Estimation: Hypothesis Testing

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Estimation: Hypothesis Testing

Dr. Patrick Toche

References :

† Douglas A. Lind, William G Marchal, Samuel A. Wathen, Statistical Tech- niques in Business and Economics, The McGraw-Hill/Irwin Series in Opera- tions and Decision Sciences, 17th edition (2017), 978-1259666360.

Other references may be given from time to time.

Learning Objectives

1. Explain the process of testing a hypothesis.

2. Interpret Type-I and Type-II errors.

3. Distinguish between a one-tailed and a two-tailed test.

4. Test a hypothesis about a population mean.

5. Compute and interpret a z -score and a t -statistic.

6. Compute and interpret a p -value.

7. Compute the probability of a Type-II error.

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Hypothesis Testing

A hypothesis is a statement about a population parameter sub- ject to verification. Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

I Hypothesis testing does not provide definitive mathematical proof, but it does provide levels of confidence about the validity of a state- ment. Specific rules are followed to gather the evidence.

I The first step in hypothesis testing is to set the null hypothesis H

0

and the alternative H

₁

. The

₀

subscript stands for “no change” — the

₁

stands for the negation.

I The null hypothesis H

₀

is said to be “true” if the alternate hypoth- esis H

₁

is “improbable”, that is less likely than a set threshold — the significance level α .

Six-Step Procedure

1. Set the null hypothesis and the alternate hypothesis.

2. Set the desired significance level.

3. Identify the test distribution.

4. Identify the acceptance/rejection regions.

5. Compute the test statistic from the sample data.

6. Interpret the test results.

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1. Set Null & Alternate Hypotheses

Null Hypothesis H

₀

A statement about the value of a population parameter, that may or may not be true. The validity of the null hypothesis is to be as- sessed on the basis of sample data. Example: µ = µ

₀

, where µ is the true population value and µ

₀

is the hypothetical value.

Alternate Hypothesis H

₁

A statement about the value of a population parameter that complements the null hypothesis. Will be considered true if evi- dence suggests the null hypothesis is false. Example: µ 6= µ

₀

. I The conclusion of a test is to either “reject” or “fail to reject” the null

hypothesis. A test cannot prove with certainty that the null is true

— only that the test failed to reject the null. The value of failing to reject a hypothesis depends on the power of the test.

1. One-Tailed Versus Two-Tailed Hypotheses

I Two-Tailed:

H

₀

: µ = µ

₀

H

₁

: µ 6= µ

₀

Evidence of µ µ

₀

or µ µ

₀

counts against H

₀

. I Left-Tailed:

H

₀

: µ ≥ µ

₀

H

₁

: µ < µ

₀

Evidence of µ µ

₀

counts against H

0

, but evidence of µ µ

₀

counts in favor of it. Sometimes H

₀

is written as an equality even when H

₁

is an inequality, which may be confusing.

I Right-Tailed: The direction of the inequalities is reversed.

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2. Set Level of Significance

Significance Level

The probability of rejecting the null hypothesis when it is true.

I The level of significance is designated α — It measures the risk of incorrectly rejecting the null hypothesis — a Type-I error.

I The level of significance is related to the level of confidence:

significance = 1 − confidence

I Significance levels of 0.01 , 0.05 , and 0.10 are popular.

– ^{α = 0.001} : high-risk medical procedures.

– ^{α = 0.01} : quality assurance.

– ^{α = 0.05} : consumer research projects.

– ^{α = 0.10} or higher: political polling.

2. Set Level of Significance

Type-I Error

Incorrectly rejecting the null hypothesis H

₀

, when the null is true.

Denoted α . Example: Medical False Positive.

Type-II Error

Incorrectly not rejecting the null hypothesis H

₀

, when the null is actually false. Denoted β . Example: Medical False Negative.

I Example of Type-I Error: A medical test suggests a patient has a disease when — in fact — the patient does not have the disease.

I Example of Type-II Error: A medical test suggests a patient does

not have a disease when — in fact — the patient does.

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3. Identify the Test Distribution

Test Statistic

A value, determined from sample information, used to determine whether to reject the null hypothesis. The sampling distribution of the test statistic, under the null hypothesis, is used to calculate critical values and p -values that form the basis of a probability- based decision to reject (or not) the null hypothesis.

I Two classes of test statistics used to test hypotheses about the population mean are:

– Standard Normal z -statistic: used when the sampling distribution is normal with standard deviation σ known and constant.

– ^Student ^t -statistic: used when the sampling distribution is normal with standard deviation s

_n−1

estimated from a single sample.

