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Economics 173Business Statistics

Lectures 3 & 4

Summer, 2001

Professor J. Petry

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Introduction to Estimation

Introduction to Estimation

Chapter 9

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9.1 Introduction

• Statistical inference is the process by which we acquire information about populations from samples.

• There are two procedures for making inferences:– Estimation.– Hypotheses testing.

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9.2 Concepts of Estimation

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

• There are two types of estimators– Point Estimator– Interval estimator

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– A point estimator draws inference about a population by estimating the value of an unknown parameter using a single value or a point.

Point estimator

Point Estimator

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Sample distribution

Point estimator

Population distributionParameter

?

• Point Estimator– A point estimator draws inference about a population

by estimating the value of an unknown parameter using a single value or a point.

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– An interval estimator draws inferences about a population by estimating the value of an unknown parameter using an interval.

– The interval estimator is affected by the sample size.

Interval estimator

Population distribution

Sample distribution

Parameter

Interval Estimator

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9.3 Estimating the Population Mean when the Population Standard Deviation is Known

• How is an interval estimator produced from a sampling distribution?– To estimate , a sample of size n is drawn from the

population, and its mean is calculated.– Under certain conditions, is normally distributed (or

approximately normally distributed.), thus

nx

Z

xx

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– We know that 1)( 22n

zxn

zP

1)n

zxn

zx(P 22

– This leads to the relationship

1 - of all the values of obtained in repeatedsampling from this distribution, construct an interval

that includes (covers) the expected value of thepopulation.

1 - of all the values of obtained in repeatedsampling from this distribution, construct an interval

that includes (covers) the expected value of thepopulation.

x

nzx,

nzx 22

10

x

nz2 2

nzx 2

n

zx 2

nzx,

nzx 22

Lower confidence limit Upper confidence limit

1 -

Confidence level

See simulation resultsdemonstrating this point

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0

50

100

150

Not all the confidence intervals cover the real expected value of 100.

1000

LCL

UCL

The selected confidence level is 90%,and 10 out of 100 intervals do not coverthe real

• The confidence interval are correct most, but not all, of the time.

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• Four commonly used confidence levelsConfidence

level 0.90 0.10 0.05 1.6450.95 0.05 0.025 1.960.98 0.02 0.01 2.330.99 0.01 0.005 2.575

Confidence level 0.90 0.10 0.05 1.6450.95 0.05 0.025 1.960.98 0.02 0.01 2.330.99 0.01 0.005 2.575

z

• Estimate the mean value of the distribution resulting from the throw of a fair die. It is known that = 1.71. Use 90% confidence level, and 100 repeated throws of the die.

• Solution: The confidence interval is

nzx 2

28.x10071.1

645.1x

The mean values obtained in repeated draws of samples of size 100 result in interval estimators of the form [sample mean - .28, Sample mean + .28]90% of which cover the real mean of the distribution.

The mean values obtained in repeated draws of samples of size 100 result in interval estimators of the form [sample mean - .28, Sample mean + .28]90% of which cover the real mean of the distribution.

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• The width of the 90% confidence interval = 2(.28) = .56 The width of the 95% confidence interval = 2(.34) = .68• Because the 95% confidence interval is wider, it is more likely to include the value of

• The width of the 90% confidence interval = 2(.28) = .56 The width of the 95% confidence interval = 2(.34) = .68• Because the 95% confidence interval is wider, it is more likely to include the value of

• Recalculate the confidence interval for 95% confidence level. • Solution:

n

zx 2 34.x10071.1

96.1x

34.x 34.x

.95

.90

28.x 28.x

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• Example 9.1– The number and the types of television programs and

commercials targeted at children is affected by the amount of time children watch TV.

– A survey was conducted among 100 North American children, in which they were asked to record the number of hours they watched TV per week.

– The population standard deviation of TV watch was known to be = 8.0

– Estimate the watch time with 95% confidence level.

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– The parameter to be estimated is the mean time of TV watch per week per child (of all American Children).

– We need to compute the interval estimator for – From the data provided in file XM09-01, the sample

mean is

761.28,621.2557.1191.27100

0.896.1191.27

100

0.8z191.27

nzx 025.2

.191.27x Since 1 - =.95, = .05. Thus /2 = .025. Z.025 = 1.96

• Solution

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• Interpreting the interval estimate– It is wrongwrong to state that the interval estimator is an

interval for which there is 1 - chance that the population mean lies between the LCL and the UCL.

– This is so because the is a parameter, not a random variable.

• Interpreting the interval estimate– It is wrongwrong to state that the interval estimator is an

interval for which there is 1 - chance that the population mean lies between the LCL and the UCL.

