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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 Statistical Methods Descriptive Statistics Inferential Statistics Estimation Hypothesis Testing Nonstatistical Hypothesis Testing A criminal trial is an example of hypothesis testing without the statistics. In a trial a jury must decide between two hypotheses. The null hypothesis is H0: The defendant is innocent The alternative hypothesis or research hypothesis is H1: The defendant is guilty The jury does not know which hypothesis is true. They must make a decision on the basis of evidence presented. Nonstatistical Hypothesis Testing In the language of statistics convicting the defendant is called rejecting the null hypothesis in favor of the alternative hypothesis. That is, the jury is saying that there is enough evidence to conclude that the defendant is guilty (i.e., there is enough evidence to support the alternative hypothesis). Nonstatistical Hypothesis Testing If the jury acquits it is stating that there is not enough evidence to support the alternative hypothesis. Notice that the jury is not saying that the defendant is innocent, only that there is not enough evidence to support the alternative hypothesis. That is why we never say that we accept the null hypothesis. Nonstatistical Hypothesis Testing There are two possible errors. A Type I error occurs when we reject a true null hypothesis. That is, a Type I error occurs when the jury convicts an innocent person. A Type II error occurs when we don’t reject a false null hypothesis. That occurs when a guilty defendant is acquitted. Nonstatistical Hypothesis Testing The probability of a Type I error is denoted as α (Greek letter alpha). The probability of a type II error is β (Greek letter beta). The two probabilities are inversely related. Decreasing one increases the other. Nonstatistical Hypothesis Testing 5. Two possible errors can be made. Type I error: Reject a true null hypothesis Type II error: Do not reject a false null hypothesis. P(Type I error) = α P(Type II error) = β Decision Results H0: Innocent Jury Trial Actual Situation Verdict Innocent Guilty Innocent Guilty H0 Test Actual Situation Decision Correct Error Accept H0 Error Correct Reject H0 H0 True H0 False 1–a Type II Error (b) Type I Power Error (a) (1 – b) a & b Have an Inverse Relationship You can’t reduce both errors simultaneously! b a Nonstatistical Hypothesis Testing In our judicial system Type I errors are regarded as more serious. We try to avoid convicting innocent people. We are more willing to acquit guilty people. We arrange to make α small by requiring the prosecution to prove its case and instructing the jury to find the defendant guilty only if there is “evidence beyond a reasonable doubt.” Nonstatistical Hypothesis Testing The critical concepts are theses: 1. There are two hypotheses, the null and the alternative hypotheses. 2. The procedure begins with the assumption that the null hypothesis is true. 3. The goal is to determine whether there is enough evidence to infer that the alternative hypothesis is true. 4. There are two possible decisions: Conclude that there is enough evidence to support the alternative hypothesis. Conclude that there is not enough evidence to support the alternative hypothesis. Concepts of Hypothesis Testing (1) There are two hypotheses. One is called the null hypothesis and the other the alternative or research hypothesis. The usual notation is: pronounced H “nought” H0: — the ‘null’ hypothesis H1: — the ‘alternative’ or ‘research’ hypothesis The null hypothesis (H0) will always state that the parameter equals the value specified in the alternative hypothesis (H1) Concepts of Hypothesis Testing Consider the following example. Suppose our operations manager wants to know whether the mean is different from 350 units. We can rephrase this request into a test of the hypothesis: H0:µ = 350 Thus, our research hypothesis becomes: This is what we are interested in H1:µ ≠ 350 determining… Concepts of Hypothesis Testing (2) The testing procedure begins with the assumption that the null hypothesis is true. Thus, until we have further statistical evidence, we will assume: H0: = 350 (assumed to be TRUE) Concepts of Hypothesis Testing (3) The goal of the process is to determine whether there is enough evidence to infer that the alternative hypothesis is true. That is, is there sufficient statistical information to determine if this statement is true? This is what we are interested in determining… H1:µ ≠ 350 Concepts of Hypothesis Testing (4) There are two possible decisions that can be made: Conclude that there is enough evidence to support the alternative hypothesis (also stated as: rejecting the null hypothesis in favor of the alternative) Conclude that there is not enough evidence to support the alternative hypothesis (also stated as: not rejecting the null hypothesis in favor of the alternative) NOTE: we do not say that we accept the null hypothesis… Concepts of Hypothesis Testing Once the null and alternative hypotheses are stated, the next step is to randomly sample the population and calculate a test statistic (in this example, the sample mean). If the test statistic’s value is inconsistent with the null hypothesis we reject the null hypothesis and infer that the alternative hypothesis is true. 