Download Chapter 7 Section 1

Chapter 7
Estimation
Point Estimate
an estimate of a population
parameter given by a
single number
Examples of
Point Estimates

x is used as a point estimate for  
 s is used as a point estimate for .
Error of Estimate
the magnitude of the
difference between the
point estimate and the
true parameter value
The error of estimate
using x as a point
estimate for  is:
x
Confidence Level
A confidence level, c, is a measure
of the degree of assurance we have
in our results.
The value of c may be any number
between zero and one.
Typical values for c include 0.90,
0.95, and 0.99.
Critical Value for a
Confidence Level, c
the value zc such that the area
under the standard normal
curve falling between – zc and
zc is equal to c.
Critical Value for a
Confidence Level, c
P(– zc < z < zc ) = c
-zc
zc
Find z0.90 such that 90% of the
area under the normal curve lies
between z-0.90 and z0.90
.90
-z0.90
z0.90
P(-z0.90 < z < z0.90 ) = 0.90
Find z0.90 such that 90% of the
area under the normal curve lies
between z-0.90 and z0.90
.4500
-z0.90
z0.90
P(0< z < z0.90 ) = 0.90/2 = 0.4500
Find z0.90 such that 90% of the
area under the normal curve lies
between z-0.90 and z0.90
According to Table 4 in Appendix I,
0.4500 lies roughly halfway between two
values in the table (.4495 and .4505).
Calculating the invNorm(0.05) gives you
the critical value of z0.90 = 1.6449.
Common Levels of Confidence
and Their Corresponding
Critical Values
Level of Confidence, c
Critical Value, zc
0.70 or 70%
1.0364
0.75 or 75%
1.1503
0.80 or 80%
1.2816
0.85 or 85%
1.4395
0.90 or 90%
1.6449
0.95 or 95%
1.9600
0.98 or 98%
2.3263
0.99 or 99%
2.5758
0.999 or 99.9%
3.2905
Confidence Interval
for the Mean of Large
Samples (n  30)
xE 
  xE
where x  Sample Mean
E  zc

n
if the population standard
deviation s is known
Confidence Interval for the Mean
of Large Samples (n  30)
The answer is expressed in a sentence.
The form of the sentence is given below.
We can say with a c% confidence level that
(whatever the problem is about) is
s
s
between x  zc
and x + zc
units.
n
n
Create a 95% confidence
interval for the mean
driving time between
Philadelphia and Boston.
Assume that the mean
driving time of 64 trips was
5.2 hours with a standard
deviation of 0.9 hours.
Key Information
x  5.2 hours
s  0.9 hours
c = 95%, so z c = 1.9600
n  64
95% Confidence Interval:
0.9
0.9
5.2  1.9600
   5.2  1.9600
64
64
1.764
1.764
5.2 
   5.2 
8
8
5.2  0.2205    5.2  0.2205
4.9795    5.4205
We can say with 95% a confidence level that the
population mean driving time from Philadelphia to
Boston is between 4.9795 and 5.4205 hours.
95% Confidence Interval:
Calculator
Computation
We can say with a 95%
confidence level that the
population mean driving
time from Philadelphia to
Boston is between 4.9795
and 5.4205 hours.
STAT
TESTS
# 7  ZInterval
Inpt : Stats
  0.9
x  5.2
n  64
C  Level : 0.95
Calculate
4.9795, 5.4205 
When estimating the mean, how
large a sample must be used in order
to assure a given level of confidence?
Use the formula:
 zc 
n

 E 
2
Determine the sample size
necessary to determine (with 99%
confidence) the mean time it takes
to drive from Philadelphia to
Boston. We wish to be within 15
minutes of the true time. Assume
that a preliminary sample of 45
trips had a standard deviation of
0.8 hours.
... determine with 99%
confidence...
z0.99  2.5758
... We wish to be within 15
minutes of the true time. ...
E = 15 minutes
or
E = 0.25 hours
...a preliminary sample of 45
trips had a standard deviation
of 0.8 hours.
Since the preliminary sample is large enough, we
can assume that the population standard
deviation is approximately equal to 0.8 hours.
  0.8
Minimum Required Sample Size
 zc 
n

