Hypothesis Testing (On a Single Parameter)

Chapter 8: Hypothesis Testing (on a Single Parameter)

Hildebrand, Ott and Gray Basic Statistical Ideas for Managers Second Edition

Learning Objectives for Ch. 8

• The concept of hypothesis testing • Type I and Type II errors • The probability of making a Type I error (α ) • The probability of making a Type II error (β ) • The steps in the hypothesis testing procedure

Learning Objectives for Ch. 8

• Hypothesis testing on a mean (known population standard deviation) using the Z-test • Assumptions • Hypothesis testing on a mean using the t-test • Assumptions • Hypothesis testing on a proportion • Assumptions • The equivalence of Confidence Interval and Hypothesis Testing procedures • Testing a hypothesis by using a Confidence Interval

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 3 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Section 8.1 A Test for a Mean, Known Standard Deviation

8.1 A Test for a Mean, Known Standard Deviation

• Two Basic Types Of Inference • Estimation – Chapter 7

• Point estimation (a single number)

• Interval estimation (an interval of probable values) • Hypothesis Testing – Chapter 8

• Null Hypothesis (H0): • Parameter = Some Value

• Alternative or Research Hypothesis (Ha): • Parameter ≠ Some Value

8.1 A Test for a Mean, Known Standard Deviation

Example: The quality control engineer in a bottling plant needs to determine if the average content of the containers is 16 oz. Or, is the process mean at the target value of 16 oz.? Past studies indicate that the content is normally distributed and that the process standard deviation is 0.25 oz. A sample of size 25 yielded a sample mean of 16.15 oz. Using a significance level of 5%, determine if the process mean is at the target value. By the way, the quality control engineer is interested in detecting deviation in either direction from 16 ounces.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 6 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.1 A Test for a Mean, Known Standard Deviation

State the null and research hypotheses.

•H0: µ = 16

•Ha: µ ≠ 16 ( Two – Sided Research Hypothesis)

<16 >16

Ha H0 Ha • Possible Values of µ 16

•Ha selected is one of 3 possibilities

• Ha: µ ≠ 16

• Ha: µ > 16

• Ha: µ < 16

8.1 A Test for a Mean, Known Standard Deviation

• Type I and Type II Errors

• Recall H0 : µ = 16 vs. Ha : µ ≠ 16

• Base decision about µ on Y

• If Y is "close" to 16, conclude that H0 is true. (Do not reject it).

• If Y is not "close" to 16, reject H0. • Since decision is based on Y and µ is unknown, the decision to reject or not reject could be incorrect. • Two possible types of errors.

8.1 A Test for a Mean, Known Standard Deviation

Null Hypothesis

True = ? False = ?

Do Not Correct Type II Error Reject H0 Decision Based Decision on Sample Mean Reject H 0 Type I Error Correct Decision

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 9 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.1 A Test for a Mean, Known Standard Deviation

• If H0 is true and we incorrectly reject it, then a Type I error occurs. • The probability of making a Type I error is symbolized by α and called the “level of significance”.

• If H0 is false and we incorrectly fail to reject it, a Type II error occurs. • The probability of making a Type II error is symbolized by β.

8.1 A Test for a Mean, Known Standard Deviation

• α = P( Type I Error ) µ = 16.0 = P( Reject H0 | True) • β = P( Type II Error) µ ≠ 16.0 = P( Not Rejecting H0 | False) • Basic Strategy:

• Do the data support the research hypothesis?

• If so, reject the null hypothesis!

• Reject H0 if Y is not close to 16.

8.1 A Test for a Mean, Known Standard Deviation

• What is "close"?

•Fix α at say 5%. Suppose the sample size is 25 and σ = .25 oz. (from past studies)

• P( Type I Error) a.k.a. the significance level of the test

•.05 = P( Rejecting H0 | H0 is true )

= P( Y > k2 | µ = 16 ) + P( Y < k1 | µ = 16 )

• .025 = P( Y > k2 | µ = 16 ) < Splitting up α evenly • .025 = P(Z > 1.96) (From Table 3)

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 12 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.1 A Test for a Mean, Known Standard Deviation Therefore, k − 16 2 = 1.96 .25/ 25

or, k 2 = 16 + 1.96 (.25 / 25 ) = 16.098 = 16.10 Similarly,

k1 = 15.90

8.1 A Test for a Mean, Known Standard Deviation

• Distribution of Y if H0: µ = 16 is true. • The critical points determine the rejection region (R.R.).

