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```Chapter 6
Confidence Intervals
1
Chapter Outline
• 6.1 Confidence Intervals for the Mean (Large
Samples)
• 6.2 Confidence Intervals for the Mean (Small
Samples)
• 6.3 Confidence Intervals for Population Proportions
• 6.4 Confidence Intervals for Variance and Standard
Deviation
2
Section 6.1
Confidence Intervals for the Mean
(Large Samples)
3
Section 6.1 Objectives
• Find a point estimate and a margin of error
• Construct and interpret confidence intervals for the
population mean
• Determine the minimum sample size required when
estimating μ
4
Point Estimate for Population μ
Point Estimate
• A single value estimate for a population parameter
• Most unbiased point estimate of the population mean
μ is the sample mean x
Estimate Population with Sample
Parameter…
Statistic
x
Mean: μ
5
Example: Point Estimate for Population μ
Market researchers use the number of sentences per
sample of the number of sentences found in 50
population mean, . (Source: Journal of Advertising
Research)
9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 25
17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7
14 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20
6
Solution: Point Estimate for Population μ
The sample mean of the data is
x 620
x

 12.4
n
50
Your point estimate for the mean length of all magazine
7
Interval Estimate
Interval estimate
• An interval, or range of values, used to estimate a
population parameter.
Point estimate
(
12.4
•
)
Interval estimate
How confident do we want to be that the interval estimate
contains the population mean μ?
8
Level of Confidence
Level of confidence c
• The probability that the interval estimate contains the
population parameter.
c is the area under the
c
standard normal curve
between the critical values.
½(1 – c)
½(1 – c)
-zc
z=0
Critical values
z
zc
Use the Standard
Normal Table to find the
corresponding z-scores.
The remaining area in the tails is 1 – c .
9
Level of Confidence
• If the level of confidence is 90%, this means that we
are 90% confident that the interval contains the
population mean μ.
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
-zc = -1.645
z=0
zc =zc1.645
z
The corresponding z-scores are +1.645.
10
Sampling Error
Sampling error
• The difference between the point estimate and the
actual population parameter value.
• For μ:
 the sampling error is the difference x – μ
 μ is generally unknown
 x varies from sample to sample
11
Margin of Error
Margin of error
• The greatest possible distance between the point
estimate and the value of the parameter it is
estimating for a given level of confidence, c.
• Denoted by E.
E  zcσ x  zc
σ
n
When n  30, the sample
standard deviation, s, can
be used for .
• Sometimes called the maximum error of estimate or
error tolerance.
12
Example: Finding the Margin of Error
confidence level to find the margin of error for the
mean number of sentences in all magazine
13
Solution: Finding the Margin of Error
• First find the critical values
0.95
0.025
0.025
-zc = -1.96
zc
z=0
zczc= 1.96
z
95% of the area under the standard normal curve falls
within 1.96 standard deviations of the mean. (You
can approximate the distribution of the sample means
with a normal curve by the Central Limit Theorem,
because n ≥ 30.)
14
Solution: Finding the Margin of Error
E  zc

n
 1.96 
 zc
s
n
You don’t know σ, but
since n ≥ 30, you can
use s in place of σ.
5.0
50
 1.4
You are 95% confident that the margin of error for the
population mean is about 1.4 sentences.
15
Confidence Intervals for the Population
Mean
A c-confidence interval for the population mean μ
• x E   x E
where E  zc

n
• The probability that the confidence interval contains μ
is c.
16
Constructing Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean
(n  30 or σ known with a normally distributed population)
In Words
1. Find the sample statistics n and
x.
2. Specify , if known.
Otherwise, if n  30, find the
sample standard deviation s and
use it as an estimate for .
In Symbols
x
x
n
(x  x )2
s
n 1
17
Constructing Confidence Intervals for μ
In Words
3. Find the critical value zc that
corresponds to the given
level of confidence.
4. Find the margin of error E.
5. Find the left and right
endpoints and form the
confidence interval.
In Symbols
Use the Standard
Normal Table.
E  zc

