chapter1 notes - applied statistics

CHAPTER 1

PARAMETER ESTIMATION

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.1

PARAMETER ESTIMATION

12-3

INTRODUCTION

Parameter estimation is the first step in inferentialstatistics. In other words, it is the process of estimating thevalue of a parameter using information obtained from asample.

The process that acquires information from samples andThe process that acquires information from samples andused the information to make conclusions about populationsis called statistical inference.

In order to do statistical inference, we require the skills andknowledge of descriptive statistics, probability distributions,and sampling distributions. The process can be simply asin figure 1.

INTRODUCTION (cont..)

The objective of estimation is to determine theapproximate value of a population parameter on the basisof a sample statistic.

There are two approaches to parameter estimation whichare

3

arei) Point estimation

q Using point estimate will obtain value that iseither 100% accurate or 100% different from thetrue value

q Note that, true value = parameter value

ii) Interval estimation

Estimator is the statistic used to obtain the pointestimate.

Estimate is a specific value or range of values used toapproximate some population parameter.


4

approximate some population parameter.

Why we estimate? Can we get the exact value from thepopulation?

Sampling process

Samples are taken at random

census survey

Population Eg: All UUM

students

Part of population unit

eg: a number of UUM students

Collect Information/ data

Collect Information/data


5

PARAMETER

Population measurement

Sample measurement

estimate STATISTICS

Figure 1.1: Relationship between parameter and statistic

POINT ESTIMATION

A point estimate is a specific numerical value of aparameter or a single value (or point) used to approximatea population parameter.

6

A point estimator draws inferences about a population byestimating the value of an unknown parameter using asingle value or point.

POINT ESTIMATION (CONT…)

Table 1.1: Symbols for parameter and statistics Parameter Statistics/ estimator

mean, µ mean,

variance, variance,

7

variance, variance,

standard deviation, standard deviation,

proportion, proportion,

Table 1.2: Formulas for statistics Statistics/ estimator Formula

Sample mean

Sample variance


8

Sample standard deviation

Sample proportion

Characteristics of Good Estimator

The objective of each characteristic good estimator is toobtain an estimator with the sampling distribution meancentered to the parameter being estimated.


9

The characteristic include:

§ un-biasness§ Consistency§ relatively efficiency

An unbiased estimator of a population parameter is anestimator whose expected value is equal to that parameter.

Ø An estimator is an unbiased estimator for parameterif E( )=θ

ØE.g. the sample mean is an unbiased estimator of


10

ØE.g. the sample mean is an unbiased estimator ofthe population mean µ , since:

An unbiased estimator is said to be consistent if thedifference between the estimator and the parametergrows smaller as the sample size grows larger.

E.g. is a consistent estimator of because: x µ


11

That is, as n grows larger, the variance of growssmaller.

x

If there are two unbiased estimators of a parameter, theone whose variance is smaller is said to be relativelyefficient.

E.g. both sample median and mean are unbiasedestimators of the population mean. However, accordingto the variances

x


12

so we choose mean, since it is relatively efficientcompared to the sample median .

x x~

Example 1:

A sample of 10 frogs has been taken at random and theweight (in grams) for each of the frog was recorded andgiven as below:

250 230 200 210 195 225 200 230 240


13

250 230 200 210 195 225 200 230 240190

1. compute the point estimate for the mean weight of thefrogs

2. estimate the standard deviation for the weight of thefrogs

3. estimate the proportion of frogs that have weight notmore than 200 grams

Solution:

i. let represent the mean weight of the frogs

ii. the estimate of the standard deviation for the weight of the frogs is given by

14

ii. the estimate of the standard deviation for the weight of the frogs is given by

iii. let be the number of frogs with weight not more than 200 grams and be the point estimate for the proportion of frog with weight not more than 200 grams

Frogs with weight not more than 200 grams are; 200, 195, 200, 190 Then,

Thus;

Solution:

15

Thus;

Note: the answer for question i) and ii) can be found directly from your calculator using the mode (SD) function. Those who use calculator model Casio can refer to Appendix 1 for the complete procedure.

