Download SBE10ch08

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Degrees of freedom (statistics) wikipedia , lookup

Taylor's law wikipedia , lookup

Foundations of statistics wikipedia , lookup

Bootstrapping (statistics) wikipedia , lookup

Student's t-test wikipedia , lookup

History of statistics wikipedia , lookup

Resampling (statistics) wikipedia , lookup

German tank problem wikipedia , lookup

Misuse of statistics wikipedia , lookup

Transcript
IS 310
Business
Statistics
CSU
Long Beach
IS 310 – Business Statistics
Slide 1
Interval Estimation
In chapter 7, we studied how to estimate a population parameter
with a sample statistic. We used a point estimate as a single
value.
To increase the level of confidence in estimation, we use a range of
values (rather than a single value) as the estimate of a
population parameter.
Let’s take a real-life example. Suppose, someone asks you how
long it takes to go from CSULB to LAX. What would be a more
reliable estimate – 30 minutes or between 25 and 40 minutes?
If you use 30 minutes as estimate, you are using a point estimate.
On the other hand, if you use between 25 and 40 minutes as
estimate, you are using an interval estimation.
IS 310 – Business Statistics
Slide 2
Interval Estimation
Interval estimation uses a range of values.
The width of the range indicates the level of confidence.
The narrower the range, lower the confidence. Wider
the range, higher the confidence.
Once a confidence level is specified, the interval
estimate can be calculated using the formula, 8.1, in
the book. If one wants 95% confidence, the interval
estimate is known as 95% Confidence Interval. For a
90% confidence, the interval is known as 90%
Confidence Interval.
IS 310 – Business Statistics
Slide 3
Interval Estimation

In this chapter, we will cover the following:

Interval Estimate for a Population Mean (known σ)

Interval Estimate for a Population Mean (unknown σ)

Determining Size of a Sample

Interval Estimate for a Population Proportion
IS 310 – Business Statistics
Slide 4
Margin of Error and the Interval Estimate
The general form of an interval estimate of a
population mean is
x  Margin of Error
IS 310 – Business Statistics
Slide 5
Margin of Error
The Margin of Error implies both
Confidence level and
Amount of error
IS 310 – Business Statistics
Slide 6
Interval Estimate of a Population Mean:
s Known

Interval Estimate of m
x  z /2
where:
s
n
x is the sample mean
1 - is the confidence coefficient
z/2 is the z value providing an area of
/2 in the upper tail of the standard
normal probability distribution
s is the population standard deviation
n is the sample size
IS 310 – Business Statistics
Slide 7
Interval Estimate for Population Mean
(Known σ)



Example:
Lloyd’s Department Store wants to determine a 95%
confidence interval on the average amount spent by
all of its customers (population mean).
Lloyd took a sample of 100 customers and found that
the average amount spent is $82 (this is sample mean
or x‾ ). Lloyd assumes that the population standard
deviation ( σ ) is $20.
IS 310 – Business Statistics
Slide 8
Interval Estimation for Population Mean
(Known σ )

Using Formula 8.1, the 95% confidence interval is:
_
x ± z
(σ/√ n)
= 0.05
0.025

82 ± 1.96 (20/√100)

82 ± 3.92

78.08 to 85.92

We are 95% sure that the average amount spent by all Lloyd
customers is between $78.08 and $85.92.



IS 310 – Business Statistics
Slide 9
Sample Problem
Problem # 5 (10-Page 306; 11-Page 315)
_
Given: x = 24.80 n = 49 σ = 5 confidence level = 95%
a.
Margin of Error = z
b.
/2
= 1.96 (5/√49) = 1.4
95% Confidence Interval is:
_
x ± Margin of Error
24.8 ± 1.4
= 0.05 or 5%
x (σ / √n)
23.4, 26.2
It means that we are 95% confident that the average amount spent by all
customers for dinner at the Atlanta restaurant is between $23.40 and
$26.20
IS 310 – Business Statistics
Slide 10
Interval Estimation of a Population Mean:
s Unknown

If an estimate of the population standard deviation s
cannot be developed prior to sampling, we use the
sample standard deviation s to estimate s .

This is the s unknown case.

In this case, the interval estimate for m is based on the
t distribution.

(We’ll assume for now that the population is
normally distributed.)
IS 310 – Business Statistics
Slide 11
t Distribution
The t distribution is a family of similar probability
distributions.
A specific t distribution depends on a parameter
known as the degrees of freedom.
Degrees of freedom refer to the sample size minus 1.
IS 310 – Business Statistics
Slide 12
t Distribution
A t distribution with more degrees of freedom has
less dispersion.
As the number of degrees of freedom increases, the
difference between the t distribution and the
standard normal probability distribution becomes
smaller and smaller.
IS 310 – Business Statistics
Slide 13
t Distribution
t distribution
(20 degrees
of freedom)
Standard
normal
distribution
t distribution
(10 degrees
of freedom)
z, t
0
IS 310 – Business Statistics
Slide 14
t Distribution
Degrees
Area in Upper Tail
of Freedom
.20
.10
.05
.025
.01
.005
.
.
.
.
.
.
.
50
.849
1.299
1.676
2.009
2.403
2.678
60
.848
1.296
1.671
2.000
2.390
2.660
80
.846
1.292
1.664
1.990
2.374
2.639
100
.845
1.290
1.660
1.984
2.364
2.626
.842
1.282
1.645
1.960
2.326
2.576

Standard normal
z values
IS 310 – Business Statistics
Slide 15
Interval Estimation of a Population Mean:
s Unknown

Interval Estimate
x  t /2
s
n
where: 1 - = the confidence coefficient
t/2 = the t value providing an area of /2
in the upper tail of a t distribution
with n - 1 degrees of freedom
s = the sample standard deviation
IS 310 – Business Statistics
Slide 16
Interval Estimate for Population Mean
(Unknown σ)