n
Left endpoint: x  E
Right endpoint: x  E
Interval:
xE  xE
18
Example: Constructing a Confidence
Interval
Construct a 95% confidence interval for the mean
Solution: Recall x  12.4 and E = 1.4
Left Endpoint:
xE
 12.4  1.4
 11.0
Right Endpoint:
xE
 12.4  1.4
 13.8
11.0 < μ < 13.8
19
Solution: Constructing a Confidence
Interval
11.0 < μ < 13.8
11.0
(
12.4 13.8
•
)
With 95% confidence, you can say that the population
mean number of sentences is between 11.0 and 13.8.
20
Example: Constructing a Confidence
Interval σ Known
A college admissions director wishes to estimate the
mean age of all students currently enrolled. In a random
sample of 20 students, the mean age is found to be 22.9
years. From past studies, the standard deviation is
known to be 1.5 years, and the population is normally
distributed. Construct a 90% confidence interval of the
population mean age.
21
Solution: Constructing a Confidence
Interval σ Known
• First find the critical values
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
z
z=0
-zc = -1.645
zc
zc = 1.645
zc = 1.645
22
Solution: Constructing a Confidence
Interval σ Known
• Margin of error:

E  zc
n
 1.645 
• Confidence interval:
Left Endpoint:
xE
 22.9  0.6
 22.3
1.5
20
 0.6
Right Endpoint:
xE
 22.9  0.6
 23.5
22.3 < μ < 23.5
23
Solution: Constructing a Confidence
Interval σ Known
22.3 < μ < 23.5
Point estimate
22.3
(
x E
22.9
23.5
x
xE
•
)
With 90% confidence, you can say that the mean age
of all the students is between 22.3 and 23.5 years.
24
Using Ti-83 or 84 to find z confidence interval:
n  20
  1.5
c  0.95
x  22.9
The same as:
22.3 < μ < 23.5
Slide 6- 25
Interpreting the Results
• μ is a fixed number. It is either in the confidence
interval or not.
• Incorrect: “There is a 90% probability that the actual
mean is in the interval (22.3, 23.5).”
• Correct: “If a large number of samples is collected
and a confidence interval is created for each sample,
approximately 90% of these intervals will contain μ.
26
Confidence Intervals
http://www.learner.org/courses/againstallodds/unitpages/unit24.html
Slide 6- 27
Interpreting the Results
The horizontal segments
represent 90% confidence
intervals for different
samples of the same size.
In the long run, 9 of every
10 such intervals will
contain μ.
μ
28
Sample Size
• Given a c-confidence level
and a margin of error E, the
minimum sample size n
needed to estimate the
population mean  is
E  zc
n
2

2
E 2  zc
n
z
E 
n
zc2 2
zc 2
n 2 (
)
E
E
2
• If  is unknown, you can
estimate it using s provided
you have a preliminary
sample with at least 30
members.

2
c
2
 zc 
n

 E 
2
29
Example: Sample Size
You want to estimate the mean number of sentences in a
want to be 95% confident that the sample mean is
within one sentence of the population mean? Assume
the sample standard deviation is about 5.0.
30
Solution: Sample Size
• First find the critical values
0.95
0.025
0.025
zc
-zc = -1.96
z=0
zczc= 1.96
z
zc = 1.96
31
Solution: Sample Size
zc = 1.96
  s = 5.0
E=1
 zc   1.96  5.0 
n

