Transcript PP_8_2

Understandable Statistics
Eighth Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Edited by: Jeff, Yann, Julie, and Olivia
Chapter 8: Estimation
Section 8.2
Estimating μ When σ Is Unknown
Focus Points
• Learn about degrees of freedom and
Student’s t distribution.
• Find critical values using degrees of
freedom and confidence level.
• Compute confidence interval for μ when σ
is unknown. What does this information tell
you?
Statistics Quote
“The death of one man is a tragedy. The
death of millions is a statistic.” —Joe
Stalin
What if it is impossible or
impractical to use a large
sample?
Apply the Student’s t distribution.
Properties of a Student’s t
Distribution
1. The distribution is symmetric about the mean
0.
2. The distribution depends on the degrees of
freedom, d.f. (d.f. = n – 1 for μ confidence
intervals).
3. The distribution is bell-shaped, but has thicker
tails than the standard normal distribution.
4. As the degrees of freedom increase, the t
distribution approaches the standard normal
distribution.
Student’s t Variable
x
t
s
n
The shape of the t distribution
depends only only the sample
size, n, if the basic variable x
has a normal distribution.
When using the t distribution, we will
assume that the x distribution is normal.
Table 6 in Appendix II gives
values of the variable t
corresponding to the number
of degrees of freedom (d.f.)
Student’s t Distribution Critical
Values (Table Excerpt)
Degrees of Freedom
d.f. = n – 1
where n = sample size
The t Distribution has a Shape
Similar to that of the the
Normal Distribution
A Normal
distribution
A “t”
distribution
The Student’s t Distribution
Approaches the Normal Curve as
the Degrees of Freedom Increase
Find the critical value tc for a
95% confidence interval if n = 8.
d.f.=7
c
'
''
d.f.
...
6
7
8
...0.90
...
0.95
...
0.98
...
0.99
...
...1.943
...1.895
...1.860
2.447
2.365
2.306
3.143
2.998
2.896
3.707
3.499
3.355
Convention for Using a Student’s t
Distribution
• If the degrees of freedom d.f. you need are
not in the table, use the closest d.f. in the
table that is smaller. This procedure
results in a critical value tc that is more
conservative in the sense that it is larger.
The resulting confidence interval will be
longer and have a probability that is
slightly higher than c.
Confidence Interval for the
Mean of Small Samples (n < 30)
from Normal Populations
xE xE
where x  Sample Mean
s
E  tc
n
c = confidence level (0 < c < 1)
tc = critical value for confidence level c, and
degrees of freedom = n - 1
The mean weight of eight fish
caught in a local lake is 15.7
ounces with a standard
deviation of 2.3 ounces.
Construct a 90% confidence
interval for the mean weight of
the population of fish in the lake.
Mean = 15.7 ounces
Standard deviation = 2.3
ounces
• n = 8, so d.f. = n – 1 = 7
• For c = 0.90, Table 6 in Appendix II gives
t0.90 = 1.895.
s
2.3
E  tc
 1.895
 1.54
n
8
Mean = 15.7 ounces
Standard deviation = 2.3 ounces.
E = 1.54
The 90% confidence interval is:
xE xE
15.7 - 1.54 <  < 15.7 + 1.54
14.16 <  < 17.24
The 90% Confidence Interval:
14.16 <  < 17.24
We are 90% sure that the true
mean weight of the fish in the
lake is between 14.16 and 17.24
ounces.
Summary: Confidence Intervals for
the Mean
• Assume that you have a random sample
of size n from an x distribution and that
you have computed x-bar and s. A
confidence interval for μ is
x  E    x E
where E is the margin of error
• How do you find E? It depends on how
much you know about the x distribution.
Situation I (most common)
• You don't know the population standard
deviation σ. In this situation you
• use the t distribution with margin of error
s
E  tc
n
• with d.f. = n – 1
• Guidelines: If n is less than 30, x should have a
distribution that is mound-shaped and
approximately symmetric. It's even better if the x
distribution is normal. If n is 30 or more, the
central limit theorem (Chapter 7) implies these
restrictions can be relaxed.
Situation II (almost never happens!)
• You actually know the population value of
σ. In addition, you know that x has a
normal distribution. If you don't know that
the x distribution is normal, then your
sample size n must be 30 or larger. In this
situation, you use the standard normal z
distribution with margin of error
E  zc