– This is so because the is a parameter, not a random variable.

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– LCL, UCL and the sample mean are the random variables, is a parameter, NOT a random variable.

– Thus, it is correct to state that there is 1 - chance that LCL will be less than and UCL will be greater than

nzxLCL 2

n

zxUCL 2

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• Example 9.2– To lower inventory costs, the Doll Computer

company wants to employ an inventory model.– Lead time demand is normally distributed with

standard deviation of 50 computers.– It is required to know the mean in order to calculate

optimum inventory levels.– Estimate the mean demand during lead time with

95% confidence.

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• Solution– The parameter to be estimated is

The interval estimator is

• Demand during 60 lead times is recorded514, 525, …., 476.

• The sample mean is calculated

• The 95% confidence interval is:

nzx,

nzx 22

75.49960985,29

n

xx i

4.512,1.48765.1275.49960

5096.175.499

nzx 2

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– Wide interval estimator provides little information.Where is

???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? Ahaaa!

Here is a much smaller interval.If the confidence level remains unchanged, the smaller interval provides more meaningful information.

Here is a much smaller interval.If the confidence level remains unchanged, the smaller interval provides more meaningful information.

Information and the Width of the Interval

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The width of the interval estimate is a function of:• the population standard deviation• the confidence level• the sample size.

The width of the interval estimate is a function of:• the population standard deviation• the confidence level• the sample size.

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90%

Confidence level

To maintain a certain level of confidence,changing to a larger standard deviation requires a longer confidence interval.

To maintain a certain level of confidence,changing to a larger standard deviation requires a longer confidence interval.

n)645.1(2

nz2 05.

/2 = .05/2 = .05

n

5.1)645.1(2

n

5.1z2 05.

Suppose the standard deviationhas increased by 50%.

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90%

Confidence level95%

Let us increase the confidence level from 90% to 95%.

n)96.1(2

nz2 025.

/2 = 2.5%/2 = 2.5%

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

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

There is an inverserelationship between the width of the intervaland the sample size

Increasing the confidence level produces a wider interval

Increasing the confidence level produces a wider interval

n)645.1(2

nz2 05.

/2 = 5%/2 = 5%

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9.4 Selecting the Sample size

• We can control the width of the interval estimate by changing the sample size.

• Thus, we determine the interval width first, and derive the required sample size.

• The phrase “estimate the mean to within W units”, translates to an interval estimate of the form

wx wx

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• The required sample size to estimate the mean is

• Example 9.3– To estimate the amount of lumber that can be harvested

in a tract of land, the mean diameter of trees in the tract must be estimated to within one inch with 99% confidence.

– What sample size should be taken? (assume diameters are normally distributed with = 6 inches.

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w

zn

22

w

zn

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• Solution– The estimate accuracy is +/-1 inch. That is w = 1.– The confidence level 99% leads to = .01, thus z/2 =

z.005 = 2.575.– We compute 239

1)6(575.2

w

zn

222

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Introduction to Hypothesis Testing

Introduction to Hypothesis Testing

Chapter 10

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10.1 Introduction

• The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter.

• Examples– Is there statistical evidence in a random sample of potential

customers, that support the hypothesis that more than p% of the potential customers will purchase a new products?

– Is a new drug effective in curing a certain disease? A sample of patient is randomly selected. Half of them are given the drug where half are given a placebo. The improvement in the patients conditions is then measured and compared.

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10.2 Concept of hypothesis testing

• The critical concepts of hypothesis testing.– There are two hypotheses (about a population parameter(s))

• H0 - the null hypothesis [ for example = 5]

• H1 - the alternative hypothesis [ > 5] This is what you want to prove

– Assume the null hypothesis is true.

• Build a statistic related to the parameter hypothesized.

• Pose the question: How probable is it to obtain a statistic value at least as extreme as the one observed from the sample? = 5 x

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• Continued– Make one of the following two decisions (based on the test):

• Reject the null hypothesis in favor of the alternative hypothesis.• Do not reject the null hypothesis in favor of the alternative hypothesis.

– Two types of errors are possible when making the decision whether to reject H0

• Type I error - reject H0 when it is true.

• Type II error - do not reject H0 when it is false.

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10.3 Testing the Population Mean When the Population Standard Deviation is Known

• Example 10.1– A new billing system for a department store will be cost-

effective only if the mean monthly account is more than $170.

– A sample of 400 monthly accounts has a mean of $178.– If the account are approximately normally distributed with

= $65, can we conclude that the new system will be cost effective?