  96.04

1

 E  
2
2
When necessary, round up to obtain a whole number.
32
Section 6.1 Summary
• Found a point estimate and a margin of error
• Constructed and interpreted confidence intervals for
the population mean
• Determined the minimum sample size required when
estimating μ
33
Section 6.2
Confidence Intervals for the Mean
(Small Samples)
34
Section 6.2 Objectives
• Interpret the t-distribution and use a t-distribution
table
• Construct confidence intervals when n < 30, the
population is normally distributed, and σ is unknown
35
The t-Distribution
• When the population standard deviation is unknown,
the sample size is less than 30, and the random
variable x is approximately normally distributed, it
follows a t-distribution.
x -
t
s
n
• Critical values of t are denoted by tc.
36
In statistics, the t-distribution was first derived in 1876 by Helmert and Lüroth. In the
English-language literature it takes its name from William Sealy Gosset's 1908 paper in
Biometrika under the pseudonym "Student", published while he worked at the Guinness
Brewery in Dublin, Ireland. One version of the origin of the pseudonym is that Gosset's
employer forbade members of its staff from publishing scientific papers, so he had to
hide his identity. Another version is that Guinness did not want their competition to know
that they were using the t-test to test the quality of raw material. The t-test and the
associated theory became well-known through the work of Ronald A. Fisher, who called
the distribution "Student's distribution".
T-distribution’s mathematical formulas:
Slide 6- 37
Properties of the t-Distribution
1. The t-distribution is bell shaped and symmetric
2. The t-distribution is a family of curves, each
determined by a parameter called the degrees of
freedom. The degrees of freedom are the number
of free choices left after a sample statistic such as x
is calculated. When you use a t-distribution to
estimate a population mean, the degrees of freedom
are equal to one less than the sample size.
 d.f. = n – 1
Degrees of freedom
38
Properties of 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 t-distribution
approaches the normal distribution. After 30 d.f., the tdistribution is very close to the standard normal zdistribution.
The tails in the tdistribution are “thicker”
than those in the standard
normal distribution.
d.f. = 2
d.f. = 5
Standard normal curve
t
0
39
Example: Critical Values of t
Find the critical value tc for a 95% confidence when the
sample size is 15.
Solution: d.f. = n – 1 = 15 – 1 = 14
Table 5: t-Distribution
tc = 2.145
40
Solution: Critical Values of t
95% of the area under the t-distribution curve with 14
degrees of freedom lies between t = +2.145.
c = 0.95
t
-tc = -2.145
tc = 2.145
41
Using the inverse t with Ti-84: the invT gives the number
such that the area to the left of it is given.
For example for 95% in the middle we’ll have to enter
.95+(.05/2) for the area in the calculator command:
invT(.95 +(.05/2), 14) to get: tc = 2.145
and invT(.05/2, 14) to get: -tc = -2.145
c = 0.95
t
-tc =
-2.145
tc = 2.145
Slide 6- 42
Confidence Intervals for the Population
Mean
A c-confidence interval for the population mean μ
s
• x  E    x  E where E  tc
n
• The probability that the confidence interval contains μ
is c.
43
Confidence Intervals and t-Distributions
In Words
1. Identify the sample
statistics n, x , and s.
2. Identify the degrees of
freedom, the level of
confidence c, and the
critical value tc.
3. Find the margin of error E.
In Symbols
x
(x  x )2
x
s
n 1
n
d.f. = n – 1
E  tc
s
n
44
Confidence Intervals and t-Distributions
In Words
4. Find the left and right
endpoints and form the
confidence interval.
In Symbols
Left endpoint: x  E
Right endpoint: x  E
Interval:
xE  xE
s
E  tc
n
45
Example: Constructing a Confidence
Interval
You randomly select 16 coffee shops and measure the
temperature of the coffee sold at each. The sample mean
temperature is 162.0ºF with a sample standard deviation
of 10.0ºF. Find the 95% confidence interval for the
mean temperature. Assume the temperatures are
approximately normally distributed.
Solution:
Use the t-distribution (n < 30, σ is unknown,
temperatures are approximately normally distributed.)
46
Solution: Constructing a Confidence
Interval
• n =16, x = 162.0 s = 10.0 c = 0.95
• df = n – 1 = 16 – 1 = 15
• Critical Value Table 5: t-Distribution
tc = 2.131
47
Solution: Constructing a Confidence
Interval
• Margin of error:
s
10
E  tc
 2.131
 5.3
n
16
• Confidence interval:
Left Endpoint:
xE
 162  5.3
 156.7
Right Endpoint:
xE
 162  5.3
 167.3
156.7 < μ < 167.3
48
Solution: Constructing a Confidence
Interval
• 156.7 < μ < 167.3
Point estimate
156.7
(
x E
162.0
•x
167.3
)
xE
With 95% confidence, you can say that the mean
temperature of coffee sold is between 156.7ºF and
167.3ºF.
49
n =16, x = 162.0 s = 10.0 c = 0.95
df = n – 1 = 16 – 1 = 15
The same as:
156.7 < μ < 167.3
Slide 6- 50
Normal or t-Distribution?
Is n  30?
Yes
No
Is the population normally,
or approximately normally,
distributed?
Use the normal distribution with
σ
E  zc
n
If  is unknown, use s instead.
No
Cannot use the normal
distribution or the t-distribution.
Yes
Use the normal distribution
with E  z σ
Yes
Is  known?
No
c
n
Use the t-distribution with
E  tc
s
n
and n – 1 degrees of freedom.
51
Example: Normal or t-Distribution?
You randomly select 25 newly constructed houses. The
sample mean construction cost is \$181,000 and the
population standard deviation is \$28,000. Assuming
construction costs are normally distributed, should you
use the normal distribution, the t-distribution, or neither
to construct a 95% confidence interval for the
population mean construction cost?
Solution:
Use the normal distribution (the population is
normally distributed and the population standard
deviation is known)
52
Section 6.2 Summary
• Interpreted the t-distribution and used a t-distribution
table
• Constructed confidence intervals when n < 30, the
population is normally distributed, and σ is unknown
53
Section 6.3
Confidence Intervals for Population
Proportions
54
Section 6.3 Objectives
• Find a point estimate for the population proportion
• Construct a confidence interval for a population
proportion
• Determine the minimum sample size required when
estimating a population proportion
55
Point Estimate for Population p
Population Proportion
• The probability of success in a single trial of a
binomial experiment.
• Denoted by p
Point Estimate for p
• The proportion of successes in a sample.
• Denoted by
x number of successes in sample
 pˆ  n 
number in sample
56
Point Estimate for Population p
Estimate Population with Sample
Parameter…
Statistic
Proportion: p
p̂
Point Estimate for q, the proportion of failures
• Denoted by qˆ  1  pˆ
57
Example: Point Estimate for p
In a survey of 1219 U.S. adults, 354 said that their
favorite sport to watch is football. Find a point estimate
for the population proportion of U.S. adults who say
their favorite sport to watch is football. (Adapted from The
Harris Poll)
Solution: n = 1219 and x = 354
x 354
pˆ  
 0.290402  29.0%
n 1219
58
Confidence Intervals for p
Think of proportion as:
x
pˆ  where x is binomial.
n
So the mean of p̂ is the mean
of x divided by n, and std is std of
x divided by n:
Remember : the mean for the
 pˆ  np  p
Binomial distribution is np
n
npq pq and variance is npq.
2
 pˆ  2 
n
n
Also, when we divide the data points by
pq
a number the new mean gets divided by
 pˆ 
n
that number, and the variance gets divided
59
by the square of the number!
Constructing Confidence Intervals for p
In Words
In Symbols
1. Identify the sample statistics n
and x.
2. Find the point estimate p̂.
3. Verify that the sampling
distribution of p̂ can be
approximated by the normal
distribution.
4. Find the critical value zc that
corresponds to the given level of
confidence c.
pˆ 
x
n
npˆ  5, nqˆ  5
Use the Standard
Normal Table
60
Constructing Confidence Intervals for p
In Words
5. Find the margin of error E.
6. Find the left and right
endpoints and form the
confidence interval.
In Symbols
E  zc
pq
ˆˆ
n
Left endpoint: p̂  E
Right endpoint: p̂  E
Interval:
pˆ  E  p  pˆ  E
61
Example: Confidence Interval for p
In a survey of 1219 U.S. adults, 354 said that their
favorite sport to watch is football. Construct a 95%
confidence interval for the proportion of adults in the
United States who say that their favorite sport to watch
is football.
Solution: Recall pˆ  0.290402
qˆ  1  pˆ  1  0.290402  0.709598
62
Solution: Confidence Interval for p
• Verify the sampling distribution of p̂ can be
approximated by the normal distribution
npˆ  1219  0.290402  354  5
nqˆ  1219  0.709598  865  5
• Margin of error:
E  zc
pq
(0.290402)  (0.709598)
ˆˆ
 1.96
 0.025
n
1219
63
Solution: Confidence Interval for p
• Confidence interval:
Left Endpoint:
pˆ  E
 0.29  0.025
 0.265
Right Endpoint:
pˆ  E
 0.29  0.025
 0.315
0.265 < p < 0.315
64
Solution: Confidence Interval for p
• 0.265 < p < 0.315
Point estimate
0.265
(
p̂  E
0.29
0.315
p̂
p̂  E
•
)
With 95% confidence, you can say that the proportion
of adults who say football is their favorite sport is
between 26.5% and 31.5%.
65
Confidence Interval for p using Ti-83 or 84:
The same as:
0.265 < p < 0.315
Slide 6- 66
Sample Size
• Given a c-confidence level and a margin of error E,
the minimum sample size n needed to estimate p is
2
 zc 
ˆ ˆ 
n  pq
E
• This formula assumes you have an estimate for p̂
and qˆ .
• If not, use pˆ  0.5 and qˆ  0.5.
67
Sample Size:
E  zc
E 
2
pq
ˆˆ
n
2
zc pq
ˆˆ
n
2
zc pq
ˆˆ
n 2
E
Slide 6- 68
If there is no preliminary estimate for the population proportion, use :
pˆ  0.5
and
qˆ  0.5
The product of pq is the largest when p and q are both 0.5.
 y  pq

