Transcript Chapter 6: Confidence Intervals
Chapter 6
Confidence Intervals
§
6.1
Confidence Intervals for the Mean (Large Samples)
Point Estimate for Population μ
A point estimate is a single value estimate for a population parameter. The most unbiased point estimate of the population mean, , is the
sample
mean,
x
.
Example : A random sample of 32 textbook prices (rounded to the nearest dollar) is taken from a local college bookstore. Find a point estimate for the population mean, .
34 56 79 94 34 65 86 95 38 65 87 96 45 66 87 98 45 67 87 98 45 67 88 101 45 68 90 110 54 74 90 121
x
74.22
The point estimate for the population mean of textbooks in the bookstore is $74.22.
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Interval Estimate
An interval estimate is an interval, or range of values, used to estimate a population parameter. Point estimate for textbooks interval estimate How confident do we want to be that the interval estimate contains the population mean,
μ
?
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Level of Confidence
The level of confidence c is the probability that the interval estimate contains the population parameter.
c c is the area beneath the normal curve between the critical values.
1 2 (1 – c ) 1 (1 – 2 c ) z c z = 0 z c z Critical values Use the Standard Normal Table to find the corresponding z scores.
The remaining area in the tails is 1 –
c
.
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Common Levels of Confidence
If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean,
μ
.
0.90
0.05
z c = z c z = 0 z c = z c The corresponding
z
scores are ± 1.645.
0.05
z
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Common Levels of Confidence
If the level of confidence is 95%, this means that we are 95% confident that the interval contains the population mean,
μ
.
0.95
0.025
z c = z c z = 0 z c z c 1.96
The corresponding
z
scores are ± 1.96.
0.025
z
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Common Levels of Confidence
If the level of confidence is 99%, this means that we are 99% confident that the interval contains the population mean,
μ
.
0.99
0.005
z c = z c z = 0 z c = z c The corresponding
z
scores are ± 2.575.
0.005
z
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Margin of Error
The difference between the point estimate and the actual population parameter value is called the sampling error .
When
μ
x
μ
is estimated, the sampling error is the difference – . Since
μ
is usually unknown, the maximum value for the error can be calculated using the level of confidence.
Given a level of confidence, the margin of error (sometimes called the maximum error of estimate or error tolerance) is the greatest possible distance between the point estimate and the value of the parameter it is estimating. E
z
c
σ n
When n deviation, 30, the sample standard s , can be used for .
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Margin of Error
Example: A random sample of 32 textbook prices is taken from a local college bookstore. The mean of the sample is = 74.22, and the sample standard deviation is
s
= 23.44.
x
Use a 95% confidence level and find the margin of error for the mean price of all textbooks in the bookstore.
E
z
c
σ n
1.96
23.44
32 8.12
Since
n
30,
s
substituted for can be
σ.
We are 95% confident that the margin of error for the population mean (all the textbooks in the bookstore) is about $8.12.
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Confidence Intervals for μ
A c
-
confidence interval for the population mean
μ
is
x E
x E
The probability that the confidence interval contains
μ
is
c
.
Example: A random sample of 32 textbook prices is taken from a 74.22, the sample standard deviation is margin of error is
E
= 8.12.
s
x
= 23.44, and the Construct a 95% confidence interval for the mean price of all textbooks in the bookstore.
Continued.
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Confidence Intervals for μ
Example continued: Construct a 95% confidence interval for the mean price of all textbooks in the bookstore.
x
= 74.22
s
= 23.44
E
= 8.12
Left endpoint = ?
Right endpoint = ?
• x •
x x
= 66.1
= 82.34
With 95% confidence we can say that the cost for all textbooks in the bookstore is between $66.10 and $82.34.
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Finding Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean ( 30 or
σ
known with a normally distributed population)
n In Words
1. Find the sample statistics n and
x
.
2. Specify , if known. Otherwise, if n find the sample standard deviation use it as an estimate for .
s 30, and
In Symbols
x
n
x
s
x
n
x
1 ) 2 3. Find the critical value z c that corresponds to the given level of confidence.
Use the Standard Normal Table.
4. Find the margin of error E .
c σ n 5. Find the left and right endpoints and form the confidence interval.