The age of 15 students who came to the recreationalclub during last weekend are as given below:

Example 2:

8 15 10 16 17

17 13 12 12 15

15 15 16 16 18

16

15 15 16 16 18

Calculate the point estimate of the:

i. average age of studentsii. variance of age of studentsiii. proportion of students with age more than 15

years old.

A research has been done to determine percentage of UUM’s staff living in Jitra, From a sample of 200 randomly chosen people, 88 of them are living in Jitra.

i. Obtain the point estimate for the percentage of

Example 3:

17

i. Obtain the point estimate for the percentage of UUM’s staff living in Jitra.

ii. Estimate the mean and the standard deviation of the proportion.

Interval Estimation

No matter how good is the point estimator is, we have toadmit that the point estimate can sometimes gives a valuewhich is 100% different from the true value.

Besides that, the point estimators don’t reflect the effectsof larger sample sizes.

18

Thus, it is recommended to use interval estimator toestimate population parameters, which is less precise butsafer.

An interval estimator draws inferences about a populationby estimating the value of an unknown parameter using aninterval.

The value of interval estimator is between lower andupper boundaries. Generally, we write the value as

Lower bound < population parameter < Upper bound

Interval Estimation (cont…)

19

If is the point estimate of parameter , then the interval estimate is given by

Where,

( )θ̂S is the standard deviation of the estimator k is the distribution of the parameter (distribution can be define based on Central Limit Theorem)

Once the interval estimate is obtained, we canconclude (with some ___% of certainty) that thepopulation parameter of interest is between some lowerand upper bounds.

In this section we will discuss the interval estimate for


20

In this section we will discuss the interval estimate forthe mean and the proportion and is summarize in Figure1.2.


21

Figure 1.2: Interval estimation

Interval estimation for mean

Generally, the interval estimator for one populationmean is given by

( ) ( )xSZxxSZx αα µ +<<−


22

Note: the Z distribution can be replace by t distributionif the condition to use Z distribution is not satisfied.

( ) ( )( )xSZxor

xSZxxSZx

2

22

α

αα µ

±

+<<−

To determine whether to use Z or t distribution, wehave to follow the Central Limit Theorem


23Figure 1.3: Central Limit Theorem

Characteristics of the Z Distribution

When the standard deviation of population isknown or the sample size taken is more than orequal to 30, the normal Z distribution can be used.


24

Figure 1.4: Condition to use normal Z distribution

Characteristics of the t Distribution

1. When the population standard deviation isunknown and the sample size is less than 30, thet distribution with degrees of freedom mustbe used instead of Z distribution.


25

be used instead of Z distribution.

2. The degrees of freedom are the number ofvalues that are free to vary after a samplestatistic has been computed.

The t distribution differs from the standard normaldistribution in the following ways.

i. The variance is greater than 1.

ii. The t distribution is actually a family of curves based


26

ii. The t distribution is actually a family of curves basedon the concept of degrees of freedom, which isrelated to sample size. As the sample size increases,the t distribution approaches the standard normaldistribution.


27

Figure 1.5: The Z Normal and t distribution


28

Figure 1.6: t distribution with different degrees of freedom.

When to use the z or t distribution?

Is population std. dev. σ known?

Use Z distribution no matter what the sample size is. Yes

No

* Variable are normally distributed when n<30


29

Is sample size, n > 30?

Use t distribution and s in the formula.

Use Z distribution and s in place of σ.

Yes

No

** variable are approximately normally distributed

Figure 1.7: Criteria for choosing Z or t distribution

Therefore; the confidence interval for a mean has 3formulas;

1. confidence interval for a mean with knownpopulation standard deviation

ZxZx σµσ +<<−


30

nZx

nZx σµσ αα

22+<<−

or

2. confidence interval for a mean with unknownpopulation standard deviation, sample sizemore than or equal to 30 .

nSZx

nSZx

22αα µ +<<−


31

or

3. confidence interval for a mean with unknownpopulation standard deviation, sample size lessthan 30 (n<30).

nStx

nStx nn 1,21,2 −− +<<− αα µ


32

ntx

ntx nn 1,21,2 −− +<<− αα µ

or

The graphical view of interval estimate:


33

Width of interval

LCL: UCL:

Figure 1.8: Graphical view of confidence interval

The probability of is called Confidence Level (ordegree of confidence).


( α−1 )

The is called significance level or the probability of Type I error will occur.

