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**Estimation and Confidence Intervals**

Chapter 9 . McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved.

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Learning Objectives Chapter 9 1. Define a point estimate. 2. Compute a confidence interval for the population mean when the population standard deviation is known. 3. Compute a confidence interval for a population mean when the population standard deviation is unknown. 4. Compute a confidence interval for a population proportion. Chapter 10 Conduct a test of hypothesis about a population proportion. 9-2

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Point Estimates A point estimate is a single value (point) derived from a sample and used to estimate a population value. Disadvantages of point estimates: Point estimates derived from a sample can be quite different from the corresponding population parameter it represents. Point estimates does not take into consideration of the sample size and standard deviation. 9-3

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**Confidence Interval Estimates**

A confidence interval estimate is a range of values constructed from sample data so that the population parameter is likely to occur within that range. The likelihood is controlled by the level of confidence, denoted 1-α. C.I. = point estimate ± margin of error 9-4

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**Confidence Intervals for a Mean – σ Known**

. The width of the interval is determined by the level of confidence; the higher the level, the wider the CI is. Sample size; the large the size, the narrow the CI is. Population standard deviation; CI is wider for larger σ. 9-5

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**Example: Confidence Interval for a Mean – σ Known**

The American Management Association wishes to have information on the mean income of middle managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions: 1. What is the population mean? 2. What is a reasonable range of values for the population mean? 3. What do these results mean? 9-6

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**Example: Confidence Interval for a Mean – σ Known**

The American Management Association wishes to have information on the mean income of middle managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions: 1. What is the population mean? In this case, we do not know. We do know the sample mean is $45,420. Hence, our best estimate of the unknown population value is the corresponding sample statistic. The sample mean of $45,420 is a point estimate of the unknown population mean. 9-7

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**Example: Confidence Interval for a Mean – σ Known**

The American Management Association wishes to have information on the mean income of middle managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions: 2. What is a reasonable range of values for the population mean? Suppose the association decides to use the 95 percent level of confidence: The confidence limit are $45,169 and $45,671 The ±$251 is referred to as the margin of error 9-8

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**How to Obtain z value for a Given Confidence Level**

The 95% confidence refers to the middle 95% of the observations. Therefore, the remaining 5% is equally divided between the two tails and each tail gets 2.5%. Thus the z value we are looking for is Z.025. Option 1: to find the z value in the table, we look for .025 (the subscript of Z) inside the table. Here we can find an exact match for Thus the corresponding z value is 1.96. Option 2: =NORMSINV(1-.025) 9-9

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**How to Obtain z value for a Given Confidence Level**

Formally the z-value we are looking for is Zα/2 Example: Recall that the level of confidence is denoted 1- α. For a 95% level of confidence, 1- α = 95% α = 5% =.05 α/2 = .025 Thus Zα/2 = Z.025 95% C.I. for the population mean: 9-10

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**Example: Confidence Interval for a Mean - Interpretation**

The American Management Association wishes to have information on the mean income of middle managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions: What do these results mean? The 95% confidence limit are $45,169 and $45,671. This means we are 95% confidence that the mean income of middle managers in the retail industry lies in between $45,169 and $45,671. 9-11

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**Interval Estimates - Interpretation**

The meaning of being 95% confidence: if we construct similar intervals using 100 samples with the same size from the same population, then 95% of the intervals will include the true population mean. 9-12

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**Population Standard Deviation (σ) Unknown**

In most sampling situations the population standard deviation (σ) is not known. Below are some examples where it is unlikely the population standard deviations would be known. In such situations, we replace the unknown with sample standard deviation s, and replace z values with t values. The formula is given below. 9-13

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**Example: Confidence Interval for a Mean – σ Unknown**

A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inch of tread remaining with a standard deviation of 0.09 inch. Construct a 95 percent confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of tread remaining is 0.30 inches? 9-14

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**Example: Confidence Interval for a Mean – σ Unknown**

Finding t value: tα/2, n-1= t .025, 9 = 2.262 Option 1: look up in the t table. Option 2: =tinv(.025*2, 9) Note: 1 – α = 95%. . 9-15

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T distribution table tα/2, n-1= t .025, 9 = 2.262

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**Confidence Interval Estimates for the Mean**

The manager of the Inlet Square Mall, near Ft. Myers, Florida, wants to estimate the mean amount spent per shopping visit by customers. A sample of 20 customers reveals the following amounts spent. What is the best estimate of the population mean? Determine a 95% confidence interval. Interpret the result. Would it be reasonable to conclude that he population mean is $50? What about $60? Data Inlet Square Mall . 9-17

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**Confidence Interval Estimates for the Mean – By Formula**

The best estimated of the population mean is the sample mean: $49.35. The 95% confidence interval is from $45.13 to $53.57. This means we are 95% confidence that the mean amount spent per shopping visit lies between $45.13 and $53.57. Since $50 is in the interval, it is reasonable to conclude that the population mean is $50. But $60 is not in the interval, thus we conclude that the population mean is unlikely to be $60. 9-18

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**Confidence Interval Estimates for the Mean – Using Excel**

Note: this only works for constructing intervals using t distribution. Excel instruction: see Lind et al., p 329, #5 for more details. 9-19

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**When to Use the z or t Distribution for Confidence Interval Computation**

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**A Confidence Interval for a Population Proportion (π)**

The examples below illustrate the situations that involve proportion. 1. The career services director at Southern Technical Institute reports that 80 percent of its graduates enter the job market in a position related to their field of study. 2. A company representative claims that 45 percent of Burger King sales are made at the drive-through window. 3. A survey of homes in the Chicago area indicated that 85 percent of the new construction had central air conditioning. Inference on population proportion is useful for nominal (categorical) data Proportion: The fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest. 9-21

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**Confidence Interval for a Population Proportion - Formula**

Where X is the number of “success,” that is, the number of observations that fall into the category we are interested in. 9-22

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**Confidence Interval for a Population Proportion- Example**

The union representing the Bottle Blowers of America (BBA) is considering a proposal to merge with the Teamsters Union. According to BBA union bylaws, at least three-fourths of the union membership must approve any merger. A random sample of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal. What is the estimate of the population proportion? Develop a 95 percent confidence interval for the population proportion. Basing your decision on this sample information, can you conclude that the necessary proportion of BBA members favor the merger? Why? 9-23

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