σY = .25/5 = .05

.025 .025

16 k1 = 15.90 k2 = 16.10 Critical points

8.1 A Test for a Mean, Known Standard Deviation

R.R.15.90 16.10 R.R. • R.R. in terms of y 16 • Suppose the sample of size 25 yielded y = 16.15

• Decision: ______Reject: H0 = 16 at the 5% level of significance.

• In general, in testing Ho: µ = µo vs. Ha: µ ≠ µo with σ known, reject Ho

⎫ if Y > µ 0 + zσ / 2 ()σ / n ⎪ ⎬ R.R or Y < µ 0 − zα / 2 ()σ / n ⎭⎪

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 15 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.1 A Test for a Mean, Known Standard Deviation

• Equivalently, calculate the standardized test statistic Z: Y - µ Z = 0 {Z-test statistic σ/ n

• Reject H0 if | Z | > z α /2 } R.R.

Example (Bottling Plant):

y - µ 16.15 -16 16.15 = estimate of population mean Z = 0 = = 3 σ / n .25/5 16 = hypothesized value for pop. mean

•y differs from µ0 by 3 standard errors.

8.1 A Test for a Mean, Known Standard Deviation

R.R.-1.96 1.96 R.R. • R.R. in terms of Z. 0

Example (Bottling Plant): The sample of size 25 yielded y = 16.15 ⇒ Z = 3

• Decision: ______Reject H0: µ = 16 at the 5% level of significance.

Section 8.2 Type II Error, β Probability, and Power of a Test

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 18 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.2 Type II Error, β Probability, and Power of a Test

Probability of Type II error (β ) µ ≠ 16 • β = P( Do not reject H0 | H0 is false )

Example (Bottling Plant): Suppose the true process mean is 16.2. What’s the chance of saying the process is on target? β = P (15.90 ≤ Y ≤16.10 | µ =16.2) ⎛15.90 -16.2 16.10 −16.2 ⎞ = P⎜ ≤ Z ≤ ⎟ ⎝ .05 .05 ⎠ = P()- 6 ≤ Z ≤ − 2 =0.0228 About 2.3% of the time, we conclude the process is on target (µ = 16 oz.) when in fact µ = 16.2 oz.

8.2 Type II Error, β Probability, and Power of a Test

Graph showing the relationship between α and β

α/2 = .025 β = .0228

15.90 16 16.10 16.2 }

H0 is true

• As P( Type I Error) decreases, P( Type II Error) increases.

8.2 Type II Error, β Probability, and Power of a Test

• β is not a single number but a collection of probabilities. • General expression to calculate β for a two- sided research hypothesis

⎛ µ0 − µa µ0 − µa ⎞ β = P⎜- zα/2 + ≤ Z ≤ zα / 2 + ⎟ ⎝ σ / n σ / n ⎠

• Given any two of the three (α, β, n), we can determine the other.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 21 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.2 Type II Error, β Probability, and Power of a Test

• Power of a statistical test is the probability of rejecting the null hypothesis when the null hypothesis is false.

Example (Bottling Plant): Find the power of the test if µ = 16.2.

• Power = 1 - β = 1 - .02285 = .9772

Section 8.3 The p-Value for a Hypothesis Test

8.3 The p-Value for a Hypothesis Test

• P-value is the chance of observing the sample result, or a more extreme result, if the null hypothesis is true. Example (Is a coin fair?):

H0: Coin is fair.

• Toss coin 10 times.

• Observe 10 heads. 10 • If coin is fair, P(10 heads in 10 tosses) = (½) = .000976 = p-value Since p-value is small, hypothesis of a fair coin must be incorrect.

Example: (Bottling Plant) Find the p-value of the test. • We were given that y = 16.15 ⎛ 16.15 -16 ⎞ P ()Y >16.15 | µ =16 = P⎜ Z > ⎟ ⎝ .05 ⎠ = P()Z > 3 = .00135

• p-value = 2(.00135) = .0027 (Use the factor of 2 only

for two-sided Ha)

• Note: If p-value < α, reject H0 [Universal Rejection Region]

8.3 The p-Value for a Hypothesis Test

Exercise 8.1 – 8.4: 8.1: The manager of a health maintenance organization has set as a target that the mean waiting time of non-emergency patients will not exceed 30 minutes. In spot checks, the manager finds the waiting times of 22 patients; the patients are selected randomly on different days. Assume that the population standard deviation of waiting times is 10 minutes: a. What is the relevant parameter to be tested? µ = mean waiting time of ALL non-emergency patients b. Formulate null and research hypotheses.