34

Confidence Level is the relative frequency of times theconfidence interval actually does contain the populationparameter, assuming that the estimation process isrepeated a large number of times.

There are four commonly used confidence levels…

Confidence Level

1- 0.90 0.10 0.05 1.6449


35

0.90 0.10 0.05 1.6449 0.95 0.05 0.025 1.9600 0.98 0.02 0.01 2.3323 0.99 0.01 0.005 2.5758

There are the critical values for t distribution.

Confidence Level (1-

df

0.90 3 0.10 0.05 2.3534

0.95 5 0.05 0.025 2.5706

36

0.95 5 0.05 0.025 2.5706

0.98 7 0.02 0.01 2.9980

0.99 9 0.01 0.005 3.2498

Example 4:

A computer company samples demand during lead time over 25 time periods:

235 374 309 499 253

421 361 514 462 369

394 439 348 344 330

37

261 374 302 466 535

386 316 296 332 334

It is known that the standard deviation of demand over lead time is 75 computers. Estimate the mean demand over lead time with 95% confidence level in order to set inventory levels.

Example 5:

The president of a large university wishes to estimate the average age of the students presently enrolled. From past studies, the standard deviation is known to be 2 years. A sample of 50 students is selected, and the mean is found to be 23.2 years. Find the 95% confidence interval of the population mean.

38

confidence interval of the population mean.

Example 6:

A survey of 30 adults found that the mean age of a person’s primary vehicle is 5.6 years. Assuming the standard deviation of the population is 0.8 year; find the 99% confidence interval of the population mean.

Example 7:

A cereal company selects twenty five 12-ounce boxes of corn flakes every 10 minutes and weighs the boxes. Suppose the weights have a normal distribution with variance is 0.04 ounces. One such sample yields calculate the 90% confidence interval of the population mean.

39

interval of the population mean.

Example 8:

Ten randomly selected automobiles were stopped, and tread depth of the right front tire was measured. The mean was 0.32 inch and the standard deviation was 0.08 inch. Find the 95% confidence interval of the mean depth. Assume that the variable is approximately normally distributed.

Example 9:

The average production of peanuts in the state of Virginia is 3000 pounds per acre. A new plant food has been developed and is tested on 60 individual plots of land. The mean yield with the new plant food is 3120 pounds of peanuts per acre with a standard deviation of 578 pounds. Find the 95% confidence interval for the

40

578 pounds. Find the 95% confidence interval for the mean amount of rainfall during the summer months for the northeast part of the United States. Interpret the interval.

Example 10:

The following daily highs were recorded in the city of Chicago on 20 randomly selected December days.

32 21 25 25 31 27 22 44 39 18 49 32 34 36 38 40 30 28 36 38

41

Find a confidence interval for the mean daily high temperature, should we use t or Z distribution? Explain.

Interval estimation for proportion

Whenever the information is given in percentage,proportion or number of success for a specific event,then the problem being investigated has somethingto do with proportion.

The procedures for drawing inferences about

42

The procedures for drawing inferences about proportion are involved the nominal and sometimes ordinal scale (i.e categorical data).

Example of categorical data: gender (male and female), job satisfaction (satisfied and unsatisfied), opinion (poor and good), attendance (absent, present), examination result (pass and failed), etc.


The point estimate for the proportion is given by

nxp=ˆ = symbol for the sample proportion

43

Where; = number of sample units that possess the characteristics of interest = sample size.

Knowing that:

v Sample size n is bigv Both and are greater than or equal to 5


then, the formula to estimate the confidence interval for a

44

then, the formula to estimate the confidence interval for aproportion is given by

( ) ( )

( ) ( )

nqpZppn

qpZp

nppZppn

ppZp

pSZpppSZp

ˆˆˆˆˆˆ

ˆ1ˆˆˆ1ˆˆ

ˆˆˆˆ

22

22

22

αα

αα

αα

+<<−=

−+<<−−=

+<<−

Where is, 1ˆˆ =+qp

Example 11:

A recent study of 100 people in Miami found 27 were obese. What is the proportion of individual living in Miami who are obese? Obtain the 95% confidence interval of the proportion of individual living in Miami who are obese and interpret.

