Statistics for Business and Economics Chapter 6 Inferences

Statistics for Business and Economics Chapter 6 Inferences Based on a Single Sample: Tests of Hypothesis

Learning Objectives 1. Distinguish Types of Hypotheses 2. Describe Hypothesis Testing Process 3. Explain p-Value Concept 4. Solve Hypothesis Testing Problems Based on a Single Sample 5. Explain Power of a Test

Statistical Methods Descriptive Statistics Inferential Statistics Estimation Hypothesis Testing

Hypothesis Testing Concepts

Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Random sample Mean X = 20 Reject hypothesis! Not close.

What’s a Hypothesis? A belief about a population parameter I believe the mean GPA of this class is 3. 5! • Parameter is population mean, proportion, variance • Must be stated before analysis © 1984 -1994 T/Maker Co.

Null Hypothesis 1. 2. 3. 4. 5. What is tested Has serious outcome if incorrect decision made Always has equality sign: , , or Designated H 0 (pronounced H-oh) Specified as H 0: some numeric value • • Specified with = sign even if or Example, H 0: 3

Alternative Hypothesis 1. Opposite of null hypothesis 2. Always has inequality sign: , , or 3. Designated Ha 4. Specified Ha: , , or some value • Example, Ha: < 3

Identifying Hypotheses Steps Example problem: Test that the population mean is not 3 Steps: • State the question statistically ( 3) • State the opposite statistically ( = 3) — Must be mutually exclusive & exhaustive • Select the alternative hypothesis ( 3) — Has the , <, or > sign • State the null hypothesis ( = 3)

What Are the Hypotheses? Is the population average amount of TV viewing 12 hours? • State the question statistically: = 12 • State the opposite statistically: 12 • Select the alternative hypothesis: Ha: 12 • State the null hypothesis: H 0: = 12

What Are the Hypotheses? Is the population average amount of TV viewing different from 12 hours? • State the question statistically: 12 • State the opposite statistically: = 12 • Select the alternative hypothesis: Ha: 12 • State the null hypothesis: H 0: = 12

What Are the Hypotheses? Is the average cost per hat less than or equal to $20? • State the question statistically: 20 • State the opposite statistically: 20 • Select the alternative hypothesis: Ha: 20 • State the null hypothesis: H 0: 20

What Are the Hypotheses? Is the average amount spent in the bookstore greater than $25? • State the question statistically: 25 • State the opposite statistically: 25 • Select the alternative hypothesis: Ha: 25 • State the null hypothesis: H 0: 25

Basic Idea Sampling Distribution It is unlikely that we would get a sample mean of this value. . . therefore, we reject the hypothesis that = 50. . if in fact this were the population mean 20 = 50 H 0 Sample Means

Level of Significance 1. Probability 2. Defines unlikely values of sample statistic if null hypothesis is true • Called rejection region of sampling distribution 3. Designated (alpha) • Typical values are. 01, . 05, . 10 4. Selected by researcher at start

Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1– Nonrejection Region Critical Value Ho Value Sample Statistic Observed sample statistic

Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1– Nonrejection Region Ho Value Critical Value Observed sample statistic Sample Statistic

Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 Rejection Region 1– Nonrejection Region Critical Value 1/2 Ho Sample Statistic Value Critical Value Observed sample statistic

Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 Rejection Region 1– Nonrejection Region 1/2 Ho Sample Statistic Critical Value Observed sample statistic

Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 Rejection Region 1– Nonrejection Region 1/2 Ho Sample Statistic Critical Value Observed sample statistic

Decision Making Risks

Errors in Making Decision 1. Type I Error • • • Reject true null hypothesis Has serious consequences Probability of Type I Error is (alpha) — Called level of significance 2. Type II Error • • Do not reject false null hypothesis Probability of Type II Error is (beta)

Decision Results H 0: Innocent H 0 Test Jury Trial Actual Situation Verdict Innocent Guilty Actual Situation Decision Correct Error Accept H 0 Error Correct Reject H 0 True H 0 False 1– Type II Error ( ) Type I Power Error ( ) (1 – )

& Have an Inverse Relationship You can’t reduce both errors simultaneously!