 E 
2
 2.5758 0.8 
n


0.25
 2.0606 
n
 0.25 
n  8.2426 
2
n  67.9398
n  68
2
2
Rounding Sample Size
Any fractional value
of n is always
rounded to the next
higher whole number.
We would need a sample of 68 trip
times to be 99% confidence that
the sample mean time it takes to
drive from Philadelphia to Boston
is within the 0.25 hours of the
population mean time it takes to
drive from Philadelphia to Boston.
THE
END
OF THE
PRESENTATION
Answers
to the
Sample
Questions
1. As part of a study on AP test results, a
local guidance counselor gathered data
on 200 tests given at local high schools.
The test results are based on scores of 1
to 5, where a 1 means a very poor test
result to a 5 which means a superior test
result. The sample mean was 3.62 with a
standard deviation of 0.84.
a. Construct a 90% confidence interval for
the population mean.
x  zc

   x  zc

n
n
 0.84 
 0.84 
3.62  1.6449 
   3.62  1.6449 

 200 
 200 
1.3817
1.3817
3.62 
   3.62 
14.1421
14.1421
3.62  0.0977    3.62  0.0977
3.5223    3.7177
We can say with a 90% confidence level that the
population mean score on AP tests at local high
schools is between 3.5223 and 3.7177.
1. As part of a study on AP test results, a
local guidance counselor gathered data
on 200 tests given at local high schools.
The test results are based on scores of 1
to 5, where a 1 means a very poor test
result to a 5 which means a superior test
result. The sample mean was 3.62 with a
standard deviation of 0.84.
b. Construct a 95% confidence interval for
the population mean.
x  zc

   x  zc

n
n
 0.84 
 0.84 
3.62  1.9600 
   3.62  1.9600 


 200 
 200 
1.6464
1.6464
3.62 
   3.62 
14.1421
14.1421
3.62  0.1164    3.62  0.1164
3.5036    3.7364
We can say with a 95% confidence level that the
population mean score on AP tests at local high
schools is between 3.5036 and 3.7364.
1. As part of a study on AP test results, a
local guidance counselor gathered data
on 200 tests given at local high schools.
The test results are based on scores of 1
to 5, where a 1 means a very poor test
result to a 5 which means a superior test
result. The sample mean was 3.62 with a
standard deviation of 0.84.
c. Perform the calculator checks for parts a
and b.
1. Part c - Calculator check for part a
ZINTERVAL
ZINTERVAL
Input : Stats
3.5223, 3.7177 
 : 0.84
x  3.62
x : 3.62
n  200
n : 200
C  Level : 0.90
Calculate
VARS  STATISTICS  TEST
H :lower  3.522300679
I :upper  3.717699321
1. Part c - Calculator check for part b
ZINTERVAL
ZINTERVAL
Input : Stats
3.5036, 3.7364 
 : 0.84
x  3.62
x : 3.62
n  200
n : 200
C  Level : 0.95
Calculate
VARS  STATISTICS  TEST
H :lower  3.503584079
I :upper  3.736415921
1. As part of a study on AP test results, a
local guidance counselor gathered data
on 200 tests given at local high schools.
The test results are based on scores of 1
to 5, where a 1 means a very poor test
result to a 5 which means a superior test
result. The sample mean was 3.62 with a
standard deviation of 0.84.
d.
How many test results would be require to be 95%
confident that the sample mean test score is within
0.05 of the population mean score?
 zc 
n

 E 
2
 1.9600 0.84  We would need to
n
acquire
1,085
AP
test



0.05
2
 1.6464 
n

 0.05 
2
n  32.928 
2
n  1084.2532
n  1085
results to have a 95%
confidence level with
an error of no more
than 0.05 for the
population mean AP
test scores.
2. The SAT results for 50 randomly
selected seniors are listed below. The
score are based only on the English and
Math portions of the SAT examination.
Use your calculator to determine a 99%
confidence interval for the mean score of
the SAT examination.
980 1240
1280
11901370
1060
1080
1380 950 870 1030 1220 750 1410 1150
1100 1070
890
930 1520
810 1090 1310 1030
1200
990 1560
810
940 1010 1140 1060
1250 1240 1130 1170 1080 1210
970
810
920
1160
940 1050 1110 1300 1230
790 1050 1240
Remember that you need perform a 1-VARSTATS calculation on the data first to get the
value for the sample standard deviation.
ZINTERVAL
ZINTERVAL
Input : Data
1033.9, 1169.9 
 :185.29634290476
List : L1
x  1101.4
n  50
Freq :1
C  Level : 0.99
Calculate
VARS  STATISTICS  TEST
H :lower  1033.900763
I :upper  1168.899247
We can say with a 99%
confidence level that the
population mean score on the
SAT examination of the seniors
at a local high school is between
1033.9008 and 1168.8992.
THE
END
OF
SECTION 1