p  q 1
y  p(1  p)
y  p p
Max at p=0.5, q=0.5
2
Slide 6- 69
Example: Sample Size
You are running a political campaign and wish to
estimate, with 95% confidence, the proportion of
registered voters who will vote for your candidate. Your
estimate must be accurate within 3% of the true
population. Find the minimum sample size needed if
1. no preliminary estimate is available.
Solution:
Because you do not have a preliminary estimate
for p̂ use pˆ  0.5 and qˆ  0.5.
70
Solution: Sample Size
• c = 0.95
zc = 1.96
2
E = 0.03
2
 zc 
 1.96 
ˆ ˆ    (0.5)(0.5) 
n  pq
  1067.11
 0.03 
E
Round up to the nearest whole number.
With no preliminary estimate, the minimum sample
size should be at least 1068 voters.
71
Example: Sample Size
You are running a political campaign and wish to
estimate, with 95% confidence, the proportion of
registered voters who will vote for your candidate. Your
estimate must be accurate within 3% of the true
population. Find the minimum sample size needed if
2. a preliminary estimate gives pˆ  0.31 .
Solution:
Use the preliminary estimate pˆ  0.31
qˆ  1  pˆ  1  0.31  0.69
72
Solution: Sample Size
• c = 0.95
zc = 1.96
2
E = 0.03
2
 zc 
 1.96 
ˆ ˆ    (0.31)(0.69) 
n  pq
  913.02
 0.03 
E
Round up to the nearest whole number.
With a preliminary estimate of pˆ  0.31, the
minimum sample size should be at least 914 voters.
Need a larger sample size if no preliminary estimate
is available.
73
Section 6.3 Summary
• Found a point estimate for the population proportion
• Constructed a confidence interval for a population
proportion
• Determined the minimum sample size required when
estimating a population proportion
74
Section 6.4
Confidence Intervals for Variance
and Standard Deviation
75
Section 6.4 Objectives
• Interpret the chi-square distribution and use a chisquare distribution table
• Use the chi-square distribution to construct a
confidence interval for the variance and standard
deviation
76
The Chi-Square Distribution
• The point estimate for 2 is s2
• The point estimate for  is s
• s2 is the most unbiased estimate for 2
Estimate Population with Sample
Parameter…
Statistic
Variance: σ2
s2
Standard deviation: σ
s
77
The Chi-Square Distribution
• You can use the chi-square distribution to construct a
confidence interval for the variance and standard
deviation.
• If the random variable x has a normal distribution,
then the distribution of
 
