A confidence interval for an unknown parameter consists of ... · PDF fileexpected proportion...

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.9-1

A confidence interval for an unknown

parameter consists of an interval of

numbers based on a point estimate.

The level of confidence represents the

expected proportion of intervals that will

contain the parameter if a large number of

different samples is obtained. The level of

confidence is denoted (1 – α)·100%.


For example, a 95% level of confidence

(α = 0.05) implies that if 100 different

confidence intervals are constructed, each

based on a different sample from the same

population, we will expect 95 of the intervals

to contain the parameter and 5 not to include

the parameter.

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Section

Estimating a Population Mean

9.2


A point estimate is the value of a statistic

that estimates the value of a parameter.

For example, the sample mean, , is a

point estimate of the population mean μ.

x


nzx

2/

x

Estimation of a Population Mean, m100(1-)% Confidence Interval for a Population Mean m using a Large Sample (n>=30)

is unknown, s may be used instead as an approximation.

is the sample mean in a simple random sample and

where

is the population standard deviation

if


Pennies minted after 1982 are made from 97.5% zinc and 2.5%

copper. The following data represent the weights (in grams) of

17 randomly selected pennies minted after 1982.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Treat the data as a simple random sample. Estimate the

population mean weight of pennies minted after 1982.

Parallel Example 1: Computing a Point Estimate


The sample mean is

The point estimate of μ is 2.464 grams.

Solution

x 2.46 2.47 2.45

17 2.464


• 1) Suppose 40 teachers from Washington have been randomly selected and their salaries recorded. The resulting sample average was $29,000 and the standard deviation was $2,000.

• Construct a 90, 95, and 99% confidence interval for μ, the mean salary for the population of teachers in Washington. What do you learn about the width of the intervals as you increase the confidence level?

• Suppose our sample size is 80 instead of 60. Construct a 95% confidence interval for μ.

• Now suppose our sample size is 100. Construct a 95% confidence interval for μ. What do you learn as you increase the sample size?

9-8


• If we have a sample of size smaller than 30,

we use student’s t distribution

9-9


Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution, the distribution of

follows Student’s t-distribution with n – 1 degrees of freedom where is the sample mean and s is the sample standard deviation.

Student’s t-Distribution

t x m

s

n

x


a) Obtain 1,000 simple random samples of size n = 5 from a

normal population with μ = 50 and σ = 10.

b) Determine the sample mean and sample standard deviation

for each of the samples.

c) Compute and for each sample.

d) Draw a histogram for both z and t.

Parallel Example 1: Comparing the Standard Normal

Distribution to the t-Distribution Using Simulation

z x m

n

t x m

sn


Histogram for z


Histogram for t


CONCLUSION:

• The histogram for z is symmetric and bell-shaped

with the center of the distribution at 0 and virtually all

the rectangles between –3 and 3. In other words, z

follows a standard normal distribution.


CONCLUSION (continued):

• The histogram for t is also symmetric and bell-shaped

with the center of the distribution at 0, but the

distribution of t has longer tails (i.e., t is more

dispersed), so it is unlikely that t follows a standard

normal distribution. The additional spread in the

distribution of t can be attributed to the fact that we

use s to find t instead of σ. Because the sample

standard deviation is itself a random variable (rather

than a constant such as σ), we have more dispersion

in the distribution of t.


1. The t-distribution is different for different degrees of freedom.

2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. The area under the curve to the right of 0 equals the area under the curve to the left of 0, which equals 1/2.

4. As t increases or decreases without bound, the graph approaches, but never equals, zero.

Properties of the t-Distribution


5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of σ, thereby introducing further variability into the t-statistic.

6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because, as the sample size n increases, the values of s get closer to the values of σ, by the Law of Large Numbers.

Properties of the t-Distribution


Objective 3

• Determine t-Values


Parallel Example 2: Finding t-values

Find the t-value such that the area under the t-distribution

to the right of the t-value is 0.2 assuming 10 degrees of

freedom. That is, find t0.20 with 10 degrees of freedom.


The figure to the left

shows the graph of the

t-distribution with 10

degrees of freedom.

