introduction to hypothesis testing

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Introduction to Hypothesis Testing

Introduction to Hypothesis Testing

Chapter 11

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11.1 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 10% of the potential customers will purchase a new products?

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

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11.2 Concepts of Hypothesis Testing

• The critical concepts of hypothesis testing.– Example:

• An operation manager needs to determine if the mean demand during lead time is greater than 350.

• If so, changes in the ordering policy are needed. – There are two hypotheses about a population mean:

• H0: The null hypothesis = 350 • H1: The alternative hypothesis > 350

This is what you want to prove

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= 350

• Assume the null hypothesis is true (= 350).

– Sample from the demand population, and build a statistic related to the parameter hypothesized (the sample mean).

– Pose the question: How probable is it to obtain a sample mean at least as extreme as the one observed from the sample, if H0 is correct?

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– Since the is much larger than 350, the mean is likely to be greater than 350. Reject the null hypothesis.

x

355x


= 350

• Assume the null hypothesis is true (= 350).

450x

– In this case the mean is not likely to be greater than 350. Do not reject the null hypothesis.

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Types of Errors

• Two types of errors may occur when deciding whether to reject H0 based on the statistic value.

– Type I error: Reject H0 when it is true.

– Type II error: Do not reject H0 when it is false.• Example continued

– Type I error: Reject H0 ( = 350) in favor of H1 ( > 350) when the real value of is 350.

– Type II error: Believe that H0 is correct ( = 350) when the real value of is greater than 350.

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Controlling the probability of conducting a type I error

• Recall:– H0: = 350 and H1: > 350.

– H0 is rejected if is sufficiently large

• Thus, a type I error is made if when = 350.

• By properly selecting the critical value we can limit the probability of conducting a type I error to an acceptable level.

xvaluecriticalx

Critical value

x= 350

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

• Example 11.1– 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 accounts has a mean of $178.– If accounts are approximately normally distributed with

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

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• Example 11.1 – Solution– The population of interest is the credit accounts at

the store.– We want to know whether 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

Testing the Population Mean ( is Known)

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Approaches to Testing

• There are two approaches to test whether the sample mean supports the alternative hypothesis (H1)– The rejection region method is mandatory for

manual testing (but can be used when testing is supported by a statistical software)

– The p-value method which is mostly used when a statistical software is available.

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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.

The Rejection Region Method

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Example 11.1 – solution continued

• Recall: H0: = 170 H1: > 170

therefore, • It seems reasonable to reject the null hypothesis and believe that > 170 if the sample mean is sufficiently large.

The Rejection Region Method – for a Right - Tail Test

Reject H0 here

Critical value of the sample mean

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• Define a critical value for that is just large enough to reject the null hypothesis.

xLx

• Reject the null hypothesis if

Lxx Lxx

The Rejection Region Method for a Right - Tail Test

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• Allow the probability of committing a Type I error be (also called the significance level).

• Find the value of the sample mean that is just large enough so that the actual probability of committing a Type I error does not exceed Watch…

Determining the Critical Value for the Rejection Region

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

170x x

= P( given that H0 is true)Lxx

40065

170xz L


… is allowed to be

)ZZ(PSince we have:

Determining the Critical Value – for a Right – Tail Test

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Determining the Critical Value – for a Right – Tail Test

.34.17540065

645.1170x

.645.1z,05.0selectweIf

.40065

z170x

L

05.

L

40065

170xz L

= 0.05

170x Lx


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Determining the Critical value for a Right - Tail Test

34.175xifhypothesisnullthejectRe

34.175xifhypothesisnullthejectRe

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence to infer that the mean monthly balance is greater than $170 at the 5% significance level.

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence to infer that the mean monthly balance is greater than $170 at the 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

zzOne tail test

The standardized test statistic

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• Example 11.1 - continued– We redo this example using the standardized test

statistic.Recall:H0: = 170

H1: > 170– Test statistic:

– Rejection region: z > z.051.645.

46.240065

170178

n

xz


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• Example 11.1 - continued


645.1ZifhypothesisnullthejectRe

645.1ZifhypothesisnullthejectRe

ConclusionSince Z = 2.46 > 1.645, reject the null hypothesis in favor of the alternative hypothesis.

ConclusionSince Z = 2.46 > 1.645, reject the null hypothesis in favor of the alternative hypothesis.

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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 11.1

P-value Method

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0069.)4615.2z(P

)40065170178

z(P

170x 178x

The probability of observing a test statistic at least as extreme as 178, given that = 170 is…

The p-value

P-value Method

)170when178x(P

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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

Interpreting the p-value

24178x

170:H x0 170:H x1

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

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


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

evidence that supports 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 the alternative hypothesis.


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– 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-value <; otherwise, do not reject the null hypothesis.