2
(n  1)s 2
σ2
forms a chi-square distribution for samples of any
size n > 1.
78
The Chi-Square Distribution
Slide 6- 79
Properties of The Chi-Square Distribution
1. All chi-square values χ2 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 χ2-distribution with
degrees of freedom equal to one less than the sample
size.
•
d.f. = n – 1
Degrees of freedom
3. The area under each curve of the chi-square distribution
equals one.
80
Properties of The Chi-Square Distribution
4. Chi-square distributions are positively skewed.
chi-square distributions
81
Critical Values for χ2
• There are two critical values for each level of
confidence.
• The value χ2R represents the right-tail critical value
• The value χ2L represents the left-tail critical value.
1 c
2
c
The area between
the left and right
critical values is c.
1 c
2
 L2
 R2
χ2
82
Example: Finding Critical Values for χ2
Find the critical values  R2 and  L2 for a 90% confidence
interval when the sample size is 20.
Solution:
• d.f. = n – 1 = 20 – 1 = 19 d.f.
• Each area in the table represents the region under the
chi-square curve to the right of the critical value.
1  c 1  0.90

 0.05
2
2
• Area to the right of
χ2R =
• Area to the right of
1 c
2
χ L= 2

1  0.90
 0.95
2
83
Solution: Finding Critical Values for χ2
Table 6: χ2-Distribution
 R2  30.144
 L2  10.117
90% of the area under the curve lies between 10.117 and
30.144
84
Slide 6- 85
Confidence Intervals for 2 and 
Confidence Interval for 2:
2
2
(
n

