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McGraw-Hill/IrwinCopyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Confidence Intervals Chapter 8

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8-2 Confidence Intervals 8.1z-Based Confidence Intervals for a Population Mean: σ Known 8.2t-Based Confidence Intervals for a Population Mean: σ Unknown 8.3Sample Size Determination 8.4Confidence Intervals for a Population Proportion 8.5Confidence Intervals for Parameters of Finite Populations (Optional) 8.6A Comparison of Confidence Intervals and Tolerance Intervals (Optional)

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8-3 z-Based Confidence Intervals for a Mean: σ Known The starting point is the sampling distribution of the sample mean –Recall that if a population is normally distributed with mean and standard deviation σ, then the sampling distribution of is normal with mean = and standard deviation –Use a normal curve as a model of the sampling distribution of the sample mean Exactly, because the population is normal Approximately, by the Central Limit Theorem for large samples

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8-4 The Empirical Rule 68.26% of all possible sample means are within one standard deviation of the population mean 95.44% of all possible observed values of x are within two standard deviations of the population mean 99.73% of all possible observed values of x are within three standard deviations of the population mean

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8-5 Example 8.1: The Car Mileage Case Assume a sample size (n) of 5 Assume the population of all individual car mileages is normally distributed Assume the population standard deviation (σ) is 0.8 The probability that with be within ±0.7 of µ is 0.9544

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8-6 Example 8.1 The Car Mileage Case Continued Assume the sample mean is 31.3 That gives us an interval of [31.3 ± 0.7] = [30.6, 32.0] The probability is 0.9544 that the interval [ ± 2σ] contains the population mean µ

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8-7 Example 8.1: The Car Mileage Case #3 That the sample mean is within ±0.7155 of µ is equivalent to… will be such that the interval [ ± 0.7115] contains µ Then there is a 0.9544 probability that will be a value so that interval [ ± 0.7115] contains µ –In other words P( – 0.7155 ≤ µ ≤ + 0.7155) = 0.9544 –The interval [ ± 0.7115] is referred to as the 95.44% confidence interval for µ

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8-8 Example 8.1: The Car Mileage Case #4 Three 95.44% Confidence Intervals for Thee intervals shown Two contain µ One does not

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8-9 Example 8.1: The Car Mileage Case #5 According to the 95.44% confidence interval, we know before we sample that of all the possible samples that could be selected … There is 95.44% probability the sample mean is such the interval [ ± 0.7155] will contain the actual (but unknown) population mean µ –In other words, of all possible sample means, 95.44% of all the resulting intervals will contain the population mean µ –Note that there is a 4.56% probability that the interval does not contain µ The sample mean is either too high or too low

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8-10 Generalizing In the example, we found the probability that is contained in an interval of integer multiples of More usual to specify the (integer) probability and find the corresponding number of The probability that the confidence interval will not contain the population mean is denoted by –In the mileage example, = 0.0456

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8-11 Generalizing Continued The probability that the confidence interval will contain the population mean is denoted by –1 – is referred to as the confidence coefficient –(1 – ) 100% is called the confidence level Usual to use two decimal point probabilities for 1 – –Here, focus on 1 – = 0.95 or 0.99

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8-12 General Confidence Interval In general, the probability is 1 – that the population mean is contained in the interval –The normal point z /2 gives a right hand tail area under the standard normal curve equal to /2 –The normal point - z /2 gives a left hand tail area under the standard normal curve equal to /2 –The area under the standard normal curve between -z /2 and z /2 is 1 –

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8-13 Sampling Distribution Of All Possible Sample Means

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8-14 z-Based Confidence Intervals for a Mean with σ Known If a population has standard deviation (known), and if the population is normal or if sample size is large (n 30), then … … a (1- )100% confidence interval for is

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8-15 95% Confidence Level For a 95% confidence level, 1 – = 0.95, so = 0.05, and /2 = 0.025 Need the normal point z 0.025 –The area under the standard normal curve between -z 0.025 and z 0.025 is 0.95 –Then the area under the standard normal curve between 0 and z 0.025 is 0.475 –From the standard normal table, the area is 0.475 for z = 1.96 –Then z 0.025 = 1.96

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8-16 95% Confidence Interval The 95% confidence interval is

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8-17 99% Confidence Interval For 99% confidence, need the normal point z 0.005 –Reading between table entries in the standard normal table, the area is 0.495 for z 0.005 = 2.575 The 99% confidence interval is

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8-18 The Effect of a on Confidence Interval Width z /2 = z 0.025 = 1.96z /2 = z 0.005 = 2.575

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8-19 Example 8.2: Car Mileage Case Given – = 31.56 –σ = 0.8 –n = 50 Will calculate –95 Percent confidence interval –99 Percent confidence interval