H0 : µ = 30

Ha : µ > 30

8.3 The p-Value for a Hypothesis Test

c. State the test statistic and the rejection region corresponding to α = .05. y - µ Test Statistic : Z = 0 σ / n

Rejection Region : Reject H0 if Z > 1.645

R.R. Z 1.645

8.2: Suppose that the mean waiting time for the 22 patients in Exercise 8.1 is 38.1 minutes. Can H0 be rejected? 38.1 - 30 Z = = 3.80 10/ 22

Conclusion: Reject H0: µ = 30 at the .05 level.

Equivalently, reject H0 if Y > µ0 + zα (σ / n )

30 + 1.645 10 / 22 ? Is 38.1 > () ( ) 33.51

Conclusion: Reject H0: µ = 30 at the .05 level

8.3 The p-Value for a Hypothesis Test

8.3: For the test procedure of Exercise 8.1, find the probability that H0 will not be rejected, assuming a true mean waiting time of 34 minutes. Do the same for other values of µ, and sketch a β curve.

• P(Not Rejecting H0 | True Mean Waiting Time = 34) ⎛ µ − µ ⎞ ⎛ 30 - 34 ⎞ β = P⎜ Z < z + 0 a ⎟ = P Z < 1.645 + ⎜ α ⎟ ⎜ ⎟ ⎝ σ / n ⎠ ⎝ 10 / 22 ⎠ = P()Z < - 0.23 = .41 • There is a 41% chance of concluding that the targeted mean waiting time of 30 minutes is being met when, in fact, patients have to wait 4 minutes longer than the target.

8.3 The p-Value for a Hypothesis Test

Values of β for other values of µ

µ β 32 .7602 34 .4086 36 .1212

8.4: We stated in Exercise 8.1 that the 22 patients were selected on different days. Why would one not want to select 22 patients on one randomly chosen day?

If there is an initial delay on the randomly chosen day, that effect carries over to all 22 patients on that day.

Section 8.4 Hypothesis Testing with the t Distribution

8.4 Hypothesis Testing with the t Distribution

• An easy extension of the Z-test y - µ Z = 0 σ / n • σ is unknown, replace it by s. • Replace Z-percentile by the t-percentile. • The t-test statistic: Y − µ t - statistic = 0 s / n

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 33 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.4 Hypothesis Testing with the t Distribution

Exercise 8.13: A dealer in recycled paper places empty trailers at various sites; these are gradually filled by individuals who bring in old newpapers and the like. The trailers are picked up (and replaced by empties) on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1600 cubic feet per two-week period. The dealer’s records for 18 two-week periods show the following volumes (in cubic feet) at a particular site:

1660 1820 1590 1440 1730 1680 1750 1720 1900 1570 1700 1900 1800 1770 2010 1580 1620 1690 ( y = 1718.3, s = 137.8)

Assume that these figures represent the results of a random sample. Do they support the research hypothesis that µ > 1600, using α = .10? Write out all parts of the hypothesis-testing procedure.

8.4 Hypothesis Testing with the t Distribution

H0 : µ = 1600 Ha : µ > 1600 y - µ T.S.: t = 0 = 1718.3 - 1600 = 3.64 s / n 137.8/ 18

R.R.: Reject H0 if t > t.10, 17 = 1.333 (Table 4) R.R. 1.33 3.64

Conclusion: Reject H0 : µ = 1600 at the 10% level. Implication: Stay with 2-week scheduled pickup since threshold (at least 1600) is being met.

8.4 Hypothesis Testing with the t Distribution

• Procedure for using Minitab for Exercise 8.13:

ÆClick on Stat > Basic Statistics > 1-Sample t. ÆEnter column where data is stored. Enter “1600” for test mean. ÆClick on Options. Change Confidence level to 90.0. ÆSelect “greater than” for alternative.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 36 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.4 Hypothesis Testing with the t Distribution

Minitab Output

T-Test of the Mean

Test of µ = 1600 vs. µ > 1600

Variable N Mean StDev SE Mean Bound T P Cubic Feet 18 1718.33 137.77 32.47 1661.85 3.64 0.001 © s 137.77 = n 18

8.4 Hypothesis Testing with the t Distribution

8.13: b. Place an upper bound on the p-value, i.e. state “p-value is less than ‘number’.” Would you say that µ > 1600 is strongly supported? • Table 4 is not needed to place upper bound. • From MTB output, p-value = .001 • 0.001 is the chance of obtaining a T-statistic > 3.64 if

H0: µ = 1600 is true.