Left endpoint: Interval: x E x E x E x E
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Confidence Intervals for μ (
Known)
Example: A random sample of 25 students had a grade point average with a mean of 2.86. Past studies have shown that the standard deviation is 0.15 and the population is normally distributed. Construct a 90% confidence interval for the population mean grade point average.
n z
x
c = 25 = 1.645
x
E = 2.86
z c σ n 2.86 ± 0.05
= 0.15
1.645
0.15
2.81 < 25
μ
0.05
< 2.91
With 90% confidence we can say that the mean grade point average for all students in the population is between 2.81 and 2.91.
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Sample Size
Given a
c
confidence level and a maximum error of estimate, estimate
E
, the minimum sample size
n
, needed to , the population mean, is
n
z c E
2 .
If is unknown, you can estimate it using
s
provided you have a preliminary sample with at least 30 members.
Example : You want to estimate the mean price of all the textbooks in the college bookstore. How many books must be included in your sample if you want to be 99% confident that the sample mean is within $5 of the population mean?
Continued.
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Sample Size
Example continued: You want to estimate the mean price of all the textbooks in the college bookstore. How many books must be included in your sample if you want to be 99% confident that the sample mean is within $5 of the population mean?
n x
= 74.22
s =
23.44
z
c = 2.575
z c
E
2 2
145.7
5
(Always round up.) You should include at least 146 books in your sample.
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§ 6.2
Confidence Intervals for the Mean (Small Samples)
The t
-
Distribution
When a sample size is less than 30, and the random variable is approximately normally distributed, it follow a t
-
x distribution .
t x μ s n Properties of the t -distribution 1. The t -distribution is bell shaped and symmetric about the mean.
2. The t -distribution is a family of curves, each determined by a parameter called the degrees of freedom. The are the number of free choices left after a sample statistic such as is calculated. When you use a the sample size.
t degrees of freedom -distribution to estimate a
x
population mean, the degrees of freedom are equal to one less than d.f. = n – 1 Degrees of freedom Continued.
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The t
-
Distribution
3. The total area under a t curve is 1 or 100%.
4. The mean, median, and mode of the t distribution are equal to zero.
5. As the degrees of freedom increase, the normal distribution. After 30 d.f., the the standard normal z distribution.
t t distribution approaches the distribution is very close to The tails in the t distribution are “thicker” than those in the standard normal distribution.
d.f. = 2 d.f. = 5 t Standard 0 normal curve
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Critical Values of
t Example : Find the critical value sample size is 5.
t
c for a 95% confidence when the Appendix B: Table 5: t -Distribution d.f.
1 2 3 4 5 Level of confidence, One tail, Two tails, c d.f. =
c n
= 0.95
– 1 = 5 – 1 = 4
t
c 0.50
0.25
0.80
0.10
0.90
0.05
0.95
0.025
0.98
0.01
0.50
0.20
0.10
0.05
0.02
1.000 3.078 6.314 12.706 31.821
.816 1.886 2.920
.765 1.638 2.353
.741 1.533 2.132
.727 1.476 2.015
4.303
3.182
2.776
2.571
6.965
4.541
3.747
3.365
= 2.776
Continued.
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Critical Values of
t Example continued : Find the critical value sample size is 5.
t
c for a 95% confidence when the 95% of the area under the t -distribution curve with 4 degrees of freedom lies between t = ±2.776.
c = 0.95
t c = 2.776
t c = 2.776
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t
21
Confidence Intervals and
t
-
Distributions
Constructing a Confidence Interval for the Mean: Distribution
In Words In Symbols t
1. Identify the sample statistics and s .
n
x
,
x
n
x
s
x
n
x
1 ) 2 2. Identify the degrees of freedom, the level of confidence c , and the critical value t c .
3. Find the margin of error E .
4. Find the left and right endpoints and form the confidence interval.
d.f. = n – 1 E t c s n Left endpoint: Right endpoint: Interval: x E x E x E x E
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Constructing a Confidence Interval
Example: In a random sample of 20 customers at a local fast food restaurant, the mean waiting time to order is 95 seconds, and the standard deviation is 21 seconds. Assume the wait times are normally distributed and construct a 90% confidence interval for the mean wait time of all customers.
n
= 20
x
95 s =
21 d.f.