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 H0 (pronounced H-oh) Specified as H0: some numeric value • • Specified with = sign even if or Example, H0: 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) — • Select the alternative hypothesis ( 3) — • Must be mutually exclusive & exhaustive 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: H0: = 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: H0: = 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: H0: 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: H0: 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 H0 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 a(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–a a Nonrejection Region Critical Value Ho Value Sample Statistic Observed sample statistic Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region a 1–a Nonrejection Region Ho Value Critical Value Observed sample statistic Sample Statistic Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region Rejection Region 1–a 1/2 a 1/2 a Nonrejection Region Critical Value Ho Value Sample Statistic Critical Value Observed sample statistic Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 a Rejection Region 1–a 1/2 a Nonrejection Region Ho Value Critical Critical Value Value Observed sample statistic Sample Statistic Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 a Rejection Region 1–a 1/2 a Nonrejection Region Ho Value Critical Value Observed sample statistic Critical Value Sample Statistic Decision Making Risks Decision Results H0: Innocent Jury Trial Actual Situation Verdict Innocent Guilty Innocent Guilty H0 Test Actual Situation Decision Correct Error Accept H0 Error Correct Reject H0 H0 True H0 False 1–a Type II Error (b) Type I Power Error (a) (1 – b) Errors in Making Decision 1. Type I Error • • • Reject true null hypothesis Has serious consequences Probability of Type I Error is a(alpha) — Called level of significance 2. Type II Error • • Do not reject false null hypothesis Probability of Type II Error is b(beta) Factors Affecting b 1. True value of population parameter • Increases when difference with hypothesized parameter decreases 2. Significance level, a • Increases when adecreases 3. Population standard deviation, • Increases when increases 4. Sample size, n • Increases when n decreases Hypothesis Testing Steps H0 Testing Steps • State H0 • Set up critical values • State Ha • Collect data • Choose a • 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 c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) Two-Tailed Z Test of Mean ( Known) One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (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 Z X x x X n Two-Tailed Z Test for Mean Hypotheses H0:= 0 Ha: ≠ 0 Reject H 0 Reject H a/2 a/2 0 Z Two-Tailed Z Test Finding Critical Z What is Z given a = .05? .500 - .025 .475 =1 a /2 = .025 Standardized Normal Probability Table (Portion) Z .05 .06 .07 1.6 .4505 .4515 .4525 1.7 .4599 .4608 .4616 1.8 .4678 .4686 .4693 -1.96 0 1.96 Z 1.9 .4744 .4750 .4756 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 • • • • • H0: = 368 Ha: 368 a .05 n 25 Critical Value(s): Reject H 0 Reject H 0 .025 .025 -1.96 0 1.96 Z Test Statistic: Decision: Conclusion: Two-Tailed Z Test Solution Test Statistic: X 372.5 368 Z 1.50 15 n 25 Decision: Do not reject at a = .05 Conclusion: No evidence average is not 368 Example 11.2 In recent years, a number of companies have been formed that offer competition to AT&T in long-distance calls. All advertise that their rates are lower than AT&T's, and as a result their bills will be lower. AT&T has responded by arguing that for the average consumer there will be no difference in billing. Suppose that a statistics practitioner working for AT&T determines that the mean and standard deviation of monthly long-distance bills for all its residential customers are $17.09 and $3.87, respectively. Example 11.2 He then takes a random sample of 100 customers and recalculates their last month's bill using the rates quoted by a leading competitor. Assuming that the standard deviation of this population is the same as for AT&T, can we conclude at the 5% significance level that there is a difference between AT&T's bills and those of the leading competitor? IDENTIFY Example 11.2 The parameter to be tested is the mean of the population of AT&T’s customers’ bills based on competitor’s rates. What we want to determine whether this mean differs from $17.09. Thus, the alternative hypothesis is H1: µ ≠ 17.09 The null hypothesis automatically follows. H0: µ = 17.09 IDENTIFY Example 11.2 The rejection region is set up so we can reject the null hypothesis when the test statistic is large or when it is small. stat is “small” stat is “large” That is, we set up a two-tail rejection region. The total area in the rejection region must sum to , so we divide this probability by 2. IDENTIFY Example 11.2 At a 5% significance level (i.e. α = .05), we have α/2 = .025. Thus, z.025 = 1.96 and our rejection region is: z < –1.96 -z.025 -or- 0 z > 1.96 +z.025 z 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* • • • • • H0: = 70 Ha: 70 a =.05 n = 36 Critical Value(s): Reject H 0 Reject H 0 .025 .025 -1.96 0 1.96 Z Test Statistic: Decision: Conclusion: Two-Tailed Z Test Solution* Test Statistic: X 69.7 70 Z .51 3.5 n 36 Decision: Do not reject at a = .