The p-value

34.175xL

= 0.05

170x

178x

The p-value and the Rejection Region Methods

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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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A Left - Tail Test

• The SSA Envelop Example.– The chief financial officer in FedEx believes that

including a stamped self-addressed (SSA) envelop in the monthly invoice sent to customers will decrease the amount of time it take for customers to pay their monthly bills.

– Currently, customers return their payments in 24 days on the average, with a standard deviation of 6 days.

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• The SSA envelop example – continued – It was calculated that an improvement of two days on the

average will cover the costs of the envelops (checks can be deposited earlier).

– A random sample of 220 customers was selected and SSA envelops were included with their invoice packs.

– The times customers’ payments were received were recorded (SSA.xls)

– Can the CFO conclude that the plan will be profitable at 10% significance level?

A Left - Tail Test

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• The SSA envelop example – Solution– The parameter tested is the population mean

payment period ()– The hypotheses are:

H0: = 22H1: < 22(The CFO wants to know whether the plan will be profitable)

A Left - Tail Test

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• The SSA envelop example – Solution continued– The rejection region:

It makes sense to believe that < 22 if the sample mean is sufficiently smaller than 22.

– Reject the null hypothesis if

A Left - Tail Test

Sxx Sxx

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• The SSA envelop example – Solution continued– The standardized one tail left hand test is:

A Left -Tail Test

28.110. zzz

91.2206

2263.21

n

xz

Since -.91 > –1.28 do not reject the null hypothesis. The p value = P(Z<-.91) = .1814

Since .1814 > .10, do not reject the null hypothesis

Left-tail test

Define the rejection region

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A Two - Tail Test

• Example 11.2– AT&T has been challenged by competitors who

argued that their rates resulted in lower bills.– A statistics practitioner determines that the mean

and standard deviation of monthly long-distance bills for all AT&T residential customers are $17.09 and $3.87 respectively.

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A Two - Tail Test

• Example 11.2 - continued– A random sample of 100 customers is selected and

customers’ bills recalculated using a leading competitor’s rates (see Xm11-02).

– Assuming the standard deviation is the same (3.87), can we infer that there is a difference between AT&T’s bills and the competitor’s bills (on the average)?

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• Solution– Is the mean different from 17.09?

H0: = 17.09

09.17:H1

– Define the rejection region

A Two - Tail Test

2/2/ zzorzz

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17.09

We want this erroneous rejection of H0 to be a rare event, say 5% chance.

x x

If H0 is true ( =17.09), can still fall far above or far below 17.09, in which case we erroneously reject H0 in favor of H1

x

)09.17(

20.025 20.025

Solution - continued

A Two – Tail Test

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20.025

17.09

0

x x20.025

20.025 20.025

19.110087.3

09.1755.17

n

xz

-z= -1.96 z= 1.96

Rejection region

Solution - continued

A Two – Tail Test

55.17x

From the sample we have:

17.55

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20.025 20.025

19.110087.3

09.1755.17

n

xz

-z= -1.96 z= 1.96

There is insufficient evidence to infer that there is a difference between the bills of AT&T and the competitor.

-1.19

Also, by the p value approach:The p-value = P(Z< -1.19)+P(Z >1.19) = 2(.1173) = .2346 > .05

1.190

A Two – Tail TestTwo-tail test

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11.4 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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• To calculate Type II error we need to…– express the rejection region directly, in terms of the

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

• Let us revisit Example 11.1

Calculation of the Probability of a Type II Error

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Express the rejection region directly, not in standardized terms

34.175xL

=.05

= 170


• Let us revisit Example 11.1– The rejection region was with = .05.34.175x

Do not reject H0

180

H1: = 180

H0: = 170

Specify the alternative value

under H1.

– Let the alternative value be = 180 (rather than just >170)

4234.175

xL

=.05

= 170


34.175x 180

H1: = 180

H0: = 170

– A Type II error occurs when a false H0 is not rejected.

A false H0……is not rejected

4334.175

xL = 170


180

H1: = 180

H0: = 170

)180thatgiven34.175x(P

)falseisHthatgiven34.175x(P 0

0764.)40065

18034.175z(P

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• Decreasing the significance level increases the value of and vice versa

Effects on of changing

= 170 180

2 >2 <

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• A hypothesis test is effectively defined by the significance level and by 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.

Judging the Test

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• Increasing the sample size reduces

Judging the Test

By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, decreases.

Lx

nzxthus,

nx

z:callRe LL

47Lx 180= 170

Judging the Test

Lx

Note what happens when n increases:

Lx LxLx Lx

does not change,but becomes smaller

• Increasing the sample size reduces

nzxthus,

nx

z:callRe LL

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• Increasing the sample size reduces • In Example 11.1, suppose n increases from 400

to 1000.

0)22.3Z(P)100065

18038.173Z(P

38.173100065

645.1170n

zxL

Judging the Test

• remains 5%, but the probability of a Type II drops dramatically.

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• Power of a test– The power of a test is defined as 1 - – It represents the probability of rejecting the null

hypothesis when it is false.

Judging the Test

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