45

Example 12:

A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone, message, faxes and etc. Estimate with 90% confidence level, the percentage of the workers who are not interrupted three or more times an hour.

Example 13:

In a random sample of 500 observations, we found the proportion of successes to be 48%. Estimate with 95% confidence the population proportion of successes.

Example 14:

46

A random sample of 1500 pine trees was tested for traces of the Bark Beetle infestation. The result showed that 153 of the trees showed such traces. Assuming the data is approximately normally distributed, calculate the point estimator of the proportion of pine trees has been infested, and find a 95% confidence interval for the proportion of pine trees have been infested

Example 15:

The quality control manager at Ameen Company claims that the production of model A telephone ‘to be out of control’ when the overall rate of defects exceed 4%. The test for a random sample of 150 telephones revealed that 9 of them are defective. Construct a 98% confidence interval for the proportion of telephone’s defect.

Example 16:

47

Example 16:

A statistics practitioner working for a major league baseball wants to supply radio and television commentators with interesting statistics. He observed several hundred games and counted the number of time runner on first base attempted to steal second base. He found there were 373 such events of which 259 were successful. Estimate with 95% confidence the population proportion of all attempted theft of second base that is successful.

Sample size

Sample size for Mean

Recall back: the interval formula for estimating populationmean is

nZx

nZx σµσ

αα +<<−

48

43421)(EError

22 nZx

nZx µ αα +<<−

note that, maximum error;

nZE σα

2=

Sample size (cont…)

using the maximum Error formula, we then can calculatethe value of sample size, n which is given by

Sample size for Mean

22 σ

49

22

2

E

Zn

=ασ

Sample size for proportion

Recall back: interval formula for estimating populationproportion is

( ) ( )ˆˆˆˆˆˆ p1pZppp1pZp −+<<−−


50

( ) ( )44 344 21

)(

ˆˆˆˆˆˆ

EError

22 np1pZppn

p1pZp −+<<−− αα

note that, maximum error; ( )

np1pZE

2

ˆˆ −= α

using the maximum Error formula, we then can calculatethe value of sample size, n which is given by

Sample size for proportion


51

the value of sample size, n which is given by

( )

2

2

22

2

2

E

Zqp

E

Zp1pn

=

−

=αα ˆˆˆˆ

Conclusion

In conclusion, the width of the confidence interval estimate isaffected by

ü the population standard deviation,

ü the confidence level

52

ü the confidence level

ü the sample size,

The width of the confidence interval is a function of the confidence level, the population standard deviation, and the sample size…

nSZx

nSZx

nSZx

2

22

α

αα µ

±=

+<<−

A larger confidence level produces a wider confidence interval:

Conclusion

53

A larger confidence level produces a wider confidence interval:

Figure 1.9: relationship between width and confidence level

Larger values of standard deviation produce widerconfidence intervals

Conclusion

54

Increasing the sample size decreases the width of the confidenceinterval while the confidence level can remain unchanged.

Figure 1.10: relationship between width and standard deviations

INTERVAL ESTIMATION FOR TWO MEANS

Previously… we have discussed the techniques to estimateparameters for one population mean

Now, consider this parameter but with two populations. With

55

Now, consider this parameter but with two populations. Withtwo populations, our interest will now be on the differencebetween two population means.

Sample, size: n1

Population 1

Parameters: and Statistics: and


56

Figure 1.11: Independent Population and Samples

Sample, size: n2

Population 2

Parameters: and

Parameters: and

There are two different types of sample which are:

Dependent Samples also called related (or paired) samples occurwhen the response of the nth person in the second sample is partly afunction of the response of the nth person in the first sample. Thereare two (2) common forms of sample dependency,


57

übefore-after and other studies in which the same people are surveyed at different points in time including panel studies.

ümatched-pairs studies in which similar people are surveyed at different points in time.

Independent Samples are samples that are completely unrelatedto one another.

Interval estimation for difference of two independent means

In order to test and estimate the difference between twomeans, we draw random samples from each of twopopulations. Initially, we will consider independent samples,that is, samples that are completely unrelated to one another.


58

that is, samples that are completely unrelated to one another.