• Since .001 < .10, reject H0 (universal rejection region) • “Very small p-values indicate strong, conclusive evidence for rejecting the null hypothesis.” (Hildebrand, Ott and Gray)

8.4 Hypothesis Testing with the t Distribution

Exercise 8.13 (continued): Were the requirements met to use t-test? • Random sample? Stated • Data from a normally distributed population?

• The NPP (next slide) is linear

• The p-value for test of normality = .921

• Do not reject hypothesis that data came from a normal distribution

Conclusion: Requirements met.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 39 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.4 Hypothesis Testing with the t Distribution

NPP for Exercise 8.13

Normal Probability Plot for Exercise 8.13 Normal

99 Mean 1718 StD ev 137.8 95 N18 A D 0.169 90 P-Value 0.921 80 70 60 50 40 Percent 30 20

1 1400 1500 1600 1700 1800 1900 2000 2100 Cubic Feet

Section 8.5 Assumptions for t Tests

8.5 Assumptions for t Tests

• The hypothesis tests in Chapter 8 require certain assumptions. • All of the tests require a random sample. • A biased sample is one that consistently yields units that differ from the true population for any number of reasons, including selection bias. • The testing procedures do not allow for bias in gathering the data.

• Another requirement is independence between the observations within the sample. • This requirement is frequently violated in time-series data, where the values of a variable are measured at successive points in time, e.g., monthly mortgage rates. • Serial dependence occurs when y1 is correlated with y2, y2 with y3, …, yn-1 with yn. • Consider a Z-test on a population mean. If the serial correlation is 0.4, n = 25 and the nominal level of significance is 5%, then the true probability of Type 1 error is approximately 14%. • For positive serial correlation, the true probability of Type 1 error is greater than the stated level of significance. This is a serious problem.

8.5 Assumptions for t Tests

• Another assumption for the t-test is that the underlying population be normally distributed.

• “In practice, no population is exactly normal.” (Hildebrand, Ott & Gray)

8.5 Assumptions for t Tests

• The consequences of nonnormality on the t-test depend on the type of nonnormality. • If the distribution of the population is skewed, the nominal α and p-value probabilities may be in error, particularly for one-tailed tests. • If the distribution of the population is symmetric, but heavy-tailed, the nominal α value and p-value are reasonably accurate. • When the distribution of the population is symmetric, but heavy-tailed, more efficient procedures are recommended, such as the median test.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 45 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Section 8.6 Testing a Proportion: Normal Approximation

8.6 Testing a Proportion: Normal Approximation

• Tests on a population proportion (π):

• H0: π = π0 • π is population proportion.

• π0 is a specific value.

8.6 Testing a Proportion: Normal Approximation

• Binomial random variable (Y) can be approximated by a normal random variable because of the Central Limit Theorem. • If Y is binomial, E(Y) = nπ and V(Y) = nπ (1 - π), Yn− π Z = nπ (1−π ) is approximately standard normal. • One version of Test Statistic

• Substitute π0 for π Yn− π Z = 0 nπ 00(1−π )

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 48 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.6 Testing a Proportion: Normal Approximation

• The rejection region (R.R.) is determined by the format of the research hypothesis. Research hypothesis Rejection criteria

Ha : π ≠ π 0 Reject H0 when z < -zα/2 or when z > zα/2 or

Ha : π > π 0 Reject H0 when z > zα or

Ha : π < π 0 Reject H0 when z < -zα

8.6 Testing a Proportion: Normal Approximation

• An equivalent version of test statistic • Base test on sample proportion (π )

π^ = Y N

• Test statistic [y − nπ ](1/ n) Z = 0 [ nπ 0 ()1 − π 0 ](1/ n)

Λ π − π = 0 π 0 ()1 − π 0 / n

8.6 Testing a Proportion: Normal Approximation

• Conditions for validity of the normal approximation:

• nπ^-5 ≥ 0 and • nπ^+ 5 ≤ n

• Use the test statistic only if these conditions are satisfied.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 51 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.6 Testing a Proportion: Normal Approximation

Exercises 8.21 and 8.23: A team of builders has surveyed buyers of their new homes for years. Consistently, only 41% of the buyers have indicated that they were “quite satisfied” or “very satisfied” with the construction quality of their homes. The builders have adopted a revised quality- inspection system to try to improve customer satisfaction. They have surveyed 104 buyers since then; these buyers seem representative, with no systematic changes from past purchasers. Of the 104 buyers, 51 indicated they were quite or very satisfied. a. Formulate the null hypothesis that there has been no real change in customer satisfaction from the past rate. π = proportion of satisfied customers after program went into effect.