= 19
t c
= 1.729
E t c s n 1.729
21 20 8.1
x
E
= 95 ± 8.1
86.9 <
μ
< 103.1
We are 90% confident that the mean wait time for all customers is between 86.9 and 103.1 seconds.
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Normal or t -Distribution?
Is n 30?
No Yes Use the normal distribution with If c σ .
n is unknown, use s instead.
Is the population normally, or approximately normally, distributed?
Yes Is known?
No No Yes You cannot use the normal distribution or the t -distribution. Use the normal distribution with c σ n .
Use the t distribution with and n E t c s n – 1 degrees of freedom.
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Normal or t -Distribution?
Example: Determine whether to use the normal distribution, the
t
distribution, or neither.
a.)
n
= 50, the distribution is skewed, sample size is 50.
s
= 2.5
The normal distribution would be used because the b.)
n
= 25, the distribution is skewed,
s
= 52.9
Neither distribution would be used because n < 30 and the distribution is skewed.
c.)
n
= 25, the distribution is normal, = 4.12
The normal distribution would be used because although n < 30, the population standard deviation is known.
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§
6.3
Confidence Intervals for Population Proportions
Point Estimate for Population p
The probability of success in a single trial of a binomial experiment is proportion .
p
. This probability is a population The point estimate for
p
, the population proportion of successes, is given by the proportion of successes in a sample and is denoted by
x
where
x
is the number of successes in the sample and the number in the sample. The point estimate for the are read as “
p
hat” and “
q
ˆ
q
hat.”
p
ˆ .
p
ˆ
n q
ˆ is
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Point Estimate for Population
p Example : In a survey of 1250 US adults, 450 of them said that their favorite sport to watch is baseball. Find a point estimate for the population proportion of US adults who say their favorite sport to watch is baseball.
n
= 1250
x
= 450
x
450 1250 0.36
The point estimate for the proportion of US adults who say baseball is their favorite sport to watch is 0.36, or 36%.
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Confidence Intervals for p
A c
-
confidence interval for the population proportion
p p p E
is where c
n
The probability that the confidence interval contains
p
is
c
.
Example: Construct a 90% confidence interval for the proportion of US adults who say baseball is their favorite sport to watch.
n
= 1250
x
= 450
p
ˆ 0.36
Continued.
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Confidence Intervals for p
Example continued:
n
= 1250
x
= 450
p
ˆ 0.36
q
ˆ 0.64
E z c pq n 1.645
(0.36)(0.64) 1250 0.022
Right endpoint = ?
Left endpoint = ?
•
0.338
p
ˆ • 0.382
With 90% confidence we can say that the proportion of all US adults who say baseball is their favorite sport to watch is between 33.8% and 38.2%.
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Finding Confidence Intervals for p
Constructing a Confidence Interval for a Population Proportion In Words In Symbols 1. Identify the sample statistics n and
x.
2. Find the point estimate p ˆ .
x 3. Verify that the sampling distribution can be approximated by the normal distribution.
4. Find the critical value confidence.
z c that corresponds to the given level of 5. Find the margin of error E .
6. Find the left and right endpoints and form the confidence interval.
5, nq ˆ 5 Use the Standard Normal Table.
c pq n Left endpoint : Right endpoint : Interval : p E p E p p E
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Sample Size
Given a
c
confidence level and a margin of error,
E
, the minimum sample size
n
, needed to estimate
p
is
n
z E c
2 .
This formula assumes you have an estimate for and If not, use and 0.5.
q
ˆ.
Example : You wish to find out, with 95% confidence and within 2% of the true population, the proportion of US adults who say that baseball is their favorite sport to watch.
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Continued.
32
Sample Size
Example continued : You wish to find out, with 95% confidence and within 2% of the true population, the proportion of US adults who say that baseball is their favorite sport to watch.
n
= 1250
x
= 450
p
ˆ 0.36
n
z c E
2 (0.36)(0.64) 2212.8
2 1.96
0.02
(Always round up.) You should sample at least 2213 adults to be 95% confident.
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§
6.4
Confidence Intervals for Variance and Standard Deviation
The Chi
-
Square Distribution
The is
s
. point estimate for
s
2 2 is
s
2 , and the point estimate for is the most unbiased estimate for 2 . You can use the deviation.
chi
-
square distribution
to construct a confidence interval for the variance and standard If the random variable
x
the distribution of has a normal distribution, then 2 (
n
σ
1) 2
s
2 forms a chi
-
square distribution for samples of any size
n
> 1.