Statistics used is or

Two assumptions need to be fulfilled in order to determinethe difference between two independent means:

q The samples must be independent of each other;

Interval estimation for difference of two independent means


59

q The samples must be independent of each other;that is, there can be no relationship between thesubjects in each sample.

q The populations from which the samples wereobtained must be normally distributed.

There are four (4) different formulas to estimate theconfidence level for the difference between twoindependent means, which are:

Interval estimation for difference of two independent means(cont..)

i. Confidence interval when both population variance(or standard deviation) are known

60

(or standard deviation) are known

ii. Confidence interval when both population variance(or standard deviation) are unknown but both samplesizes are more or equal to 30


61

iii. Confidence interval when both population variance (orstandard deviation) are unknown, any one or bothsample sizes less than 30 and both populationvariances are assume equal

iv. Confidence interval when both population variance (orstandard deviation) are unknown, any one or both samplesizes less than 30 and both population variances areassume unequal


62

As in interval estimator for one mean, same situation shouldbe consider in deciding the formula to use to determine thedifference between two means

Are both and

known?

Are both n1 & n2 > 30?

Use t values

Use zα/2 values no matter what the sample size is.

Use zα/2 values and s in place of σ.

Yes

Yes

No

No

* Variable must be normally distributed when n<30

Figure 1.12: Flow diagram for choosing the correct distribution

63

Use tα/2 values and s in the formula.

** variable must be approximately normally distributed

Conduct equal variances t-test. Is ?

No

Yes

Use tα/2 values with pooled variance estimator,

Use tα/2 values with

Are both and

known?

Are both n1 & n2 > 30?

Yes

Yes

No

No

* Variable must be normally distributed when n<30

Figure 1.13: Flow diagrams for choosing the correct confidence interval formula

64

Use tα/2 values and s in the formula.

** variable must be approximately normally distributed

Conduct equal variances t-test. Is ?

No

Yes

Example 17:

Two random samples of 40 students were drawn independently from two normal populations. The following statistics regarding their scores in a final exam were obtained;

65

Construct a 95% confidence interval for the difference between the means.

Solution The populations’ standard deviations are unknown. However, since both sample sizes are large enough (both ), according to Central Limit Theorem, the means follow Normal distribution.

66

The 95% confidence interval for the difference between the means is

Example 18:

A random sample of 22 male customers who shopped at this supermarket showed that they spent an average of RM80 with standard deviation of RM17.50. While a random sample of 20 female customers who shopped at the same supermarket showed that they spent an average of RM96 with standard deviation RM14.40.

67

average of RM96 with standard deviation RM14.40. Assume that the amount spent at this supermarket by all the male and female customers are normally distributed with equal but unknown standard deviation.

Construct a 99% confidence interval for the difference between the mean amount spent by all male and all female customers at this supermarket and interpret the interval.

Example 19:

Because of the rising costs of industrial accidents, many chemical, mining, and manufacturing firms have instituted safety courses. Employees are encouraged to take these courses designed to heighten safety awareness. A company is trying to decide which one of two courses to institute. To help make a decision eight employees take Course 1 and another eight take Course 2. Each employee takes a test,

68

another eight take Course 2. Each employee takes a test, which is graded out of a possible 25. The safety test results are shown below. Assume that the scores are normally distributed. Construct a 90% confidence interval for different of mean.

Course 1 14 21 17 14 17 19 20 16

Course 2 20 18 22 15 23 21 19 15

Example 20:

Random samples of children sent to kindergarten aged 4 to 6 years in Bandar A and B were taken to find the number of hours spend for outdoor activities in the kindergarten daily. A sample of 321 children in Bandar B and 94 children in Bandar A give the mean of 3.01 hours and 2.88 hours, respectively. From past studies the

69

and 2.88 hours, respectively. From past studies the population standard deviation for the children in Bandar B is assumed to be 1.09, while the population standard deviation for the children in Bandar A is 1.01. Find a 95% confidence interval for the difference between the two population means.

Interval estimation for the difference between twoproportions

We will now look at procedures for drawing inferences about thedifference between populations whose data are nominal (i.e.categorical).

70

With nominal data, we can calculate the proportions ofoccurrences of each type of outcome. Thus, the parameter to beestimated in this section is the difference between two populationproportions: p1–p2.