Factors Affecting 1. True value of population parameter • Increases when difference with hypothesized parameter decreases 2. Significance level, • Increases when decreases 3. Population standard deviation, • Increases when increases 4. Sample size, n • Increases when n decreases

Hypothesis Testing Steps

H 0 Testing Steps • State H 0 • Set up critical values • State Ha • Collect data • Choose • Compute test statistic • Choose n • Make statistical decision • Choose test • Express decision

One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test 2 Test (1 & 2 tail)

Two-Tailed Z Test of Mean ( Known)

One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test 2 Test (1 & 2 tail)

Two-Tailed Z Test for Mean ( Known) 1. Assumptions • • Population is normally distributed If not normal, can be approximated by normal distribution (n 30) 2. Alternative hypothesis has sign 3. Z-Test Statistic

Two-Tailed Z Test for Mean Hypotheses H 0: = 0 Ha: ≠ 0 Reject H / 2 0 Z

Two-Tailed Z Test Finding Critical Z What is Z given =. 05? . 500 -. 025. 475 =1 / 2 =. 025 -1. 96 0 1. 96 Z Standardized Normal Probability Table (Portion) Z . 05 . 06 . 07 1. 6. 4505. 4515. 4525 1. 7. 4599. 4608. 4616 1. 9. 4744. 4750. 4756 1. 8. 4678. 4686. 4693

Two-Tailed Z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372. 5. The company has specified to be 25 grams. Test at the. 05 level of significance. 368 gm.

Two-Tailed Z Test Solution • • • H 0: = 368 Ha: 368 . 05 n 25 Critical Value(s): Reject H 0 . 025 -1. 96 0 1. 96 Z Test Statistic: Decision: Do not reject at =. 05 Conclusion: No evidence average is not 368

Two-Tailed Z Test Thinking Challenge You’re a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with = 3. 5 lb. You take a sample of 36 cords & compute a sample mean of 69. 7 lb. At the. 05 level of significance, is there evidence that the machine is not meeting the average breaking strength?

Two-Tailed Z Test Solution* • • • H 0: = 70 Ha: 70 =. 05 n = 36 Critical Value(s): Reject H 0 . 025 -1. 96 0 1. 96 Z Test Statistic: Decision: Do not reject at =. 05 Conclusion: No evidence average is not 70

One-Tailed Z Test of Mean ( Known)

One-Tailed Z Test for Mean ( Known) 1. Assumptions • • Population is normally distributed If not normal, can be approximated by normal distribution (n 30) 2. Alternative hypothesis has < or > sign 3. Z-test Statistic

One-Tailed Z Test for Mean Hypotheses H 0: = 0 Ha: < 0 H 0: = 0 Ha: > 0 Reject H 0 0 Must be significantly below Z 0 Z Small values satisfy H 0. Don’t reject!

One-Tailed Z Test Finding Critical Z What Is Z given =. 025? . 500 -. 025. 475 =1 =. 025 0 1. 96 Z Standardized Normal Probability Table (Portion) Z . 05 . 06 . 07 1. 6. 4505. 4515. 4525 1. 7. 4599. 4608. 4616 1. 9. 4744. 4750. 4756 1. 8. 4678. 4686. 4693

One-Tailed Z Test Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372. 5. The company has specified to be 25 grams. Test at the. 05 level of significance. 368 gm.

One-Tailed Z Test Solution • • • H 0: = 368 Ha: > 368 =. 05 n = 25 Critical Value(s): Reject . 05 0 1. 645 Z Test Statistic: Decision: Do not reject at =. 05 Conclusion: No evidence average is more than 368

One-Tailed Z Test Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3. 8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30. 7 mpg. At the. 01 level of significance, is there evidence that the miles per gallon is at least 32?

One-Tailed Z Test Solution* • • • H 0: = 32 Ha: < 32 =. 01 n = 60 Critical Value(s): Reject . 01 -2. 33 0 Z Test Statistic: Decision: Reject at =. 01 Conclusion: There is evidence average is less than 32

Observed Significance Levels: p-Values

p-Value 1. Probability of obtaining a test statistic more extreme ( or than actual sample value, given H 0 is true 2. Called observed level of significance • Smallest value of for which H 0 can be rejected 3. Used to make rejection decision • • If p-value , do not reject H 0 If p-value < , reject H 0

Two-Tailed Z Test p-Value Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372. 5. The company has specified to be 25 grams. Find the p-Value. 368 gm.

Two-Tailed Z Test p-Value Solution 0 1. 50 Z Z value of sample statistic (observed)

Two-Tailed Z Test p-Value Solution p-value is P(Z -1. 50 or Z 1. 50) 1/2 p-Value. 4332 -1. 50 0 From Z table: lookup 1. 5000 -. 4332. 0668 Z Z value of sample statistic (observed)

Two-Tailed Z Test p-Value Solution p-value is P(Z -1. 50 or Z 1. 50) =. 1336 1/2 p-Value. 0668 -1. 50 1/2 p-Value. 0668 0 From Z table: lookup 1. 5000 -. 4332. 0668 Z Z value of sample statistic

Two-Tailed Z Test p-Value Solution (p-Value =. 1336) ( =. 05). Do not reject H 0. 1/2 p-Value =. 0668 Reject H 0 1/2 =. 025 -1. 50 0 1. 50 Test statistic is in ‘Do not reject’ region Z

One-Tailed Z Test p-Value Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372. 5. The company has specified to be 25 grams. Find the p. Value. 368 gm.