Solution

The unknown value of t is labeled, and the area under

the curve to the right of t is shaded. The value of t0.20

with 10 degrees of freedom is 0.879.


Objective 4

• Construct and Interpret a Confidence Interval

for the Population Mean


Constructing a (1 – α)100% Confidence Interval for μ

A (1 – α)·100% confidence interval for μ is given by

Lower Upper bound: bound:

where is the critical value with n – 1 df.

x t

2

s

nx t

2

s

n

t

2


We will use Minitab to simulate obtaining 30 simple

random samples of size n = 8 from a population that is

normally distributed with μ = 50 and σ = 10. Construct

a 95% confidence interval for each sample. How many

of the samples result in intervals that contain μ = 50 ?

Parallel Example 2: Using Simulation to Demonstrate the

Idea of a Confidence Interval


Sample Mean 95.0% CI

C1 47.07 ( 40.14, 54.00)

C2 49.33 ( 42.40, 56.26)

C3 50.62 ( 43.69, 57.54)

C4 47.91 ( 40.98, 54.84)

C5 44.31 ( 37.38, 51.24)

C6 51.50 ( 44.57, 58.43)

C7 52.47 ( 45.54, 59.40)

C9 43.49 ( 36.56, 50.42)

C10 55.45 ( 48.52, 62.38)

C11 50.08 ( 43.15, 57.01)

C12 56.37 ( 49.44, 63.30)

C13 49.05 ( 42.12, 55.98)

C14 47.34 ( 40.41, 54.27)

C15 50.33 ( 43.40, 57.25)

C8 59.62 ( 52.69, 66.54)


SAMPLE MEAN 95% CI

C16 44.81 ( 37.88, 51.74)

C17 51.05 ( 44.12, 57.98)

C18 43.91 ( 36.98, 50.84)

C19 46.50 ( 39.57, 53.43)

C20 49.79 ( 42.86, 56.72)

C21 48.75 ( 41.82, 55.68)

C22 51.27 ( 44.34, 58.20)

C23 47.80 ( 40.87, 54.73)

C24 56.60 ( 49.67, 63.52)

C25 47.70 ( 40.77, 54.63)

C26 51.58 ( 44.65, 58.51)

C27 47.37 ( 40.44, 54.30)

C29 46.89 ( 39.96, 53.82)

C30 51.92 ( 44.99, 58.85)

C28 61.42 ( 54.49, 68.35)


Note that 28 out of 30, or 93%, of the confidence intervals contain the population mean μ = 50.

In general, for a 95% confidence interval, any sample mean that lies within 1.96 standard errors of the population mean will result in a confidence interval that contains μ.

Whether a confidence interval contains μ

depends solely on the sample mean, .

x


Construct a 99% confidence interval about the

population mean weight (in grams) of pennies minted

after 1982. Assume μ = 0.02 grams.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Parallel Example 3: Constructing a Confidence Interval


•

• Lower bound:

= 2.464 – 1.746

= 2.464 – 0.008 = 2.456

• Upper bound:

= 2.464 + 2.575

= 2.464 + 0.008 = 2.472

We are 99% confident that the mean weight of pennies

minted after 1982 is between 2.456 and 2.472 grams.

0.0217

0.0217

2 1.746t

x t 2 s

n

x t 2 s

n


For a fixed sample size, as the confidence level

increases, the margin of error increases.

As the sample size increases, the margin of error

decreases.

So, how to decide the sample size for a fixed

margin of error and confidence level ??


Determining the Sample Size n

The sample size required to estimate the population mean, µ, with a level of confidence (1– α)·100% with a specified margin of error, E,is given by

where n is rounded up to the nearest whole number.

n

z

2

s

E

2


Back to the pennies. How large a sample would be

required to estimate the mean weight of a penny

manufactured after 1982 within 0.005 grams with 99%

confidence? Assume = 0.02.

Parallel Example 7: Determining the Sample Size


•

• s = 0.02

• E = 0.005

Rounding up, we find n = 107.

z 2 z0.005 2.575

n

z

2

s

E

2

2.575(0.02)

0.005

2

106.09

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