So we will create a margin of error by basing it on a normal distribution and the standard error of the mean. Remember that the standard error of the mean is given by the standard deviation of the population divided by the square root of n. So we will create a margin of error by basing it on a normal distribution and the standard error of the mean. Remember that the standard error of the mean is given by the standard deviation of the population divided by the square root of n.

    14. MOE for Mean from large sample MOE The margin of error will be an amount that we add and subtract. So we will create a margin of error by basing it two parts.The margin of error will be an amount that we add and subtract. So we will create a margin of error by basing it two parts.

    15. MOE for Mean from large sample MOE It will include the standard deviation of the sample mean. Recall that the standard deviation of the mean is made up of the population standard deviation divided by the square root of the sample size.It will include the standard deviation of the sample mean. Recall that the standard deviation of the mean is made up of the population standard deviation divided by the square root of the sample size.

    16. MOE for Mean from large sample MOE It will also include the Z value from the normal distribution. This will be the confidence coefficient and determine the proportion of possible samples in which the margin of error will contain the true mean.It will also include the Z value from the normal distribution. This will be the confidence coefficient and determine the proportion of possible samples in which the margin of error will contain the true mean.

    17. Problem We have a sample. Population standard deviation is unknown However, we have the problem: We only have a sample, the population standard deviation is unknown. We will use our best guess about the population value which is the sample standard deviation. Since we dont know sigma we will use our sample standard deviation s. However, we have the problem: We only have a sample, the population standard deviation is unknown. We will use our best guess about the population value which is the sample standard deviation. Since we dont know sigma we will use our sample standard deviation s.

    18. Confidence Interval Statistic Margin of Error Putting it all together we use our statistics ybar plus or minus the z-score times s over square root of n.Putting it all together we use our statistics ybar plus or minus the z-score times s over square root of n.

    19. Confidence Interval Statistic Margin of Error The sample standard deviation is s and the sample size is n. Lets take a look at an example to see how we put this to work.The sample standard deviation is s and the sample size is n. Lets take a look at an example to see how we put this to work.

    20. Example The Department of Justice funded a study entitled CONSEQUENCES OF A CRIMINAL RECORD FOR EMPLOYMENT OPPORTUNITY IN MILWAUKEE, WISCONSIN, 2002 As part of this study a random sample of 130 employers were contacted. One of the questions asked of these employers was What is the starting hourly wage for your most recently hired employee? We will look at the results of a study done in Milwaukee Wisconsin a few years ago. In this study the researchers talked with 130 employers. They were interested in a variety of issues about employment in Milwaukee. The researchers asked many questions including the starting hourly wage of the most recently hired employee. We will look at the results of a study done in Milwaukee Wisconsin a few years ago. In this study the researchers talked with 130 employers. They were interested in a variety of issues about employment in Milwaukee. The researchers asked many questions including the starting hourly wage of the most recently hired employee.

    21. Example From this study we found a sample mean of $7.515 and a standard deviation of $2.72. Find a 95% confidence interval for the true mean. Some of the things found about the most recently hired employee included the average of just over $7.51 and a standard deviation of 2.72. Some of the things found about the most recently hired employee included the average of just over $7.51 and a standard deviation of 2.72.

    22. Example From this study we found a sample mean of $7.515 and a standard deviation of $2.72. Find a 95% confidence interval for the true mean. n=130 mean=$7.515 standard deviation s= $2.72 So if we list what we know about this situation we find that n is 130, the mean (y-bar) is 7.515. The standard deviation s is 2.72. We know that we want 95% confidence. That means we know we will have Z=1.96.So if we list what we know about this situation we find that n is 130, the mean (y-bar) is 7.515. The standard deviation s is 2.72. We know that we want 95% confidence. That means we know we will have Z=1.96.

    23. Example 95% confidence =>1.96 In this example we want 95% confidence. That means we need to find the point from a normal distribution that has 2.5% above it. Using table A we see that is 1.96.In this example we want 95% confidence. That means we need to find the point from a normal distribution that has 2.5% above it. Using table A we see that is 1.96.

    24. Example Lets take our formula for a confidence interval and plug in the appropriate values.Lets take our formula for a confidence interval and plug in the appropriate values.

    25. Example

    26. Example This produces a margin of error of plus or minus 0.47.This produces a margin of error of plus or minus 0.47.

    27. Example Adding and subtracting the Adding and subtracting the

    28. Example We are 95% confident that the true mean average hourly wage in Milwaukee will be between $7.045 and $7.985. So we can say We are 95% confident that the true mean average hourly wage in Milwaukee will be between $7.045 and $7.985. Now you can try calculating the margin of error for yourself. So we can say We are 95% confident that the true mean average hourly wage in Milwaukee will be between $7.045 and $7.985. Now you can try calculating the margin of error for yourself.

    29. Class Problem Assuming that n is large, determine the confidence level for each of the following two-sided confidence intervals: A. B. C.

    30. Class Problem A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of .81 sec and a sample standard deviation of .34 sec. Calculate a 99% two-sided confidence interval for the true average echo duration and interpret the resulting interval.

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