One-Tailed Z Test p-Value Solution 0 1. 50 Z Z value of sample statistic

One-Tailed Z Test p-Value Solution p-Value is P(Z 1. 50) Use alternative hypothesis to find direction p-Value. 4332 0 From Z table: lookup 1. 5000 -. 4332. 0668 Z Z value of sample statistic

One-Tailed Z Test p-Value Solution p-Value is P(Z 1. 50) =. 0668 p-Value. 0668 Use alternative hypothesis to find direction . 4332 0 From Z table: lookup 1. 5000 -. 4332. 0668 Z Z value of sample statistic

One-Tailed Z Test p-Value Solution (p-Value =. 0668) ( =. 05). Do not reject H 0. p-Value =. 0668 Reject H 0 =. 05 0 1. 50 Test statistic is in ‘Do not reject’ region Z

p-Value Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3. 8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30. 7 mpg. What is the value of the observed level of significance (p-Value)?

p-Value Solution* p-Value is P(Z -2. 65) =. 004. p-Value < ( =. 01). Reject H 0. Use alternative hypothesis to find direction p-Value. 004 . 5000 -. 4960. 0040 . 4960 -2. 65 Z value of sample statistic 0 Z From Z table: lookup 2. 65

Two-Tailed t Test of Mean ( Unknown)

One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test 2 Test (1 & 2 tail)

t Test for Mean ( Unknown) 1. Assumptions • Population is normally distributed • If not normal, only slightly skewed & large sample (n 30) taken 2. Parametric test procedure 3. t test statistic

Two-Tailed t Test Finding Critical t Values Given: n = 3; =. 10 df = n - 1 = 2 /2 =. 05 Critical Values of t Table (Portion) v t. 10 t. 05 t. 025 1 3. 078 6. 314 12. 706 2 1. 886 2. 920 4. 303 -2. 920 0 2. 920 t 3 1. 638 2. 353 3. 182

Two-Tailed t Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372. 5 and a standard deviation of 12 grams. Test at the. 05 level of significance. 368 gm.

Two-Tailed t Test Solution • • • H 0: = 368 Ha: 368 =. 05 df = 36 - 1 = 35 Critical Value(s): Reject H 0 . 025 -2. 030 0 2. 030 t Test Statistic: Decision: Reject at =. 05 Conclusion: There is evidence population average is not 368

Two-Tailed t Test Thinking Challenge You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3. 25 lb. You take a random sample of 64 containers. You calculate the sample average to be 3. 238 lb. with a standard deviation of. 117 lb. At the. 01 level of significance, is the manufacturer correct? 3. 25 lb.

Two-Tailed t Test Solution* • • • H 0: = 3. 25 Ha: 3. 25 . 01 df 64 - 1 = 63 Critical Value(s): Reject H 0 . 005 -2. 656 0 2. 656 t Test Statistic: Decision: Do not reject at =. 01 Conclusion: There is no evidence average is not 3. 25

One-Tailed t Test of Mean ( Unknown)

One-Tailed t Test Example Is the average capacity of batteries at least 140 amperehours? A random sample of 20 batteries had a mean of 138. 47 and a standard deviation of 2. 66. Assume a normal distribution. Test at the. 05 level of significance.

One-Tailed t Test Solution • • • H 0: = 140 Ha: < 140 =. 05 df = 20 - 1 = 19 Critical Value(s): Reject H 0 . 05 -1. 729 0 t Test Statistic: Decision: Reject at =. 05 Conclusion: There is evidence population average is less than 140

One-Tailed t Test Thinking Challenge You’re a marketing analyst for Wal. Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3 At the. 05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?

One-Tailed t Test Solution* • • • H 0: = 5 Ha: > 5 =. 05 df = 10 - 1 = 9 Critical Value(s): Reject H 0. 05 0 1. 833 t Test Statistic: Decision: Do not reject at =. 05 Conclusion: There is no evidence average is more than 5

Z Test of Proportion

Data Types Data Quantitative Discrete Continuous Qualitative

Qualitative Data 1. Qualitative random variables yield responses that classify • e. g. , Gender (male, female) 2. Measurement reflects number in category 3. Nominal or ordinal scale 4. Examples • • Do you own savings bonds? Do you live on-campus or off-campus?

