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Sta220 - Statistics Mr. Smith Room 310 Class #16 Section 5-1 and 5-2 Notes Our goal in this chapter is to estimate the value of an unknown population parameter, such as the population mean. Example • The mean gas mileage for a new car model • The average expected life of a flat-screen computer monitor. The unknown population parameter (e.g., mean or proportion) that we are interested in estimating is called the target parameter. Procedure Copyright © 2013 Pearson Education, Inc.. All rights reserved. A point estimator of a population parameter is a rule or formula that tells us how to use the sample data to calculate a single number that can be used as an estimate of the target parameter. An interval estimator (or confidence interval) is a formula that tells us how to use the sample data to calculate an interval that estimates the target parameter. Example: We’ll use the sample mean to estimate the population mean . Consequently, , is a point estimator. We then attach a measure of reliability, similar to a standard deviation, to our estimate by obtaining an interval estimator, also called confidence interval. 5-2: Confidence Interval for a Population Mean: Normal (z) Statistic Suppose a large hospital wants to estimate the average length of time patients remain in the hospital. • The hospital’s target parameter is the population mean . • Hospital administrators plan to randomly sample 100 of all previous patients’ records. • The sample mean of the lengths of stay to estimate , the mean of all patients’ visits. So the sample mean represents a point estimator. According to the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal for a large samples. (n > 30) Now we can calculate the interval estimator: ± 1.96 This means we form an interval from 1.96 standard deviations (probability = .95) below the sample mean to 1.96 standard deviations above the mean. Sampling distribution of x Copyright © 2013 Pearson Education, Inc.. All rights reserved. Example 5.1: Consider the large hospital that wants to estimate the average length of stay of its patients, . The hospital randomly samples n = 100 of its patients and finds that the sample mean length of stay is = 4.5 days. Also, suppose it is known that the standard deviation of the length of stay for all hospital patients is = 4 days. Use the interval estimator ± 1.96 to calculate a confidence interval for the target parameter, . Solution Substitution = 4.5 and = 4 into the interval estimator formula, we obtain: ± 1.96 = ± 1.96 = ± 1.96 4 100 = 4.5 ± .78 or (3.72, 5.28). The confidence coefficient is the probability that an interval estimator encloses the population parameter – that is, the relative frequency with which the interval estimator encloses the population parameter when the estimator is used repeatedly a very large number of times. The confidence level is the confidence coefficient expressed as a percentage. ± /2 Confidence intervals for m: 10 samples Copyright © 2013 Pearson Education, Inc.. All rights reserved. Locating za/2 on the standard normal curve Copyright © 2013 Pearson Education, Inc.. All rights reserved. The value is defined as the value of the standard normal random variable z such that the area will lie to its right. In other words, ( > ) = . Example 5-2: Find /2 for = .80. Solution To illustrate, for confidence coefficient of .80, we have (1 − ) = .80, = .20 and /2 = .10. .10 is z value that locates area .10 in the upper tail of the sampling distribution. Since the total area to the right of the mean is .10, we find the z value corresponding to an area of .5- .1 = .4 to the right of the mean. This z value is .10 = 1.28 Table 7.2 Copyright © 2013 Pearson Education, Inc.. All rights reserved. Procedure Copyright © 2013 Pearson Education, Inc.. All rights reserved. Definition Copyright © 2013 Pearson Education, Inc.. All rights reserved. Example 5-3 Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. Descriptive statistics for the data are displayed in the MINITAB printout below. Estimate μ, the mean number of unoccupied seats per flight during the past year, using 90% confidence interval. Solution The form of a large-sample 90% confidence interval for a population mean is: ± /2 = ± .05 = ± 1.645 = 11.6 ± 1.645 4.1 225 = 11.6 ± .45 or the interval 11.15 to 12.05. That is, at the 90% confident level, we estimate the mean number of unoccupied seats per flight to be between 11.15 and 12.5 during the sampled year. Example 5-4 Many middle schools have initiated a program that provides every student with a free laptop (notebook) computer. Student usage of laptops at a middle school that participates in the initiative was investigated in American Secondary Education (fall 2009). In a sample of 106 students, the researchers reported the following statistics on how many minutes per day each student used his or her laptop for taking notes: = 13.2 and s = 19.5. Now the researchers want to estimate the average amount of time per day laptops are used for taking notes for all middle school students across the country. a. Calculate a 90% confidence interval for the target parameter. Interpret the results. b. Explain what the phrase “90% confidence” implies in part a. Solution a. For confidence coefficient .90, = .10 and /2 = .05. From the table, .05 = 1.645. The confidence interval is: ± /2 = ± 1.645 19.5 106 = 13.2 ± 3.116 (10.084, 16. 316) We are 90% confident that the true average amount of time per day laptops are used for taking notes for all middle school students across the country is between 10.084 and 16.316 minutes. b. “90% confidence” means that in a repeated sampling, 90% of all confidence intervals constructed in this manner will contain the true mean. Sometime, the we produce a confidence interval that is too wide. In this case, we want to reduce the width of the interval to obtain a more precise estimate of . Go back to the hospital example. A 90% confidence interval for is (3.92, 5.14). The interval is narrower than the previously calculated 95% confidence interval (3.81, 5.25). However, we also have “less confidence” in the 90% confidence interval. 5-2 Homework Due Friday (also 5-3 and 5-5 will be due Friday, so I encourage you to work tomorrow).