Chapter 7 Estimation

Chapter 7Chapter 7EstimationEstimation

Point EstimatePoint Estimate

an estimate of a population parameter given by a

single number

is used as a point estimate for μ

s is used as a point estimate for σ.

Examples of Examples of Point EstimatesPoint Estimates

x

Error of EstimateError of Estimate

the magnitude of the difference between the point estimate and the true parameter value

The error of estimate The error of estimate using as a point using as a point estimate for estimate for μμ is: is:

x −μ

x

Confidence LevelConfidence 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 Critical Value for a Confidence Level, cConfidence 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 Critical Value for a Confidence Level, cConfidence Level, c

P(– zc < z < zc ) = c

-zc zc

Find zFind z0.900.90 such that 90% of the such that 90% of the

area under the normal curve lies area under the normal curve lies between zbetween z-0.90 -0.90 and zand z0.900.90

P(-z0.90 < z < z0.90 ) = 0.90

-z0.90 z0.90

.90.90

Find zFind z0.900.90 such that 90% of the such that 90% of the

area under the normal curve lies area under the normal curve lies between zbetween z-0.90 -0.90 and zand z0.900.90

P(0< z < z0.90 ) = 0.90/2 = 0.4500

-z0.90 z0.90

.4500.4500

Find zFind z0.900.90 such that 90% of the such that 90% of the

area under the normal curve lies area under the normal curve lies between zbetween z-0.90 -0.90 and zand z0.900.90

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 Common Levels of Confidence and Their Corresponding and Their Corresponding

Critical ValuesCritical ValuesLevel 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 IntervalConfidence Intervalfor the Mean of Large for the Mean of Large

Samples (n Samples (n ≥≥ 30) 30)x −E < μ < x + E

where x =SampleMean

E =zcσn

  if thepopulationstandard

                 deviationsisknown

Confidence Interval for the Mean Confidence Interval for the Mean of Large Samples (n of Large Samples (n ≥≥ 30) 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 + zcsn

 units.

Create a 95% confidence Create a 95% confidence interval for the mean interval for the mean driving time between driving time between

Philadelphia and Boston. 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 hours

s =0.9 hours

n =64

Key InformationKey Information

c = 95%, so zc = 1.9600

95% Confidence Interval:95% Confidence Interval:

We can say with 95% a confidence level that the We can say with 95% a confidence level that the population mean driving time from Philadelphia to population mean driving time from Philadelphia to

Boston is between 4.9795 and 5.4205 hours.Boston is between 4.9795 and 5.4205 hours.

5.2 −1.96000.964

< μ < 5.2 +1.96000.964

5.2 −1.764

8< μ < 5.2 +

1.7648

5.2 −0.2205 < μ < 5.2 + 0.2205

4.9795 < μ < 5.4205

95% Confidence Interval:95% Confidence Interval:

We can say with a 95% We can say with a 95% confidence level that the confidence level that the population mean driving population mean driving time from Philadelphia to time from Philadelphia to Boston is between 4.9795 Boston is between 4.9795

and 5.4205 hours.and 5.4205 hours.

STAT

TESTS

# 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 Size

n=zcσE

⎛⎝⎜

⎞⎠⎟

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 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.

THETHEENDEND

OF THEOF THEPRESENTATIOPRESENTATIO

NN

Answers Answers to the to the

Sample Sample QuestionsQuestions

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+ zcσn 

3.62− 1.6449( )0.84200

⎛⎝⎜

⎞⎠⎟< μ < 3.62 + 1.6449( )

0.84200

⎛⎝⎜

⎞⎠⎟  

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+ zcσn 

3.62− 1.9600( )0.84200

⎛⎝⎜

⎞⎠⎟< μ < 3.62 + 1.9600( )

0.84200

⎛⎝⎜

⎞⎠⎟  

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.

ZINTERVAL

3.5223, 3.7177( )x=3.62n=200

VARS−STATISTICS−TESTH : lower =3.522300679I :upper =3.717699321

ZINTERVAL

Input :Stats

σ :0.84x : 3.62n:200C−Level :0.90Calculate

1. Part c - Calculator check for part a

ZINTERVAL

3.5036, 3.7364( )x=3.62n=200

VARS−STATISTICS−TESTH : lower =3.503584079I :upper =3.736415921

ZINTERVAL

Input :Stats

σ :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 11501280 1100 1070 890 930 1520 810 1090 1310 10301190 1370 1200 990 1560 810 940 1010 1140 10601060 1250 1240 1130 1170 1080 1210 970 810 9201080 1160 940 1050 1110 1300 1230 790 1050 1240

ZINTERVAL

1033.9, 1169.9( )x=1101.4n=50

VARS−STATISTICS−TESTH : lower =1033.900763I :upper =1168.899247

ZINTERVAL

Input :Data

σ :185.29634290476List :L1

Freq: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.

THETHEENDENDOFOF

SECTION 1SECTION 1