McGraw-Hill/Irwin

Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.

Confidence Intervals

Chapter 8

8-2

Confidence Intervals

8.1 z-Based Confidence Intervals for a Population Mean: σ Known

8.2 t-Based Confidence Intervals for a Population Mean: σ Unknown

8.3 Sample Size Determination

8.4 Confidence Intervals for a Population Proportion

8.5 Confidence Intervals for Parameters of Finite Populations (Optional)

8.6 A Comparison of Confidence Intervals and Tolerance Intervals (Optional)

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

nx

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

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

358.05

8.0

nx

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

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

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

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

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

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

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 –

nzxzx x 22

8-13

Sampling Distribution Of All Possible Sample Means

8-14

z-Based Confidence Intervals for aMean 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

nzx,

nzx

nzx 222

8-15

95% Confidence Level

• For a 95% confidence level, 1 – = 0.95, so = 0.05, and /2 = 0.025

• Need the normal point z0.025

– The area under the standard normal curve between -z0.025 and z0.025 is 0.95

– Then the area under the standard normal curve between 0 and z0.025 is 0.475

– From the standard normal table, the area is 0.475 for z = 1.96

– Then z0.025 = 1.96

8-16

95% Confidence Interval

• The 95% confidence interval is

n.x,

n.x

n.xzx x.

961961

9610250

8-17

99% Confidence Interval

• For 99% confidence, need the normal point z0.005

– Reading between table entries in the standard normal table, the area is 0.495 for z0.005 = 2.575

• The 99% confidence interval is

n.x,

n.x

n.xzx x.

57525752

57520250

8-18

The Effect of a on Confidence Interval Width

z/2 = z0.025 = 1.96 z/2 = z0.005 = 2.575

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

8-20

Example 8.2: 95 Percent Confidence Interval

78.31,34.31

222.056.3150

8.096.156.31025.0

n

zx

8-21

Example 8.2: 99 Percent Confidence Interval

85.31,27.31

294.056.3150

8.0575.256.31005.0

n

zx

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

8-23

t-Based Confidence Intervals for aMean: 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

ns

xt

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

8-25

Degrees of Freedom and thet-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

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

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

8-28

t and Right Hand Tail Areas

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

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

8-31

Using the t Distribution Continued

8-32

t-Based Confidence Intervals for aMean: 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 n 1 degrees of freedom

n

stx 2

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

8-34

Example 8.4 Debt-to-Equity RatiosContinued

• 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 = t0.025 = 2.145

• The 95% confidence interval:

4497.1,2369.11064.03433.115

192.0145.2343.1025.0

n

stx

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

22

B

zn

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

2

2

B

stn

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 = z0.025 = 1.96

– t/2 = t0.025 = 2.776 based on n-1 = 4 d.f.

8-38

Example 8.6: The Car Mileage Case Continued

5024.49

3.0

7583.0776.222

2/

E

stn

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

79.31,33.31227.056.31

50

798.0010.256.31025.0

n

stx