4. Identify Acceptance/Rejection Region

Critical value

A value that separates acceptance and rejection regions. For symmetric, two-tailed tests, there are two critical values set at equal distance to the center of the distribution and the rejection regions consists of the left and right tails.

I If the null hypothesis is true, the probability of sampling values in the rejection region is smaller than the chosen level of significance.

I The size of the regions depends on the selected significance level.

The smaller the significance level, the smaller the rejection region

and the more “demanding” the test is. For instance, a significance

level of α = 0.01 (confidence level of 99% ) is associated with a

greater critical value — and a smaller and more distant rejection

region — than a level of α = 0.10 (confidence level of 90% ).

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4. Identify Acceptance/Rejection Region

-1.960 1.960

Reject H₀ Type-I error

in left tail α/2 = 0.025

Reject H₀ Type-I error

in right tail α/2 = 0.025 Cannot

Reject H₀

0

-3 -2 -1 0 1 2 3

density

Standard normal z-distribution two-tailed, α = 0.05

Acceptance/Rejection Region

Example: A two-tailed test for the standard normal z distribution: The null hypothesis H

₀

is rejected if the sample statistic falls into either one of the tails. Each tail contains half of the possible Type-I errors, α/2 . The critical values for a 95% confidence interval for the z distribution are z

_α/2

≈ −1.960 and z

_1−α/2

≈ 1.960 .

5. Compute the Test Statistic

I ^Example: H

0

: µ = µ

0

, H

1

: µ 6= µ

0

, σ is known, α = 0.05 . This is a two-tailed test. Under H

₀

, the sample mean x follows the normal distribution with mean µ

₀

and standard error σ/ √

n . The test statistic is the z -score:

z = x − µ

₀

σ/ √

n

and follows the z -distribution N (0, 1) .

I A test statistic of z = 4.0 lies in the rejection region. If the null is true, the probability that a z -score at least as large is drawn is very small. Thus, if the z -score were z = 4.0 , we would reject H

0

. I A test statistic of z = 0.5 lies outside the rejection region, In this

case, we cannot reject H

0

— maybe the null is true, maybe our

test lacks power.

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6. Interpret the Results

I Humility is the key to a correct interpretation.

I If the sample data leads you to reject the null hypothesis, there are several possible explanations:

1.

The null is indeed actually false.

2.

The null is true, but unfortunately the sample used was an outlier.

I If the sample data fails to reject the null hypothesis, there are sev- eral possible explanations, including:

1.

The null is indeed actually true.

2.

The null is false, but the test does not have enough power.

Example: The sample size is too small.

I Another possibility is that the selected significance level is too small (more likely to reject) or too large (more likely not to reject).

I Yet another possibility is that the sampling procedure was flawed, e.g. dependent samples, finite population (sampling without re- placement), sampling bias.

Practice: Two-Tailed Hypothesis Tests

I ^Example: Output at a plant typically follows a normal probability distribution with a mean of 200 and a standard deviation of 16 . New techniques are adopted that could have improved/disrupted production. A sample of size 50 is drawn with mean of 203.5 . Test whether mean output has changed, at the 0.01 significance level.

1. State the null & alternate hypotheses:

H

₀

: µ = 200 H

₁

: µ 6= 200

This is a two-tailed test because the alternate hypothesis does not

explicitly state a direction — any evidence that mean production

is significantly greater than 200 or significantly less than 200 will

stack up against the null hypothesis.

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Practice: Two-Tailed Hypothesis Tests

2. Set the level of significance:

α = 0.01

The probability of committing a Type-I error — the probability of incorrectly rejecting a true null hypothesis. The significance level α = 0.01 sets a 99% confidence interval.

3. Identify the test distribution: The standard deviation is known:

σ = 16 . By the central limit theorem, the test statistic z has a standard normal distribution N (0, 1) .

4. Identify the rejection region:

Half of the probability α is located in each tail. With α = 0.01 , the area where H

₀

is not rejected is 0.99 . For a two-tailed test, the z -distribution yields the critical value z

_α/2

≈ 2.576 .

Confidence Intervals,c

80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, !

0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, !

