# Chapter 8 Interval Estimation by 513E0crv

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```									              Chapter 8 Interval Estimation
I. Basic Definitions
II. Confidence Interval Estimation
III. Sample Size and Precision

I. Basic Definitions
• Point estimate: a single value of sample statistic as
an estimator of population parameter.
• Sampling error = | Sample Statistic – Population Parameter |
• Confidence interval: An interval with the confidence
level. It is constructed on the basis of a sample: sample statistic +
margin of error.
• Confidence level: % of intervals formed by samples
with sample size of n will contain the population
parameter. (1-)
• Margin of error: half-width of the confidence interval.
II. Confidence Interval Estimation
Outline:

Procedure:
1. Sample  point estimator ( X or p )
2. Confidence level and Table  Z or tn-1
3. Formulas  compute UCL and LCL:
point estimator  margin of error
Confidence Interval Estimation
of Population Mean (σ Known)
UCL  x  m.o.e.
1-α=.99         z=2.575
LCL  x  m.o.e.
1-α=.95         z=1.96
                        1-α
margin of error  z x  z                                                                            1-α=.90         z=1.645
n
1
2

x
UCL                       x                     LCL
[---------------------       ---------------------]
[--------------------- x ---------------------]
[---------------------
x ---------------------]
1. Interval Estimation for Population Mean
Example 1: p.306 #5 ( known case)
In an effort to estimate the mean amount spent per
customer for dinner at a major Atlanta restaurant.
Data were collected for a sample of 49 customers over a
three-week period. Assume a population standard
deviation of \$5.
a. At the 95% confidence, what is the margin error?
b. If the sample mean is \$24.80, What is the 95%
confidence interval for the population mean?

HWK: p.306 #8
Example 1: p.306 #5
Answer:
•        n = 49
X  \$24.8
=\$5
• Z: (1-)/2 = 0.95/2 = 0.475  Table 1: Z = 1.96
x               5
• 1.      Z / 2
n
 (1.96 )
49
 1.4

x
2.    UCL  X  Z / 2                   24.8  1.4  26.2
n
x
LCL  X  Z / 2                   24.8  1.4  23.4
n
: [23.4, 26.2]
Example 2: p.313 #16 ( unknown case)
The mean flying time for pilots at Continental Airlines
is 49 hours per month. This mean was based on a
sample of 100 pilots and the sample standard
deviation was 8.5 hours.
a. At 95% confidence, what is the margin of error?
b. What is the 95% confidence interval estimate of
the population mean flying time?
c. The mean flying time for pilots at United Airlines is
36 hours per month. Discuss difference between the
flying times at two airlines.

HWK: p.315 #16
Example 2: p.315 #16
Given: n = 100, X = 49, S = 8.5, 1- = .95
Think: What to estimate? Use Z or t?
Answer:
• Sample info (given): n = 100, X = 49, S = 8.5
• t: 1- =0.95, so /2=0.025, d.f.=n-1=99 
Table 2: d.f.=100, /2=0.025  t=1.984
d.f.=80, /2=0.025  t=1.990
*Interpolation: t  1.984  (1.990  1.984) 100  99  1.9843
100  80
S          8.5
a. m.o.e.:    m.o.e.  t / 2      1.9843      1.69
n          100

b. UCL = 49+1.69 =50.69             LCL = 49 – 1.69 = 47.31
: [47.31, 50.69]
c. 36 < LCL. The mean flying time is lower at United.
Aside:
Why t?

t distribution:
• flatter than Z - more spread out
• n , t  Z
• Degree of freedom: a measure of precision.
The d.f.   more precise.
2. Interval Estimation for Population Proportion
Example 3: p.323 #36
A survey asked 346 job seekers. The answer selected
most (152 times) was “higher compensation.”
a. What is the point estimate of the proportion of job
seekers who would select “higher compensation” as
the reason of changing jobs?
b. What is the 95% confidence interval estimate of the
population proportion?

HWK: p.323 #35
Example 3: p.321 #36
Answer:
x 152
a. Point estimate of p:      p         .4393
n 346
b. Confidence interval:
• Z: (1-)/2=0.475, Table 1  Z=1.96
p (1  p )           (0.4393 )(1  0.4393 )
• Margin of error =   Z
n
 (1.96 )
346
= 0.0523

• Confidence interval
UCL  p  m.o.e.  .4393  .0523  .4916
LCL  p  m.o.e.  .4393  .0523  .3870

p: [.3870, .4916]
III. Sample Size and Precision
Quality of estimation:
• Confidence level: 1 - 
• Precision: margin of error
Confidence level: 1 -  is guaranteed by procedure.
P.303 Figure 8.3:
sampling distribution for sample mean.
Probability that X falls between   Z  n and   Z  n
/2
x
/2
x

is 1 - . In general, any sample mean that is within this
range will provide an interval that contains the
population mean .
Margin of error:
Given n, then (1 - )  margin of error .
Given 1 - , then n   margin of error .
Determine sample size to meet requirements for both
confidence level and margin of error:
1. Determine sample size for estimation of  (p.317)
 Z / 2 2 
2

n       2   
 E
           

E: desired margin of error
[ ]: round up (“Be conservative”)
2. Determine sample size for estimation of p (p.321)
 Z / 2 p(1  p) 
2
n                 
        E2       
p = .5 (p.319 Table 8.5 “Be conservative”)
E: desired margin of error
[ ]: round up (“Be conservative”)

Homework: p.318 #25, p.323 #39
Example 4:
Bride’s magazine reported that the mean cost of a
wedding is \$19,000. Assume that the population
standard deviation is \$9,400. Use 95% confidence,
a. What is the recommended sample size if the desired
margin of error is \$1,000?
b. What is the recommended sample size if the desired
margin of error is \$500?
Answer:
a.          1
Z
 /2:      .475  Z   1.96
 /2
2
 (1.96) 2 (9400) 2 
n            2        339.44  340
      1000         
 (1.96) 2 (9400) 2 
b.    n            2        1357.78  1358
      500          
Example 6:
The League of American Theatres and Producers uses an
ongoing audience tracking survey that provides up-to-
date information about Broadway theater audiences.
Every week, the League distributes a one-page survey on
random theater seats at a rotation roster of Broadway
shows.
a. How large a sample should be taken if the desired
margin of error on any proportion is 0.04? Use 95%
confidence.
Answer:
1
Z / 2 :      .475, Table 1  Z / 2  1.96
2
 (1.96) 2 .5(1  .5) 
n             2         600.25  601
       (.04)         

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