n
Which distribution should you use
for x?
Examine Problem Statement
σ is known
σ is unknown
Use normal distribution
with margin of error
Use Student’s t distribution
with margin of error
E  zc

n
s
E  tc
n
d.f. = n – 1
Calculator Instructions
CONFIDENCE INTERVALS FOR A POPULATION MEAN
The TI-83 Plus and TI-84 Plus fully support confidence
intervals. To access the confidence interval choices,
press Stat and select TESTS. The confidence interval
choices are found in items 7 through B.
Example (σ is unknown)
• A random sample of 16 wolf dens showed
the number of pups in each to be 5, 8, 7,
5, 3, 4, 3, 9, 5, 8, 5, 6, 5, 6, 4, and 7.
• Find a 90% confidence interval for the
population mean number of pups in such
dens.
Example (σ is unknown)
• In this case we have raw data, so enter
the data in list using the EDIT option of the
Stat key. Since σ is unknown, we use the t
distribution. Under Tests from the STAT
menu, select item 8:TInterval. Since we
have raw data, select the DATA option for
Input. The data is in list and occurs with
frequency 1. Enter 0.90 for the C-Level.
Example (σ is unknown)
• Highlight Calculate and press Enter. The result
is the interval from 4.84 pups to 6.41 pups.
Section 8.2, Problem 11
Diagnostic Tests: Total Calcium Over the past several months, an
adult patient has been treated for tetany (severe muscle spasms).
This condition is associated with an average total calcium level
below 6 mg/dl (Reference: Manual of Laboratory and Diagnostic
Tests, F. Fischbach). Recently, the patient's total calcium tests gave
the following readings (in mg/dl).
9.3 8.8
10.1
8.9
9.4
9.8
10.0
9.9 11.2
12.1
(a) Use a calculator to verify that x-bar = 9.95 and s = 1.02.
(b) Find a 99.9% confidence interval for the population mean of total
calcium in this patient's blood.
(c) Based on your results in part (b), do you think this patient still has a
calcium deficiency? Explain.
Solution
Section 8.2, Problem 17
Finance: P/E Ratio The price of a share of stock divided by the
company's estimated future earnings per share is called the PIE
ratio. High PIE ratios usually indicate “growth” stocks or maybe
stocks that are simply overpriced. Low P/E ratios indicate “value”
stocks or bargain stocks. A random sample of 51 of the largest
companies in the United States gave the following P/E ratios
(Reference: Forbes).
11 35 19 13 15 21 40 18 60 72 9 20
29 53 16 26 21 14 21 27 10 12 47 14
33 14 18 17 20 19 13 25 23 27 5 16
8
49 44 20 27
8
19 12 31 67 51 26
19 18 32
(a) Use a calculator with mean and sample standard deviation keys to
verify that x-bar = 25.2 and s = 15.5.
Section 8.2, Problem 17
(b) Find a 90% confidence interval for the P/E population mean μ of all
large U.S. companies.
(c) Find a 99% confidence interval for the P/E population mean μ of all
large U.S. companies.
(d) Bank One (now merged with J. P. Morgan) had a P/E of 12, AT&T
Wireless had a P/E of 72, and Disney had a P/E of 24. Examine the
confidence intervals in parts (b) and (c). How would you describe
these stocks at this time?
(e) In previous problems, we assumed the x distribution was normal or
approximately normal. Do we need to make such an assumption in
this problem? Why or why not? Hint: See the central limit theorem in
Section 7.2.
Solution