H0 : π = .41

8.6 Testing a Proportion: Normal Approximation

b. Before taking such a survey, would you use a one- sided or two-sided research hypothesis?

Purpose of new quality system is to increase customer satisfaction. Use one-sided research hypothesis.

Ha : π > .41

c. Calculate a z-statistic for testing the null hypothesis. Λ π −π .4904−.41 T.S.: Z = 0 = =1.667 π0 ()1−π0 / n .41(.59)/104

8.6 Testing a Proportion: Normal Approximation

d. Can the null hypothesis be rejected, using α = .05?

R.R. : At the .05 level, reject H0 if Z > 1.645 (Table 3) R.R. 1.645 1.67

Conclusion: Reject H0 : π = .41 at the .05 level.

Interpretation: There is evidence that new quality program is moderately successful.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 54 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 8.6 Testing a Proportion: Normal Approximation

• Procedure for using Minitab for Exercise 8.21:

ÆClick on Stat >Basic Statistics >1Proportion ÆClick on "Summarized Data"; enter number of trials “104”; enter number of successes “51” ÆClick on Options; enter .41 for test proportion ÆChoose alternative "greater than" ÆClick on "use test based on normal distribution"

8.6 Testing a Proportion: Normal Approximation

Minitab Output

Test and Confidence Interval for One Proportion

Test of p = 0.41 vs. p > 0.41

Sample X N Sample p Bound Z-Value P-Value 1 51 104 0.490385 0.409754 1.67 0.048

P-value = .048 < α = .05

Conclusion: Reject H0.

8.6 Testing a Proportion: Normal Approximation

8.23: In Exercise 8.21 is the normal approximation reasonably accurate?

Conditions for validity of normal approximation: nπ^-5 ≥0 and nπ^ + 5 ≤ n

Is 104(51/104) – 5 ≥0? Is 104(51/104) + 5 ≤104? Yes Yes

Test assumes π^ is normally distributed.

Conditions were satisfied.

Normal approximation is reasonably accurate.

Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 8 57 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Section 8.7 Hypothesis Tests and Confidence Intervals

8.7 Hypothesis Tests and Confidence Intervals

• The Equivalence of Confidence Interval and Hypothesis Testing Procedures • Two Types of statistical inference: • Confidence Interval Estimation • Hypothesis Testing • Both use the following quantities: • Sample statistic ( Y or π ˆ ) • Standard error of the statistic • A significance level (α) or a confidence level (1-α) • A Z- or t-percentile corresponding to α, or a confidence level of (1-α)

8.7 Hypothesis Tests and Confidence Intervals

• The two procedures are very closely related. • It is in their use that they differ. • Confidence Interval Estimation is used when we do not know the parameter and want to estimate it. • Hypothesis Testing is used when we state the value for a parameter based on a scenario.

• Confidence Intervals can be used to perform Hypothesis Tests. • Hildebrand, Ott and Gray state the relationship as follows: “The equivalence of hypothesis tests and confidence intervals works for any sort of statistical test. Whether we’re doing a z test of a mean, a z test of a proportion, a t test of a mean, or any of the other procedures we will discuss in later chapters, we can reject a null hypothesis at a specified α (and have a p-value < α) if and only if a

100(1-α)% confidence interval does not include the H0 value.”

8.7 Hypothesis Tests and Confidence Intervals

Example (Bottling plant): • The hypotheses were:

H0: µ = 16.0

Ha: µ ≠ 16.0 • This test was based on the following values: Y = 16.15 n = 25 σ = 0.25 α = 0.05

8.7 Hypothesis Tests and Confidence Intervals

• The sample mean and σ are used to calculate the Z statistic.

• Values of Z that are within ±1.96 will fail to reject H0.