0.20 0.10 0.05 0.02 0.01 0.001

71 1.294 1.667 1.994 2.380 2.647 3.433

72 1.293 1.666 1.993 2.379 2.646 3.431

73 1.293 1.666 1.993 2.379 2.645 3.429

74 1.293 1.666 1.993 2.378 2.644 3.427

75 1.293 1.665 1.992 2.377 2.643 3.425

76 1.293 1.665 1.992 2.376 2.642 3.423

77 1.293 1.665 1.991 2.376 2.641 3.421

78 1.292 1.665 1.991 2.375 2.640 3.420

79 1.292 1.664 1.990 2.374 2.640 3.418

80 1.292 1.664 1.990 2.374 2.639 3.416

81 1.292 1.664 1.990 2.373 2.638 3.415

82 1.292 1.664 1.989 2.373 2.637 3.413

83 1.292 1.663 1.989 2.372 2.636 3.412

84 1.292 1.663 1.989 2.372 2.636 3.410

85 1.292 1.663 1.988 2.371 2.635 3.409

86 1.291 1.663 1.988 2.370 2.634 3.407

87 1.291 1.663 1.988 2.370 2.634 3.406

88 1.291 1.662 1.987 2.369 2.633 3.405

89 1.291 1.662 1.987 2.369 2.632 3.403

90 1.291 1.662 1.987 2.368 2.632 3.402

91 1.291 1.662 1.986 2.368 2.631 3.401

92 1.291 1.662 1.986 2.368 2.630 3.399

93 1.291 1.661 1.986 2.367 2.630 3.398

94 1.291 1.661 1.986 2.367 2.629 3.397

95 1.291 1.661 1.985 2.366 2.629 3.396

96 1.290 1.661 1.985 2.366 2.628 3.395

97 1.290 1.661 1.985 2.365 2.627 3.394

98 1.290 1.661 1.984 2.365 2.627 3.393

99 1.290 1.660 1.984 2.365 2.626 3.392

100 1.290 1.660 1.984 2.364 2.626 3.390

120 1.289 1.658 1.980 2.358 2.617 3.373

140 1.288 1.656 1.977 2.353 2.611 3.361

160 1.287 1.654 1.975 2.350 2.607 3.352

180 1.286 1.653 1.973 2.347 2.603 3.345

200 1.286 1.653 1.972 2.345 2.601 3.340

! 1.282 1.645 1.960 2.326 2.576 3.291

Student’s t Distribution (concluded )

(degreesdf freedom)of (continued )

Areas under the Normal Curve

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359

0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224

0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549

0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852

0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133

0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389

1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621

1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830

1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177

1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441

1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545

1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633

1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706

1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857

2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890

2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916

2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964

2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974

2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981

2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

Example:

If z = 1.96, then P(0 to z) = 0.4750.

z 0 1.96

0.4750 Lin01803_endsheet.qxd 10/2/10 8:07 AM Page 3

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Practice: Two-Tailed Hypothesis Tests

-2.576 2.576

0

-4 -3 -2 -1 0 1 2 3 4

density

Density

Probability Reject (1%) Accept (99%) Standard normal z-distribution

Significance level: α = 0.01

Acceptance/Rejection Region

Practice: Two-Tailed Hypothesis Tests

5. Compute the test statistic from the sample data:

The z -score is:

z = x − µ

₀

σ/ √

n = 203.5 − 200 16/ √

50 ≈ 1.547 6. Interpret the results

The z -score lies inside the “acceptance” region:

−2.576 < 1.547 < 2.576

We cannot reject the null H

₀

that µ = 200 against the alternate

µ 6= 200 . There isn’t enough evidence to conclude that the pop-

ulation mean has changed — The observed difference between

the population mean of 200 and the sample mean of 203.5 could

simply be due to chance.

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Practice: One-Tailed Hypothesis Tests

I Example: Test for an increase in the population mean.

I State the null & alternate hypotheses:

H

₀

: µ ≤ 200 H

₁

: µ > 200

I Identify the distribution & acceptance region: The variance is known, σ = 16 , so the test statistic follows the standard normal z -distribution. The critical value for α = 0.01 is z

_1−α

≈ 2.326 . I Interpret the results: Since z ≈ 1.547 < 2.326 , we cannot reject

the null H

₀

that µ ≤ 200 against the alternate H

₁

that µ > 200 . I One-Tailed Versus Two-Tailed Tests: In a one-tailed test, the en-

tire rejection region lies in one tail. In a two-tailed test, the rejection region is split equally between the two tails, with significance α/2 in each tail. The critical values for one-tailed and two-tailed tests are therefore different, and could lead to different conclusions. In this example, however, both tests fail to reject the null.

Practice: One-Tailed Hypothesis Tests

2.326 0

-4 -3 -2 -1 0 1 2 3 4

density

Density

Probability Reject (1%) Accept (99%) Standard normal z-distribution

Significance level: α = 0.01

Acceptance/Rejection Region

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Population Variance Unknown

I Identify the test distribution: When the population standard de- viation is unknown, the sample estimate s

_n−1

is used to estimate the standard error. Under the null hypothesis, the test statistic has a Student t -distribution with n − 1 degrees of freedom.