Assumptions for doing Inferences about two proportions

i. We have proportions from two independent simplerandom samples.

ii. In order to use Normal Z distribution, for both

Interval estimation for the difference between twoproportions (cont…)

71

ii. In order to use Normal Z distribution, for bothsample the conditions

( ) ( ) 5ˆ1 5ˆ1 ,5ˆ ,5ˆ 22112211 ≥−≥−≥≥ pnandpnpnpn

must be satisfied.


To draw inferences about the parameter , we takesamples of population, calculate the sample proportions andlook at their difference.

1

11ˆ n

xp = and

2

22ˆ

nx

p =

72

1n 2n

( )21 ˆˆ pp − is an unbiased estimator for ( )21 pp − .

The confidence interval estimator for (p1–p2) is given by:


1 1 2 2 1 1 2 21 2 1 2 1 2

1 2 1 22 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ( ) ( )

pq p q pq p qp p z p p p p z

n n n nα α

− − + ≤ − ≤ − + +

( ) ˆˆˆˆ qpqp

73

( )2

22

1

11

221

ˆˆˆˆˆˆ

nqp

nqp

zpp +±− α

Example 21:

A Consumer Packaged Goods (CPG) company has testing the marketing of two new versions of soap packaging. Version one (bright colors) was distributed in one supermarket, while version two (simple colors) was in another. Construct a 95% confidence interval for the difference between the two proportions of successes of

74

difference between the two proportions of successes of packaged soap sales.

Example 22:

A random sample of 500 respondents was selected in a large city to determine information concerning consumer behavior. Among the questions asked was, “Do you enjoy shopping?” Of 240 male respondents, 136 answered yes. Of 260 female respondents, 224 answered yes. Construct a 95% confidence interval

75

answered yes. Construct a 95% confidence interval estimate of the difference between the proportion of males and females who enjoy shopping.

SPSS NOTES FOR OBTAINING THE CONFIDENCE INTERVAL OF MEAN

Step 1 : Select Analyze Menu → Select Descriptive Statistics

76

Step 2 : Click on Explore → Select the appropriate variable Step 3 : Click on the button into Dependent List box

List of Variable(s)

77

Make your choice

Make your choice

Variable(s)

Step 4 : Click on Statistics → Select the appropriate statistics, eg: Descriptive

You can change the degree of confidence

(Usually use 90%

78

Step 5 : Then, click on Continue→ Click on OK

(Usually use 90% and above)

A random sample of 10 university students was surveyed to determine the amount of time spent weekly using a personal computer. The times are: 13, 14, 5, 6, 8, 10, 7, 12, 15, and 3. If the times are normally distributed with a standard deviation of 5.2

Example

79

hours, estimate with 90% confidence the mean weekly time spent using a personal computer by all university students.

Descriptives

9.30 1.3006.92

11.68

9.339.00

16.900

MeanLower BoundUpper Bound

90% ConfidenceInterval for Mean

5% Trimmed MeanMedianVariance

timesStatistic Std. Error

80

16.9004.111

315128

-.040 .687-1.396 1.334

VarianceStd. DeviationMinimumMaximumRangeInterquartile RangeSkewnessKurtosis

At 90% confidence level, the mean weekly time spent using a personal computer by all university students is between 6.92 and 11.68.

SUMMARY

A point estimator is a good estimator if it has the qualities ofgood estimator which are un-biasness, consistent andrelatively efficient.

Unlike point estimation, interval estimation involves aninterval constructed around the point estimate with a

81

interval constructed around the point estimate with aprobability of .

To construct interval, information regarding the samplingdistribution of the statistics is important.

.

The Central Limit Theorem enables us to determine thesampling distribution for the sample statistics based on sampleinformation of the sample size and knowledge of thepopulation variance.

SUMMARY (CONT…)

82

If we want to know whether the population means/ proportionequals to certain value, k and the confidence interval formeans/ proportion includes the k value, we can conclude thatthere is evidence to conclude that the mean/ proportion equalsto k, at a given level of confidence.

SUMMARY (CONT…)

If the confidence interval for the difference between twomeans/proportions includes 0 we can say that there is nosignificant difference (failed to reject) between the means ofthe two populations, at a given level of confidence.

83

END OF CHAPTER 1

chapter1 notes - applied statistics

Documents