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The Chi
-
Square Distribution
Four properties of the chi square distribution are as follows.
1. All chi-square values χ 2 are greater than or equal to zero.
2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for 2 , use the χ 2 distribution with degrees of freedom. To form a confidence interval for 2 , use the χ 2 distribution with degrees of freedom equal to one less than the sample size.
3. The area under each curve of the chi-square distribution equals one.
4. Find the critical value z c of confidence.
that corresponds to the given level 5. Chi-square distributions are positively skewed.
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Critical Values for
X 2 There are two critical values for each level of confidence. The value χ 2 R represents the right tail critical value and χ 2 L represents the left tail critical value. 1 2 c X 2 R X 2 Area to the right of X 2 R 1 1 2 c X 2 X 2 L Area to the right of X 2 L 1 2 c c 1 2 c The area between the left and right critical values is X 2 L X 2 R X 2
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c .
37
Critical Values for
X 2 Example : Find the critical values
χ
2 R when the sample size is 18.
and
χ
2 L for an 80% confidence Because the sample size is 18, there are d.f. =
n
– 1 = 18 – 1 = 17 degrees of freedom, Area to the right of
χ
2 R = 1 2
c
Area to the right of
χ
2 L = 1 2
c
Use the Chi square distribution table to find the critical values.
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Continued.
38
Critical Values for
X 2 Example continued : Appendix B: Table 6: χ 2 Distribution Degrees of freedom 1 2 3 0.995
0.010
0.072
0.99
0.975
- 0.001
0.020 0.051
0.115 0.216
0.95
0.004
0.103
0.352
0.90
0.016
0.211
0.584
0.10
2.706
4.605
6.251
0.05
3.841
5.991
7.815
16 17 18 5.142
5.697
6.265
5.812 6.908
6.408 7.564
7.015 8.231
7.962
9.312 23.542 26.296
8.672 10.085 24.769 27.587
9.390 10.865 25.989 28.869
χ
2 R = 24.769
χ
2 L = 10.085
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Confidence Intervals for
2
and
A c
-
confidence interval for a population variance and standard deviation is as follows.
Confidence Interval for 2
:
( n X 1) 2 R s 2 σ 2 ( n X 1) 2 L s 2 Confidence Interval for
:
( n X 1) 2 R s 2 ( n X 1) 2 L s 2 The probability that the confidence intervals contain 2 or is c
.
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Confidence Intervals for
2
and
Constructing a Confidence Interval for a Variance and a Standard Deviation In Words 1. Verify that the population has a normal distribution.
In Symbols 2. Identify the sample statistic n the degrees of freedom.
and d.f. = n 1 3. Find the point estimate 4. Find the critical value confidence.
s 2 . χ 2 R and χ 2 L that correspond to the given level of s 2 (x x n 1 ) 2 Use Table 6 in Appendix B.
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Continued.
41
Confidence Intervals for
2
and
Constructing a Confidence Interval for a Variance and a Standard Deviation In Words 5. Find the left and right endpoints and form the confidence interval.
In Symbols ( n X 1) 2 R s 2 σ 2 ( n X 1) 2 L s 2 6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint.
( n X 1) 2 R s 2 ( n X 1) 2 L s 2
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Constructing a Confidence Interval
Example : You randomly select and weigh 41 samples of 16-ounce bags of potato chips. The sample standard deviation is 0.05 ounce. Assuming the weights are normally distributed, construct a 90% confidence interval for the population standard deviation.
d.f. =
n
– 1 = 41 – 1 = 40 degrees of freedom, Area to the right of
χ
2 R = 1 Area to the right of
χ
2 L = 1 2
c
2
c
The critical values are
χ
2 R = 55.758 and
χ
2 L = 26.509.
Continued.
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Constructing a Confidence Interval
Example continued :
χ
2 L = 26.509
χ
2 R = 55.758
Left endpoint = ?
Right endpoint = ?
( n 1) 2 R s 2 0.04
• 55.758
2 0.04
0.06
• ( n 1) 2 L s 2 0.06
26.509
2 With 90% confidence we can say that the population standard deviation is between 0.04 and 0.06 ounces.
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