1)
s
(
n

1)
s
2
•

σ

2
R
 L2
Confidence Interval for :
•
(n  1)s 2
 R2
σ 
Note :
 R2   L2 
(n  1)s 2
 R2

(n  1)s 2
 L2
(n  1)s 2
 L2
• The probability that the confidence intervals contain
σ2 or σ is c.
86
Confidence Intervals for 2 and 
In Words
In Symbols
1. Verify that the population has a
normal distribution.
2. Identify the sample statistic n and
the degrees of freedom.
3. Find the point estimate s2.
4. Find the critical value χ2R and χ2L
that correspond to the given level
of confidence c.
d.f. = n – 1
2

(
x

x
)
s2 
n 1
Use Table 6 in
Appendix B
87
Confidence Intervals for 2 and 
In Words
5. Find the left and right
endpoints and form the
confidence interval for the
population variance.
6. Find the confidence
interval for the population
standard deviation by
taking the square root of
each endpoint.
In Symbols
(n  1)s 2
 R2
(n  1)s 2
 R2
σ2 
σ 
(n  1)s 2
 L2
(n  1)s 2
 L2
88
Example: Constructing a Confidence
Interval
You randomly select and weigh 30 samples of an allergy
medicine. The sample standard deviation is 1.20
milligrams. Assuming the weights are normally
distributed, construct 99% confidence intervals for the
population variance and standard deviation.
Solution:
• d.f. = n – 1 = 30 – 1 = 29 d.f.
89
Solution: Constructing a Confidence
Interval
1  c 1  0.99

 0.005
R= 2
2
• Area to the right of
χ2
• Area to the right of
1 c
2
χ L= 2
1  0.99

 0.995
2
• The critical values are
χ2R = 52.336 and χ2L = 13.121
90
Solution: Constructing a Confidence
Interval
Confidence Interval for 2:
Left endpoint:
(n  1)s 2
 R2
(30  1)(1.20)2

 0.80
52.336
2
2
(
n

1)
s
(30

1)(1.20)
Right endpoint:

 3.18
2
13.121
L
0.80 < σ2 < 3.18
With 99% confidence you can say that the population
variance is between 0.80 and 3.18 milligrams.
91
Solution: Constructing a Confidence
Interval
Confidence Interval for :
(n  1)s 2
2
R
σ 
(n  1)s 2
2
L
(30  1)(1.20)2
(30  1)(1.20)2
 
52.336
13.121
0.89 < σ < 1.78
With 99% confidence you can say that the population
standard deviation is between 0.89 and1.78 milligrams.
92
Slide 6- 93
Section 6.4 Summary
• Interpreted the chi-square distribution and used a chisquare distribution table
• Used the chi-square distribution to construct a
confidence interval for the variance and standard
deviation
94
Control Charts
http://www.learner.org/courses/againstallodds/unitpages/unit23.html
Slide 6- 95
Chapter 6: Confidence Intervals
Elementary Statistics:
Picturing the World
Fifth Edition
by Larson and Farber
Slide 4- 96
A random sample of 42 textbooks has a
mean price of \$114.50 and a standard
deviation of \$12.30.
Find a point estimate for the mean price of
all textbooks.
A. \$42
B. \$12.30
C. \$114.50
D. \$2.73
Slide 6- 97
A random sample of 42 textbooks has a
mean price of \$114.50 and a standard
deviation of \$12.30.
Find a point estimate for the mean price of
all textbooks.
A. \$42
B. \$12.30
x 
C. \$114.50
D. \$2.73
Slide 6- 98
Find the critical value zc necessary to form
a 98% confidence interval.
A. 2.33
B. 2.05
C. 0.5040
D. 0.8365
Slide 6- 99
Find the critical value zc necessary to form
a 98% confidence interval.
A. 2.33
B. 2.05
C. 0.5040
D. 0.8365
Slide 6- 100
A random sample of 42 textbooks has a
mean price of \$114.50 and a standard
deviation of \$12.30.
Find a 98% confidence interval for the
mean price of all textbooks.
A. (111.38, 117.62)
B. (110.08, 118.92)
C. (110.78, 118.22)
D. (109.61, 119.39)
Slide 6- 101
A random sample of 42 textbooks has a
mean price of \$114.50 and a standard
deviation of \$12.30.
Find a 98% confidence interval for the
mean price of all textbooks.
A. (111.38, 117.62)
B. (110.08, 118.92)
C. (110.78, 118.22)
D. (109.61, 119.39)
Slide 6- 102
Determine the minimum sample size needed
to construct a 95% confidence interval for the
mean age of employees at a company. The
estimate must be accurate to within 0.5 year.
Assume the standard deviation is 4.8 years.
A. 18
B. 19
C. 354
D. 355
Slide 6- 103
Determine the minimum sample size needed
to construct a 95% confidence interval for the
mean age of employees at a company. The
estimate must be accurate to within 0.5 year.
Assume the standard deviation is 4.8 years.
A. 18
B. 19
 zc   (1.96)(4.8) 
n