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 Z X x x X n One-Tailed Z Test for Mean Hypotheses H0:= 0 Ha: < 0 H0:= 0 Ha: > 0 Reject H 0 Reject H0 a a 0 Must be significantly below Z 0 Z Small values satisfy H0 . Don’t reject! One-Tailed Z Test Finding Critical Z What Is Z given a = .025? .500 - .025 .475 =1 a = .025 Standardized Normal Probability Table (Portion) Z .05 .06 .07 1.6 .4505 .4515 .4525 1.7 .4599 .4608 .4616 1.8 .4678 .4686 .4693 0 1.96 Z 1.9 .4744 .4750 .4756 Example 11.1 The manager of a department store is thinking about establishing a new billing system for the store's credit customers. She determines that the new system will be costeffective only if the mean monthly account is more than $170. A random sample of 400 monthly accounts is drawn, for which the sample mean is $178. The manager knows that the accounts are approximately normally distributed with a standard deviation of $65. Can the manager conclude from this that the new system will be cost-effective? Example 11.1 IDENTIFY The system will be cost effective if the mean account balance for all customers is greater than $170. We express this belief as our research hypothesis, that is: H1: µ > 170 (this is what we want to determine) Thus, our null hypothesis becomes: H0: µ = 170 (this specifies a single value for the parameter of interest) Example 11.1 IDENTIFY What we want to show: H0: µ = 170 (we’ll assume this is true) H1: µ > 170 We know: n = 400, = 178, and σ = 65 What to do next?! COMPUTE Example 11.1 Rejection region It seems reasonable to reject the null hypothesis in favor of the alternative if the value of the sample mean is large relative to 170, that is if > . α = P(Type I error) = P( reject H0 given that H0 is true) α = P( > ) COMPUTE Example 11.1 All that’s left to do is calculate and compare it to 170. we can calculate this based on any level of significance ( ) we want… COMPUTE Example 11.1 At a 5% significance level (i.e. =0.05), we get Solving we compute = 175.34 Since our sample mean (178) is greater than the critical value we calculated (175.34), we reject the null hypothesis in favor of H1, i.e. that: µ > 170 and that it is cost effective to install the new billing system 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 15 grams. Test at the .05 level of significance. 368 gm. One-Tailed Z Test Solution • • • • • H0: = 368 Ha: > 368 a = .05 n = 25 Critical Value(s): Test Statistic: Decision: Reject .05 0 1.645 Z Conclusion: One-Tailed Z Test Solution Test Statistic: Z X 372.5 368 1.50 15 n 25 Decision: Do not reject at a = .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 less than 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 less than 32? One-Tailed Z Test Solution* • • • • • H0: = 32 Ha: < 32 a = .01 n = 60 Critical Value(s): Test Statistic: Decision: Reject .01 -2.33 0 Conclusion: Z One-Tailed Z Test Solution* Test Statistic: X 30.7 32 Z 2.65 3.8 n 60 Decision: Reject at a = .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 H0 is true 2. Called observed level of significance • Smallest value of a for which H0 can be rejected 3. Used to make rejection decision • • If p-value a, do not reject H0 If p-value < a, reject H0 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 15 grams. Find the p-Value. 368 gm. Two-Tailed Z Test p-Value Solution X 372.5 368 Z 1.50 15 n 25 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 1/2 p-Value .4332 -1.50 0 From Z table: lookup 1.50 1.50 .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 1.50 From Z table: lookup 1.50 .5000 - .4332 .0668 Z Z value of sample statistic Two-Tailed Z Test p-Value Solution (p-Value = .1336) (a = .05). Do not reject H0. 1/2 p-Value = .0668 1/2 p-Value = .0668 Reject H0 Reject H0 1/2 a = .025 1/2 a = .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 pValue. 368 gm. One-Tailed Z Test p-Value Solution X 372.5 368 Z 1.50 15 n 25 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 1.50 1.50 .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 1.50 1.50 .5000 - .4332 .0668 Z Z value of sample statistic One-Tailed Z Test p-Value Solution (p-Value = .0668) (a = .05). Do not reject H0. p-Value = .0668 Reject H0 a = .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 less than 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 < (a = .01). Reject H0. 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 c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (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 X t S n Two-Tailed t Test Finding Critical t Values Given: n = 3; a = .10 df = n - 1 = 2 a /2 = .05 a /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 • • • • • H0: = 368 Ha: 368 a = .05 df = 36 - 1 = 35 Critical Value(s): Reject H0 Reject H0 .025 .025 -2.030 0 2.030 t Test Statistic: Decision: Conclusion: Two-Tailed t Test Solution Test Statistic: X 372.5 368 t 2.25 S 12 n 36 Decision: Reject at a = .