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8-20 Example 8.2: 95 Percent Confidence Interval

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8-21 Example 8.2: 99 Percent Confidence Interval

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8-22 Notes on the Example The 99% confidence interval is slightly wider than the 95% confidence interval –The higher the confidence level, the wider the interval Reasoning from the intervals: –The target mean should be at least 31 mpg –Both confidence intervals exceed this target –According to the 95% confidence interval, we can be 95% confident that the mileage is between 31.33 and 31.78 mpg We can be 95% confident that, on average, the mean exceeds the target by at least 0.33 and at most 0.78 mpg

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8-23 t-Based Confidence Intervals for a Mean: Unknown If is unknown (which is usually the case), we can construct a confidence interval for based on the sampling distribution of If the population is normal, then for any sample size n, this sampling distribution is called the t distribution

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8-24 The t Distribution The curve of the t distribution is similar to that of the standard normal curve –Symmetrical and bell-shaped –The t distribution is more spread out than the standard normal distribution –The spread of the t is given by the number of degrees of freedom Denoted by df For a sample of size n, there are one fewer degrees of freedom, that is, df = n – 1

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8-25 Degrees of Freedom and the t-Distribution As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve

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8-26 The t Distribution and Degrees of Freedom As the sample size n increases, the degrees of freedom also increases As the degrees of freedom increase, the spread of the t curve decreases As the degrees of freedom increases indefinitely, the t curve approaches the standard normal curve –If n ≥ 30, so df = n – 1 ≥ 29, the t curve is very similar to the standard normal curve

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8-27 t and Right Hand Tail Areas Use a t point denoted by t –t is the point on the horizontal axis under the t curve that gives a right hand tail equal to a –So the value of t in a particular situation depends on the right hand tail area a and the number of degrees of freedom df = n – 1 = 1 – a, where 1 – a is the specified confidence coefficient

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8-28 t and Right Hand Tail Areas

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8-29 Using the t Distribution Table Rows correspond to the different values of df Columns correspond to different values of a See Table 8.3, Tables A.4 and A.20 in Appendix A and the table on the inside cover –Table 8.3 and A.4 gives t points for df 1 to 30, then for df = 40, 60, 120 and ∞ On the row for ∞, the t points are the z points –Table A.20 gives t points for df from 1 to 100 For df greater than 100, t points can be approximated by the corresponding z points on the bottom row for df = ∞ –Always look at the accompanying figure for guidance on how to use the table

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8-30 Using the t Distribution Example: Find t for a sample of size n=15 and right hand tail area of 0.025 –For n = 15, df = 14 – = 0.025 Note that a = 0.025 corresponds to a confidence level of 0.95 –In Table 8.3, along row labeled 14 and under column labeled 0.025, read a table entry of 2.145 –So t = 2.145

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8-31 Using the t Distribution Continued

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8-32 t-Based Confidence Intervals for a Mean: Unknown If the sampled population is normally distributed with mean , then a (1 )100% confidence interval for is t /2 is the t point giving a right-hand tail area of /2 under the t curve having n1 degrees of freedom

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8-33 Example 8.4 Debt-to-Equity Ratios Estimate the mean debt-to-equity ratio of the loan portfolio of a bank Select a random sample of 15 commercial loan accounts – = 1.34 – s = 0.192 – n = 15 Want a 95% confidence interval for the ratio Assume all ratios are normally distributed but σ unknown

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8-34 Example 8.4 Debt-to-Equity Ratios Continued Have to use the t distribution At 95% confidence, 1 – = 0.95 so = 0.05 and /2 = 0.025 For n = 15, df = 15 – 1 = 14 Use the t table to find t /2 for df = 14, t /2 = t 0.025 = 2.145 The 95% confidence interval:

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8-35 Sample Size Determination (z) If is known, then a sample of size so that is within B units of , with 100(1- )% confidence

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8-36 Sample Size Determination (t) If σ is unknown and is estimated from s, then a sample of size so that is within B units of , with 100(1- )% confidence. The number of degrees of freedom for the t /2 point is the size of the preliminary sample minus 1

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8-37 Example 8.6: The Car Mileage Case What sample size is needed to make the margin of error for a 95 percent confidence interval equal to 0.3? Assume we do not know σ Take prior example as preliminary sample –s = 0.7583 –z /2 = z 0.025 = 1.96 –t /2 = t 0.025 = 2.776 based on n-1 = 4 d.f.

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8-38 Example 8.6: The Car Mileage Case Continued

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8-39 Example 8.7: The Car Mileage Case Want to see that the sample of 50 mileages produced a 95 percent confidence interval with a margin or error of ±0.3 The margin of error is 0.227, which is smaller than the 0.3 desired

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