I Compute the test statistic: The t -statistic is:

t = x

_n

− µ

₀

s

_n−1

/ √

n

I The Student t -distribution is continuous, hump-shaped, and sym- metrical. It has thicker tails than the standard normal distribution

— the density is lower in the middle of the distribution and declines less rapidly towards the tails. The number of degrees of freedom, n − 1 , is a parameter that defines a family of t -distributions.

I While the sample mean x is indexed by n , the sample standard deviation s is indexed by n − 1 to emphasize the importance of the Bessel-correction and the role of degrees of freedom.

Practice: Population Variance Unknown

I ^Example: The Claims Department reported the mean cost to pro- cess a claim as $60 . After a restructuring, a random sample of 26 claims is taken to test for a reduction in mean cost:

45 49 62 40 43 61 48 53 67 63 78 64 48 54 51 56 63 69 5 8 51 58 59 56 57 38 76 I State the null & alternate hypotheses:

H

0

: µ ≥ 60 H

₁

: µ < 60 I Set the level of significance: α = 0.01

I Identify the test distribution: The population variance is un-

known, so the sample estimate will be used to compute the test

statistic. The test statistic has a t -distribution with ν = n − 1 = 25

degrees of freedom. The critical value is t

_n−1,1−α

≈ 2.485 .

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Practice: Population Variance Unknown

I Compute the test statistic: The sample mean and standard deviation are:

x

₂₆

= 56.423 s

₂₅

= 10.041 The t -statistic is therefore:

t = x

₂₆

− µ

₀

s

₂₅

/ √

n = 56.423 − 60 10.041/ √

26 ≈ −1.816

I Interpret the results: The t -statistic lies inside the (one-sided)

“acceptance” region:

−2.485 < −1.816

We cannot therefore reject the null H

₀

that µ ≥ 60 against the alternate H

₁

that µ < 60 .

I You may check that we cannot reject µ = 60 against µ 6= 60 .

Practice: Population Variance Unknown

-2.485 Reject H₀ Type-I error

in left tail α = 0.01

Cannot Reject

H₀

0

-4 -3 -2 -1 0 1 2 3 4

density

Student t-distribution (df = 25) one-tailed, α = 0.01

Acceptance/Rejection Region

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p -values in Hypothesis Testing

p -value

The probability of drawing a sample value as extreme as — or more extreme than — the sample value, under the assumption that the null hypothesis is true.

I In traditional hypothesis testing, we compare the test statistic to a critical value and either reject or fail to reject the null hypothesis.

I If the p -value is smaller than the significance level, the null is re- jected. In addition, the p -value gives insight into the probability of a Type-I error. A very small p -value suggests the null has a very small probability of being true — that is, either a rare event has happened or the null hypothesis is false. A large p -value suggests that the null is very probably false.

I The two-tailed p -value is twice as large as the one-tailed p -value.

p -values in Hypothesis Testing

The p -value gives a way to express the probability that H

₀

is false.

p < 0.10 some evidence H

₀

false p < 0.05 strong evidence H

₀

false p < 0.01 very strong evidence H

₀

false p < 0.001 extremely strong evidence H

₀

false

The two-tailed p -value is:

P(x

_n

≤ −x

_n

) + P(x

_n

≥ x

_n

)

If the null hypothesis is actually false, Type-I errors cannot occur and

the interpretation of the p -value is unclear.

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p -values in Hypothesis Testing

Suppose a sample z -score under the null hypothesis H

₀

: µ = µ

₀

is:

z

_n

= x

_n

− µ

₀

σ/ √

n ≈ −1.282

for known σ . The p -value associated with this sample estimates the probability that other sample means be located in the tails:

-1.282 1.282

p-value in left tail p/2 = 0.10

p-value in right tail

p/2 = 0.10 Expect 80%

of samples to be less

extreme

standardized sample mean

-3 -2 -1 0 1 2 3

density

Two-Sided p-value for Standard Normal

Hypothesis Testing: p-value

Power in Hypothesis Testing

I Type-I error α :

The null hypothesis H

0

is true and — incorrectly — rejected.

I Type-II error β :

The null hypothesis H

₀

is false and — incorrectly — not rejected.

I Confidence 1 − α :

The probability of — correctly — not rejecting a true H

₀

— The probability of not making a Type-I error.

I Power 1 − β :

The probability of — correctly — rejecting a false H

0

— The prob- ability of not making a Type-II error.

In practice, the power of a test cannot be computed since the true

distribution is unknown. Notwithstanding, the following analysis

shows that a one-tailed test (if appropriately directed) is more pow-

erful than a two-tailed test: Under a one-tailed test, the area asso-

ciated with β is reduced and thus power 1 − β is increased.