  354.04

0.5

 E  
2
2
C. 354
D. 355
Slide 6- 104
Find the critical value tc necessary to form
a 95% confidence interval with a sample
size of 15.
A. 1.960
B. 2.145
C. 2.131
D. 2.120
Slide 6- 105
Find the critical value tc necessary to form
a 95% confidence interval with a sample
size of 15.
A. 1.960
B. 2.145
C. 2.131
D. 2.120
Slide 6- 106
A random sample of 15 DVD players has a
mean price of \$64.30 and a standard
deviation of \$5.60.
Find a 95% confidence interval for the
mean price of all DVD players.
A. (61.20, 67.40)
B. (61.47, 67.13)
C. (61.22, 67.38)
D. (61.10, 67.51)
Slide 6- 107
A random sample of 15 DVD players has a
mean price of \$64.30 and a standard
deviation of \$5.60.
Find a 95% confidence interval for the
mean price of all DVD players.
A. (61.20, 67.40)
B. (61.47, 67.13)
C. (61.22, 67.38)
D. (61.10, 67.51)
Slide 6- 108
In a survey of 250 Internet users, 195 have
high-speed Internet access at home.
Find a point estimate for the proportion of
all Internet users who have high-speed
Internet access at home.
A. 1.28
B. 0.78
C. 0.22
D. 195
Slide 6- 109
In a survey of 250 Internet users, 195 have
high-speed Internet access at home.
Find a point estimate for the proportion of
all Internet users who have high-speed
Internet access at home.
A. 1.28
B. 0.78
C. 0.22
195
pˆ  p 
 0.78
250
D. 195
Slide 6- 110
In a survey of 250 Internet users, 195 have
high-speed Internet access at home.
Find a 90% confidence interval for the
proportion of all Internet users who have
high-speed Internet access at home.
A. (1.19, 1.38)
B. (0.728, 0.832)
C. (0.731, 0.829)
D. (0.737, 0.823)
Slide 6- 111
In a survey of 250 Internet users, 195 have
high-speed Internet access at home.
Find a 90% confidence interval for the
proportion of all Internet users who have
high-speed Internet access at home.
A. (1.19, 1.38)
B. (0.728, 0.832)
C. (0.731, 0.829)
D. (0.737, 0.823)
Slide 6- 112
You want to estimate, with 95% confidence,
the proportion of households with pets.
Your estimate must be accurate within 3%
of the population proportion. No
preliminary estimate is available. Find the
minimum sample size needed.
A. 1141
B. 3267
C. 1068
D. 1067
Slide 6- 113
You want to estimate, with 95% confidence,
the proportion of households with pets.
Your estimate must be accurate within 3%
of the population proportion. No
preliminary estimate is available. Find the
minimum sample size needed.
A. 1141
B. 3267
C. 1068
D. 1067
Let
pˆ  0.5
and
qˆ  0.5
2
2
 zc 
 1.96 
ˆ ˆ    (0.5)(0.5) 
n  pq
  1067.11
 0.03 
E
Slide 6- 114
Find the critical values  and  for a 98%
confidence interval when the sample size
is 25.
2
R
A.  R2  2.492
 L2  2.492
2

B. R  2.326
 L2  2.326
2
L
2
2
C.  R  44.314  L  11.524
2
2



42
.
980
D. R
L  10.856
Slide 6- 115
Find the critical values  and  for a 98%
confidence interval when the sample size
is 25.
2
R
2
L
 R2 has (1-0.98)/2 area to the right of it.
 L2 has 0.98 + (1-0.98)/2 to the right of it.
Remember in both cases you are
looking for areas to their right!
Slide 6- 116
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