Proportions 1. Involve qualitative variables 2. Fraction or percentage of population in a category 3. If two qualitative outcomes, binomial distribution • Possess or don’t possess characteristic ^ 4. Sample Proportion (p)

Sampling Distribution of Proportion 1. Approximated by Normal Distribution Sampling Distribution ^ P(P ) – Excludes 0 or n 2. Mean . 3. 2. 1. 0 ^ P. 0 . 2 . 4 . 6 . 8 1. 0 3. Standard Error where p 0 = Population Proportion

Standardizing Sampling Distribution of Proportion Z p^ p^ p 0 (1 p 0) n Sampling Distribution P^ Standardized Normal Distribution z = 1 P^ ^ P Z= 0 Z

One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test 2 Test (1 & 2 tail)

One-Sample Z Test for Proportion 1. Assumptions • Random sample selected from a binomial population • Normal approximation can be used if 2. Z-test statistic for proportion Hypothesized population proportion

One-Proportion Z Test Example The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had 11 defects. Does the new system produce fewer defects? Test at the. 05 level of significance.

One-Proportion Z Test Solution • • • H 0: p =. 10 Ha: p <. 10 =. 05 n = 200 Critical Value(s): Reject H 0 . 05 -1. 645 0 Z Test Statistic: Decision: Reject at =. 05 Conclusion: There is evidence new system < 10% defective

One-Proportion Z Test Thinking Challenge You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the. 05 level of significance?

One-Proportion Z Test Solution* • • • H 0: p =. 04 Ha: p . 04 =. 05 n = 500 Critical Value(s): Reject H 0 . 025 -1. 96 0 1. 96 Z Test Statistic: Decision: Do not reject at =. 05 Conclusion: There is evidence proportion is not 4%

Calculating Type II Error Probabilities

Power of Test 1. Probability of rejecting false H 0 • Correct decision 2. Designated 1 - 3. Used in determining test adequacy 4. Affected by • • • True value of population parameter Significance level Standard deviation & sample size n

Finding Power Step 1 Hypothesis: H 0: 0 368 Ha: 0 < 368 Reject H 0 =. 05 Do Not Draw Reject H 0 0 = 368 X

Finding Power Steps 2 & 3 Reject H 0 Hypothesis: H 0: 0 368 Ha: 0 < 368 Do Not Draw Reject H 0 =. 05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify Draw 1 - a = 360 X X

Finding Power Step 4 Reject H 0 Hypothesis: H 0: 0 368 Ha: 0 < 368 Do Not Draw Reject H 0 =. 05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify X Draw 1 - a = 360 363. 065 X

Finding Power Step 5 Reject H 0 Hypothesis: H 0: 0 368 Ha: 0 < 368 Do Not Draw Reject H 0 =. 05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify Z Table X Draw =. 154 1 - =. 846 a = 360 363. 065 X

Power Curves Power H 0: 0 Possible True Values for a Power Possible True Values for a H 0: = 0 Possible True Values for a = 368 in Example

Chi-Square ( 2) Test of Variance

One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test 2 Test (1 & 2 tail)

2 ( ) Chi-Square Test for Variance 1. Tests one population variance or standard deviation 2. Assumes population is approximately normally distributed 3. Null hypothesis is H 0: 2 = 02 4. Test statistic 2 (n 1) S 2 2 0 Sample variance Hypothesized pop. variance

Chi-Square ( 2) Distribution Population Select simple random sample, size n. Compute s 2 Sampling Distributions for Different Sample Sizes Compute 2 = (n-1)s 2 / 2 0 Astronomical number of 2 values 1 2 3 2

Finding Critical Value Example What is the critical 2 value given: Ha: 2 > 0. 7 Reject n=3 =. 05? =. 05 df = n - 1 = 2 2 Table (Portion) 0 DF. 995 1. . . 2 0. 010 5. 991 2 Upper Tail Area …. 95 … … 0. 004 … … 0. 103 … . 05 3. 841 5. 991

Finding Critical Value Example What is the critical 2 value given: Ha: 2 < 0. 7 n=3 What do you do =. 05? if the rejection region is on the left?

Finding Critical Value Example What is the critical 2 value given: Ha: 2 < 0. 7 Upper Tail Area Reject H 0 for Lower Critical n=3 Value = 1 -. 05 =. 95 =. 05? df = n - 1 = 2 2 Table (Portion) 0. 103 DF. 995 1. . . 2 0. 010 2 Upper Tail Area …. 95 … … 0. 004 … … 0. 103 … . 05 3. 841 5. 991

Chi-Square ( 2) Test Example Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25 boxes had a standard deviation of 17. 7 grams. Test at the. 05 level of significance.

2 ( ) Chi-Square Test Solution • • • H 0: 2 = 15 Ha: 2 15 =. 05 df = 25 - 1 = 24 Critical Value(s): /2 =. 025 0 12. 401 39. 364 2 Test Statistic: = 33. 42 Decision: Do not reject at =. 05 Conclusion: There is no evidence 2 is not 15

Conclusion 1. Distinguished Types of Hypotheses 2. Described Hypothesis Testing Process 3. Explained p-Value Concept 4. Solved Hypothesis Testing Problems Based on a Single Sample 5. Explained Power of a Test