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• Solution– The population of interest is the credit accounts at

the store.– We want to show that the mean account for all

customers is greater than $170.

H1 : > 170– The null hypothesis must specify a single value of

the parameter

H0 : = 170

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Is a sample mean of 178 sufficiently greater than 170 to infer that the population mean is greater than 170?

178

If is really equal to 170, then . The distribution of the sample mean should look like this.

170x

Is it likely to have under the null hypothesis (= 170)?178x

170x

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The rejection region is a range of values such that if the test statistic falls into that range, the null hypothesis is rejected in favor of the alternative hypothesis.

The rejection region is a range of values such that if the test statistic falls into that range, the null hypothesis is rejected in favor of the alternative hypothesis.

Define the value of that is just large enough to reject the null hypothesis as . The rejection region is

xLx

Lxx Lxx

The rejection region method

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Do no reject the null hypothesis

Reject the null hypothesis

Lxx The Rejection region is:

Lxx LxLxx

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= P(commit a type I error) = P(reject H0 given that H0 is true)

Lx Reject the null hypothesishere

Lxx The Rejection region is:

170x x

= P( given that H0 is true)Lxx

)ZZ(P

40065

170xz L

37.34.175

400

65645.1170x

Then

.645.1z,05.0selectweIf

.400

65z170x

L

05.

L

Lxx The Rejection region is:

40065

170xz L

= 0.05

170x Lx

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34.175xL

Lxx The rejection region is:

34.175xifhypothesisnullthejectRe

34.175xifhypothesisnullthejectRe

= 0.05

170x 178

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence in the sample to reject H0 in favor of H1, at 5% significance level.

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence in the sample to reject H0 in favor of H1, at 5% significance level.

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– Instead of using the statistic , we can use the standardized value z.

– Then, the rejection region becomes

x

n

xz

zz One tail test

The standardized test statistic

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• Example 10.1 - continued– We redo this example using the standardized test

statistic.H0: = 170

H1: > 170– Test statistic:

– Rejection region: z > z.051.645.– Conclusion: Since 2.46 > 1.645, reject the null

hypothesis in favor of the alternative hypothesis.

46.240065

170178

n

xz

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– The p - value provides information about the amount of statistical evidence that supports the alternative hypothesis.

– The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed, given that the null hypothesis is true.

– Let us demonstrate the concept on example 10.1

P-value method

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0069.)4615.2z(P

)40065

170178z(P

)178x(P

170x 178x

The probability of observing a test statistic at least as extreme as 178, given that the null hypothesis is true is:

The p-value

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• Interpreting the p-value– Because the probability that the sample mean will

assume a value of more than 178 when = 170 is so small (.0069), there are reasons to believe that > 170.

178x

170:H x0 170:H x1

…it becomes more probable under H1, when 170x

Note how the event is rare under H0

when but...178x

,170x

We can conclude that the smaller the p-value the more statistical evidence exists to support thealternative hypothesis.

We can conclude that the smaller the p-value the more statistical evidence exists to support thealternative hypothesis.

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• Describing the p-value– If the p-value is less than 1%, there is overwhelming

evidence that support the alternative hypothesis.– If the p-value is between 1% and 5%, there is a strong

evidence that supports the alternative hypothesis.– If the p-value is between 5% and 10% there is a weak

evidence that supports the alternative hypothesis.– If the p-value exceeds 10%, there is no evidence that

supports of the alternative hypothesis.

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• The p-value and rejection region methods– The p-value can be used when making decisions

based on rejection region methods as follows:• Define the hypotheses to test, and the required

significance level • Perform the sampling procedure, calculate the test statistic

and the p-value associated with it.• Compare the p-value to Reject the null hypothesis only

if p <; otherwise, do not reject the null hypothesis.

The p-value

34.175xL

= 0.05

170x

178x

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– If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative

hypothesis is true.– If we do not reject the null hypothesis, we

conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true.

– If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative

hypothesis is true.– If we do not reject the null hypothesis, we

conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true. The alternative hypothesis

is the more importantone. It represents whatwe are investigating.

The alternative hypothesisis the more importantone. It represents whatwe are investigating.

Conclusions of a test of Hypothesis

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• Example 10.2– A government inspector samples 25 bottles of catsup

labeled “Net weight: 16 ounces”, and records their weights.

– From previous experience it is known that the weights are normally distributed with a standard deviation of 0.4 ounces.

– Can the inspector conclude that the product label is unacceptable?

Catsup

15.8

16.0

16.2

15.7

.

.

.

Catsup

15.8

16.0

16.2

15.7

.

.

.