• Value of Z that are outside of ±1.96 will reject H0. • Equivalently, the nonrejection region can be stated as: Y − µ − 1.96 ≤ 0 ≤ +1.96 σ n

• This expression can be solved for µ: σ σ Y − 1.96 ≤ µ ≤ Y + 1.96 n n

Example (Bottling plant):

⎛ 0.25⎞ ⎛ 0.25⎞ 16.15−1.96⎜ ⎟ ≤ µ ≤16.15+1.96⎜ ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ or, 16.05 ≤ µ ≤16.25

8.7 Hypothesis Tests and Confidence Intervals

• The resulting confidence interval, 16.05 ≤ µ ≤16.25 does not include the hypothesized value,

µ0 =16.0

• As a result, reject H0 and conclude that the true mean is significantly higher than 16.0.

Keywords: Chapter 8

• Hypothesis test • Probability of Type I error (α) • Null hypothesis • Probability of Type II • Research hypothesis error (β ) • One-sided hypothesis • Power of the test (1-β ) • Two-sided hypothesis • p-value • Test statistic • Statistical significance • Critical value • z-test of a mean • Rejection region • t-test of a mean • Test of a proportion

• The concept of hypothesis testing • Type I and Type II errors • The probability of making a Type I error (α ) • The probability of making a Type II error (β ) • The power of a test

• The universal rejection region: reject H0 if p-value < α • The steps in the hypothesis testing procedure: • State the null and research hypothesis • State the test statistic and its value • State the rejection region • State your conclusion

Summary of Chapter 8

• Hypothesis testing on a mean (known population standard deviation) using the Z-test • Assumptions for this test • Hypothesis testing on a mean using the t-test • Assumptions for this test • Hypothesis testing on a proportion • Assumptions for this test • The equivalence of Confidence Interval and Hypothesis Testing procedures • Testing a hypothesis by using a Confidence Interval

Summary of Chapter 8

Tests of Hypotheses on a Mean When Sampling from a Normal Distribution One-Sample Tests • Testing a hypothesis on the mean of a normal distribution with known variance.

Null hypothesis Value of test statistic under H0

y - µ0 H0 : µ = µ0 Z = σ / n

Alternative hypothesis Rejection region

Ha : µ ≠ µ0 Reject H0 when z < -zα/2 or when z > zα/2

Ha : µ > µ0 Reject H0 when z > zα

Ha : µ < µ0 Reject H0 when z < -zα

Summary of Chapter 8

Alternative hypothesis Probability of Type II Error ( β )

µ −µ µ −µ H : µ ≠ µ ⎛ 0 a 0 a ⎞ a 0 β =P⎜-zα/2 + ≤Z≤ zα/ 2 + ⎟ ⎝ σ/ n σ/ n ⎠

⎛ µ0 −µa ⎞ Ha : µ > µ0 β =P⎜Z ≤ zα + ⎟ ⎝ σ / n ⎠

µ −µ H : µ < µ ⎛ 0 a ⎞ a 0 β =P⎜Z ≥ -zα + ⎟ ⎝ σ / n ⎠

Summary of Chapter 8

A 100(1 - α)% confidence interval for µ is: σ y ± z α /2 n

• Testing a hypothesis on the mean of a normal distribution with unknown variance.

Null hypothesis Value of test statistic under H0

y − µ0 H0 : µ = µ0 t = s / n

Ha : µ ≠ µ0 Reject H0 when t < -tα/2, n-1 or when t > tα/2, n-1

Ha : µ > µ0 Reject H0 when t > tα, n-1

Ha : µ < µ0 Reject H0 when t < -tα, n-1

Summary of Chapter 8

A 100(1 - α)% confidence interval for µ is: s y ±t α /2,n−1 n

Summary of Chapter 8

Test of Hypotheses on a Proportion

Null hypothesis Value of test statistic under H0

y − nπ 0 H0 : π = π 0 z = nπ 0 ()1 − π 0

Λ π − π = 0 π 0 ()1 − π 0 / n

y is the number of successes in a sample of size n

Alternative hypothesis Rejection region

Ha : π ≠ π0 Reject H0 when z < -zα/2 or when z > zα/2

Ha : π > π0 Reject H0 when z > zα

Ha : π < π0 Reject H0 when z < -zα

Summary of Chapter 8

A 100(1 - α)% confidence interval for π is:

Λ Λ ⎛ Λ ⎞ π ± z π ⎜1 − π ⎟ / n , α / 2 ⎝ ⎠

Λ where π = y/n, the sample proportion.