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Power in Hypothesis Testing

α/2 β

μ₀ z μ₁

-z 0.0

0.1 0.2 0.3 0.4

Density

Fact: μ = μ₁ (unknown)

Hypothesis: H₀: μ = μ₀ vs H₁: μ ≠ μ₀

Power of a Test

Significance: α (Probability of Type-I error) Power: 1 - β (1- Probability of Type-II error)

In blue, the hypothetical distribution inferred from a sample. In red, the true — unknown

— distribution. Any sample mean between the critical values

−z

andz would lead to not rejecting H0. The left-tail of the true distribution measures the probability of a Type-II error,β. The power of the test is the complement

1 − β

.

Power in Hypothesis Testing

α β

μ₀ z μ₁

0.0 0.1 0.2 0.3 0.4

Density

Fact: μ = μ₁ (unknown)

Hypothesis: H₀: μ ≤ μ₀ vs H₁: μ > μ₀

Power of a Test

Significance: α (Probability of Type-I error) Power: 1 - β (1- Probability of Type-II error)

In blue, the hypothetical distribution. In red, the true distribution. Any sample mean smaller than the critical value z would lead to not rejecting H0. The left-tail of the true distribution measures the probability of a Type-II error,β. The power of the test is the complement

1 − β

. One-tailed tests can be more powerful than two-tailed tests.

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Summary

1. The objective of hypothesis testing is to verify the validity of a statement about a population parameter.

2. Hypothesis testing can be broken down into steps:

1

State the null hypothesis (H

₀

) and the alternate hypothesis (H

₁

).

2

Set the desired significance level.

3

Identify the test distribution.

4

Identify the acceptance/rejection regions.

5

Compute the test statistic from the sample.

6

Interpret the results.

3. In a two-tailed test, the rejection region is evenly split between the left and right tails. In a one-tailed test, the entire rejection region is either on the left or on the right.

4. A p -value measures the probability that, under the null hypothesis H

₀

, a sample could be drawn that would be “more extreme” than the observed sample value.

Summary

5. To test a hypothesis about a population mean using a single sample, two cases arise: If the population standard deviation σ is known, the z -score follows the standard normal distribution:

z = x − µ σ/ √

n

6. If the population standard deviation is not known, the Bessel-corrected sample estimate s is used to compute a t -statistic, which follows the Student t -distribution:

t = x − µ s/ √

n

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Summary

7. The major characteristics of the Student t -distribution are:

a.

It is a continuous distribution.

b.

It is hump-shaped and symmetrical.

c.

It has denser tails than the standard normal distribution.

d.

The degrees of freedom n − 1 defines a family of t -distributions.

8. A Type-I error occurs when a true null hypothesis is incorrectly rejected. The probability of making a Type-I error is the level of significance α . The complement 1 − α is the level of confidence.

9. A Type-II error occurs when a false null hypothesis is not rejected.

The probability of making a Type-II error is denoted β . The complement 1 − β is the power of the test.

Problems and Applications

1. A random variable follows the normal probability distribution with a mean of 45, 000 and standard deviation 3, 000 . A sample of 120 observations is taken. The sample mean is 45, 500 . At the 10% significance level, test the hypothesis that the sample mean is drawn from the population. Calculate the p -value.

2. A survey found interest on credit card debt was, on average,

$6, 658 . A sample of 12 households has:

7077 5744 6753 7381 7625 6636 7164 7348 8060 5848 9275 7052

At the 0.05 significance level, is it reasonable to conclude that

these households paid more interest?

(18)

Problems and Applications

3. A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of senior citizens reported the following amounts:

3.1 3.3 3.5 2.6 2.6 4.3 4.4 3.8 3.1 4.1 3.1 3.2

At the 0.05 significance level, is there a difference between the national average and the sample mean from senior citizens?

4. Major League Baseball has been criticized for the length of the games. A sample of 17 games revealed the following durations:

2.98 2.40 2.70 2.25 3.23 3.17 2.93 3.18 2.80 2.38 3.75 3.20 3.27 2.52 2.58 4.45 2.45 (units: hour) At the 0.05 significance level, is the mean game time less than 3.50 hours?

Problems and Applications

5. The United Nations reported the mean family income for Mexican migrants to the United States as $27, 000 per year. A Farm Labor Organizing Committee (FLOC) evaluation of 25 Mexican family units found a mean of $30, 000 , with a sample standard deviation of $10, 000 . At the 0.01 significance level, does the FLOC study disagree with the U.N. report?

Estimation: Hypothesis Testing