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• Solution– We need to draw a conclusion about the mean

weights of all the catsup bottles.– We investigate whether the mean weight is less

than 16 ounces (bottle label is unacceptable).

H1: < 16H0: = 16

– The test statistic is

n

xz

– Select a significance level: = 0.05

– Define the rejection regionz < - z1.645

Then

One tail test

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we want this mistake to happen not more than 5% of the time.

16

0.05

A sample mean far below 16,should be a rare event if = 16.

So, if in reality =16, but we reject this hypothesis in favor of < 16 because was very small, x

x

25.1254.0

1690.15

n

xz

Rejection region -1.25

0.05

0-z= -1.645

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0-z= -1.645

0.05

25.1254.0

1690.15

n

xz

Rejection region-1.25

Since the value of the test statistic does not fall in the rejection region, we do not reject the null hypothesis in favor of the alternative hypothesis.

There is insufficient evidence to inferthat the mean is less than 16 ounces.

The p-value = P(Z < - 1.25) = .1056 > .05

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• Example 10.3

– The amount of time required to complete a critical part of a production process on an assembly line is normally distributed. The mean was believed to be 130 seconds.

– To test if this belief is correct, a sample of 100 randomly selected assemblies was drawn, and the processing time recorded. The sample mean was 126.8 seconds.

– If the process time is really normal with a standard deviation of 15 seconds, can we conclude that the belief regarding the mean is incorrect?

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• Solution– Is the mean different than 130?

H0: = 130Then

130:H1

– Define the rejection regionz < - zor z > z/2

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130

0

A sample mean far below 130or far above 130, should be a rare event if = 130.

x x

we want this mistake to happen not more than 5% of the time.

So, if in reality =130, but we mistakenlyreject this hypothesis in favor ofbecause was very small or very large, x

130

20.025

20.025

20.025 20.025

13.210015

1308.126

n

xz

-z= -1.96 z= 1.96

Rejection region

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0

20.025 20.025

13.210015

1308.126

n

xz

-z= -1.96 z= 1.96

Since the value of the test statistic falls in the rejection region, we reject the null hypothesis in favor of the alternative hypothesis.

There is sufficient evidence to inferthat the mean is not 130.

-2.13

The p-value = P(Z < - 2.13)+P(Z > 2.13) = 2(.0166) = .0332 < .05

2.13

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– Interval estimators can be used to test hypotheses.– Calculate the 1 - confidence level interval

estimator, then• if the hypothesized parameter value falls within the

interval, do not reject the null hypothesis, while• if the hypothesized parameter value falls outside the

interval, conclude that the null hypothesis can be rejected ( is not equal to the hypothesized value).

Testing hypotheses and intervals estimators

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• Drawbacks – Two-tail interval estimators may not provide the right

answer to the question posed in one-tail hypothesis tests.

– The interval estimator does not yield a p-value.

There are cases where only tests producethe information needed to make decisions.

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Calculating the Probability of a Type II Error• To properly interpret the results of a test of

hypothesis, we need to– specify an appropriate significance level or judge the

p-value of a test;– understand the relationship between Type I and

Type II errors.– How do we compute a type II error?

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• Calculation of a type II error requires that– the rejection region be expressed directly, in terms of

the parameter hypothesized (not standardized).– the alternative value (under H1) be specified.

H0: 0

H1: 1 (0 is not equal to 1)Lx0

1

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• Let us revisit example 10.1– The rejection region was with = .05.– A type II error occurs when a false H0 is not rejected.

34.175x

34.175xL

170

180

.05

175.34

175.34

34.175x

…butH0 is false

Do not reject H0

)180thatgiven34.175x(P

)falseisHthatgiven34.175x(P 0

0764.)40065

18034.175z(P

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• Effects on of changing – Decreasing the significance level increases the

the value of and vice versa

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• Judging the test

– A hypothesis test is effectively defined by the significance level and by the the sample size n.

– If the probability of a type II error is judged to be too large, we can reduce it by

• increasing , and/or• increasing the sample size.

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nzx

thus,n

xz

L

L

By increasing the sample sizethe standard deviation of the sampling distribution of themean decreases. Thus,decreases.

Lx

LxLxLxLxLx LxLx

LxLxLxLxLxAs a result decreases

– In example 10.1, suppose n increases from 400 to 1000.

0)22.3Z(P)100065

18038.173Z(P

38.1731000

65645.1170

nzxL

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• In summary,– By increasing the sample size, we reduce the

probability of type II error.– Hence, we shall accept the null hypothesis when it is

false less frequently.• Power of a test

– The power of a test is defined as 1 - – It represents the probability to reject the null

hypothesis when it is false.