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Chapter 7 Estimation
Chapter 7Estimation
Point Estimate
an estimate of a population parameter given by a
single number
ü is used as a point estimate for µ.
ü s is used as a point estimate for σ.
Examples of Point Estimates
x
Error of Estimatethe magnitude of the difference between the point estimate and the true parameter value
The error of estimate using as a point estimate for µ is:
x − µ
x
Confidence LevelA 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, cthe 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
P(-z0.90 < z < z0.90 ) = 0.90-z0.90 z0.90
.90
Find z0.90 such that 90% of the area under the normal curve lies
between z-0.90 and z0.90
P(0< z < z0.90 ) = 0.90/2 = 0.4500
-z0.90 z0.90
.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 ValuesLevel of Confidence, c Critical Value, zc
0.70 or 70% 1.03640.75 or 75% 1.15030.80 or 80% 1.28160.85 or 85% 1.43950.90 or 90% 1.64490.95 or 95% 1.96000.98 or 98% 2.32630.99 or 99% 2.5758
0.999 or 99.9% 3.2905
Confidence Intervalfor the Mean of Large
Samples (n ≥ 30)x − E < µ < x + E
where x = Sample Mean
E = zcσn 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
between x − zcsn and x + zc
sn
units.
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.
x = 5.2 hourss = 0.9 hours
n = 64
Key Information
c = 95%, so zc = 1.9600
95% Confidence Interval:
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.
5.2 − 1.9600 0.964
< µ < 5.2 + 1.9600 0.964
5.2 − 1.7648
< µ < 5.2 + 1.7648
5.2 − 0.2205 < µ < 5.2 + 0.2205
4.9795 < µ < 5.4205
95% Confidence Interval:
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.
STATTESTS
#7⇒ ZIntervalInpt : Statsσ = 0.9x = 5.2n = 64
C − Level : 0.95Calculate
4.9795, 5.4205( )
Calculator Computation
When estimating the mean, how large a sample must be used in order to assure a given level of
confidence?
Use the formula:
n = zcσE
⎛⎝⎜
⎞⎠⎟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 minutesor
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.
σ = 0.8
Minimum Required Sample Sizen = zcσ
E⎛⎝⎜
⎞⎠⎟2
n = 2.5758( ) 0.8( )0.25
⎛⎝⎜
⎞⎠⎟
2
n = 2.06060.25
⎛⎝⎜
⎞⎠⎟2
n = 8.2426( )2
n = 67.9398n = 68
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 have a 99% confidence level for the population mean time it takes to drive from Philadelphia to Boston with an error of 0.25 hours.
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 − zcσn< µ < x + zc
σn
3.62 − 1.6449( ) 0.84200
⎛⎝⎜
⎞⎠⎟< µ < 3.62 + 1.6449( ) 0.84
200⎛⎝⎜
⎞⎠⎟
3.62 − 1.381714.1421
< µ < 3.62 + 1.381714.1421
3.62 − 0.0977 < µ < 3.62 + 0.09773.5223< µ < 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 − zcσn< µ < x + zc
σn
3.62 − 1.9600( ) 0.84200
⎛⎝⎜
⎞⎠⎟< µ < 3.62 + 1.9600( ) 0.84
200⎛⎝⎜
⎞⎠⎟
3.62 − 1.646414.1421
< µ < 3.62 + 1.646414.1421
3.62 − 0.1164 < µ < 3.62 + 0.11643.5036 < µ < 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.
ZINTERVAL3.5223, 3.7177( )x = 3.62n = 200
VARS − STATISTICS −TESTH : lower = 3.522300679I :upper = 3.717699321
ZINTERVALInput :Statsσ :0.84x : 3.62n :200C − Level :0.90Calculate
1. Part c - Calculator check for part a
ZINTERVAL3.5036, 3.7364( )x = 3.62n = 200
VARS − STATISTICS −TESTH : lower = 3.503584079I :upper = 3.736415921
ZINTERVALInput :Statsσ :0.84x : 3.62n :200C − Level :0.95Calculate
1. Part c - Calculator check for part b
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?
n = zcσE
⎛⎝⎜
⎞⎠⎟2
n = 1.9600( ) 0.84( )0.05
⎛⎝⎜
⎞⎠⎟
2
n = 1.64640.05
⎛⎝⎜
⎞⎠⎟2
n = 32.928( )2
n = 1084.2532n = 1085
We would need to acquire 1,085 AP test 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 1380 950 870 1030 1220 750 1410 1150 1280 1100 1070 890 930 1520 810 1090 1310 1030 1190 1370 1200 990 1560 810 940 1010 1140 1060 1060 1250 1240 1130 1170 1080 1210 970 810 920 1080 1160 940 1050 1110 1300 1230 790 1050 1240
ZINTERVAL1033.9, 1169.9( )x = 1101.4n = 50
VARS − STATISTICS −TESTH : lower = 1033.900763I :upper = 1168.899247
ZINTERVALInput :Dataσ :185.29634290476List :L1Freq :1C − Level :0.99Calculate
Remember that you need perform a 1-VAR-STATS calculation on the data first to get the value for the sample standard deviation.
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