Example:
We want to estimate the mean credit card debt of all
customers in the U.S. with a 95% confidence.
A sample of 70 households provided the credit card
balances shown in Table 8.3. The sample standard
deviation (s) is $4007.
Using formula 8.2,
_
x ±t
(s/√n)
= 0.05
0.025
9312 ± 1.995 (4007/√70)
IS 310 – Business Statistics
Slide 17
Interval Estimate (Continued)



9312 ± 955
8357 to 10,267
We are 95% confident that the average credit card
balance of all customers in the U.S. are between
$8,357 and $10,267.
IS 310 – Business Statistics
Slide 18
Summary of Interval Estimation Procedures
for a Population Mean
Can the
population standard
deviation s be assumed
known ?
Yes
No
Use the sample
standard deviation
s to estimate s
s Known
Case
Use
x  z /2
s
n
IS 310 – Business Statistics
s Unknown
Case
Use
x  t /2
s
n
Slide 19
Sample Problem
Problem # 16 (10-Page 315; 11-Page 324)
_
Given: x = 49 n = 100 s = 8.5 Confidence Level = 90%
a.
At 90% confidence, the Margin of Error is:
t
b.
. (s/√n)
1.660 x (8.5/10) = 1.411
/2
The 90% Confidence Interval is:
_
x ± 1.411
49 ± 1.411
47.59, 50.41
It means that we are 90% confident that the average hours of flying
for Continental pilots is between 47.59 and 50.41
IS 310 – Business Statistics
Slide 20
Sample Size for an Interval Estimate
of a Population Mean
Let E = the desired margin of error.
E is the amount added to and subtracted from the
point estimate to obtain an interval estimate.
IS 310 – Business Statistics
Slide 21
Sample Size for an Interval Estimate
of a Population Mean

Margin of Error
E  z /2

s
n
Necessary Sample Size
( z / 2 ) 2 s 2
n
E2
IS 310 – Business Statistics
Slide 22
Determination of Sample Size










Example Problem (10-Page 317; 11-Page 326):
We want to know the average daily rental rate of all midsize
automobiles in the U.S. We want our estimate with a margin of
error of $2 and a 95% level of confidence.
What should be the size of our sample?
Given for this problem: E =2, = 0.05, σ = 9.65
Using Formula 8.3,
2 2
2
n = (z
) (σ ) / E
0.025
= 89.43
The sample size needs to be at least 90.
IS 310 – Business Statistics
Slide 23
Sample Problem
Problem #26 (10-Page 318; 11-Page 327)
Given: µ = 2.41
σ = 0.15
Margin of Error = 0.07
Confidence Level = 95%
Margin of Error = z
. (σ /√n)
/2
2
2
2
0.07 = 1.96 (0.15/√n)
n = (1.96) (0.15) / 0.07)
= 17.63 or 18
A sample size of 18 is recommended for The Cincinnati Enquirer
for its study
IS 310 – Business Statistics
Slide 24
Interval Estimation
of a Population Proportion
The general form of an interval estimate of a
population proportion is
p  Margin of Error
IS 310 – Business Statistics
Slide 25
Interval Estimation
of a Population Proportion
The sampling distribution of p plays a key role in
computing the margin of error for this interval
estimate.
The sampling distribution of p can be approximated
by a normal distribution whenever np > 5 and
n(1 – p) > 5.
IS 310 – Business Statistics
Slide 26
Interval Estimation
of a Population Proportion

Normal Approximation of Sampling Distribution of p
Sampling
distribution
of p
/2
p(1  p)
sp 
n
1 -  of all
p values
z /2s p
IS 310 – Business Statistics
p
/2
p
z /2s p
Slide 27
Interval Estimation
of a Population Proportion

Interval Estimate
p  z / 2
where:
p (1  p )
n
1 - is the confidence coefficient
z/2 is the z value providing an area of
/2 in the upper tail of the standard
normal probability distribution
p is the sample proportion
IS 310 – Business Statistics
Slide 28
Interval Estimate for a Population Proportion










Example (10-Page 320; 11-Page 329):
We want to estimate the proportion of all women golfers in the
U.S. who are satisfied with the availability of tee times with a
95% confidence level.
We take a sample of 900 women golfers and find that 396 are
satisfied with the tee times.
_
Given: p = 396/900 = 0.44
= 0.05
The 95% confidence interval is:
_
_
_
p ± z √ [p (1 – p)]/n = 0.44 ± 1.96√ [0.44 (1 – 0.44)]/900
0.44 ± 0.0324
0.4076 to 0.4724 or 40.76% to 47.24%
We are 95% confident that the proportion of all women golfers
who are satisfied with tee times is between 40.76% and 47.24%.
IS 310 – Business Statistics
Slide 29
Sample Problem
#38 (10-Page 323; 11-Page 332))
a.
b.
Point estimate of the proportion of companies that
fell short of estimates = 29/162 = 0.179
_
Given : p = 104/162 = 0.642
_
_
Margin of error = z
√ [(p (1 – p ) / n)]
0.025
= 1.96 √ [(0.642) (1 – 0.642) / 162]
= 0.0738
IS 310 – Business Statistics
Slide 30
Sample Problem (contd)







95% confidence interval:
0.642 ± 0.0738
0.5682 and 0.7158
We are 95% confident that the proportion of companies that
beat estimates of their profits is between 56.82% and 71.58%.
2
2
c. Required sample size, n = [(1.96) (0.642) (0.358)]/(0.05)
= 353.18 use 354
IS 310 – Business Statistics
Slide 31
End of Chapter 8
IS 310 – Business Statistics
Slide 32