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* • • • • • H0: = 3.25 Ha: 3.25 a .01 df 64 - 1 = 63 Critical Value(s): Reject H 0 Reject H0 .005 .005 -2.656 0 2.656 t Test Statistic: Decision: Conclusion: Two-Tailed t Test Solution* Test Statistic: X 3.238 3.25 t .82 S .117 n 64 Decision: Do not reject at a = .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 • • • • • H0: = 140 Ha: < 140 a = .05 df = 20 - 1 = 19 Critical Value(s): Test Statistic: Decision: Reject H0 Conclusion: .05 -1.729 0 t One-Tailed t Test Solution Test Statistic: X 138.47 140 t 2.57 S 2.66 n 20 Decision: Reject at a = .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* • • • • • H0: = 5 Ha: > 5 a = .05 df = 10 - 1 = 9 Critical Value(s): Test Statistic: Decision: Reject H0 .05 0 1.833 Conclusion: t One-Tailed t Test Solution* Test Statistic: X 6.4 5 t 1.31 S 3.373 n 10 Decision: Do not reject at a = .05 Conclusion: There is no evidence average is more than 5 Summary of One- and Two-Tail Tests… One-Tail Test (left tail) Two-Tail Test One-Tail Test (right tail) 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) x number of successes pˆ n sample size Sampling Distribution of Proportion 1. Approximated by Normal Distribution npˆ 3 npˆ 1 pˆ – Excludes 0 or n 2. Mean P̂ p Sampling Distribution ^ P(P ) .3 .2 .1 .0 ^ P .0 .2 .4 .6 .8 1.0 3. Standard Error p0 (1 p0 ) pˆ where p0 = Population Proportion n Standardizing Sampling Distribution of Proportion Z p^ p^ p^ p^ p0 p0 (1 p0) 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 c 2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) One-Sample Z Test for Proportion 1. Assumptions • Random sample selected from a binomial population • Normal approximation can be used if npˆ 0 15 and nqˆ0 15 2. Z-test statistic for proportion p̂ p0 Hypothesized population Z proportion p0 q0 n 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 • • • • • H0: p = .10 Ha: p < .10 a = .05 n = 200 Critical Value(s): Test Statistic: Decision: Reject H0 .05 -1.645 0 Conclusion: Z One-Proportion Z Test Solution Test Statistic: 11 .10 pˆ p0 200 Z 2.12 p0 q0 .10 .90 200 n Decision: Reject at a = .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* • • • • • H0: p = .04 Ha: p .04 a = .05 n = 500 Critical Value(s): Reject H 0 Reject H 0 .025 .025 -1.96 0 1.96 Z Test Statistic: Decision: Conclusion: One-Proportion Z Test Solution* Test Statistic: 25 .04 pˆ p0 500 Z 1.14 p0 q0 .04 .96 500 n Decision: Do not reject at a = .05 Conclusion: There is evidence proportion is not 4% Calculating Type II Error Probabilities Power of Test 1. Probability of rejecting false H0 • Correct decision 2. Designated 1 - b 3. Used in determining test adequacy 4. Affected by • • • True value of population parameter Significance level a Standard deviation & sample size n Finding Power Step 1 Hypothesis: H0: 0 368 Ha: 0 < 368 n 15 25 Reject H0 a = .05 Do Not Draw Reject H0 0 = 368 X Finding Power Steps 2 & 3 Hypothesis: H0: 0 368 Ha: 0 < 368 n Reject H0 Do Not Draw Reject H0 15 25 a = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify Draw 1-b a = 360 b X X Finding Power Step 4 Hypothesis: H0: 0 368 Ha: 0 < 368 n Reject H0 Do Not Draw Reject H0 15 25 a = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify X 15 368 1.64 n 25 363.065 X L 0 Z Draw 1-b a = 360 363.065 X Finding Power Step 5 n Hypothesis: H0: 0 368 Ha: 0 < 368 Reject H0 Do Not Draw Reject H0 15 25 a = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Draw Specify Z Table X 15 368 1.64 n 25 363.065 X L 0 Z b = .154 1-b =.846 a = 360 363.065 X Effects on β of Changing α Decreasing the significance level α, increases the value of β and vice versa. Consider this diagram again. Shifting the critical value line to the right (to decrease α) will mean a larger area under the lower curve for β… (and vice versa) Probability of a Type II Error It is important that that we understand the relationship between Type I and Type II errors; that is, how the probability of a Type II error is calculated and its interpretation. Recall Example 11.1… H0: µ = 170 H1: µ > 170 At a significance level of 5% we rejected H0 in favor of H1 since our sample mean (178) was greater than the critical value of (175.34). Probability of a Type II Error β A Type II error occurs when a false null hypothesis is not rejected. In example 11.1, this means that if is less than 175.34 (our critical value) we will not reject our null hypothesis, which means that we will not install the new billing system. Thus, we can see that: β = P( < 175.34 given that the null hypothesis is false) Example 11.1 (revisited) β = P( < 175.34 given that the null hypothesis is false) The condition only tells us that the mean ≠ 170. We need to compute β for some new value of µ. For example, suppose that if the mean account balance is $180 the new billing system will be so profitable that we would hate to lose the opportunity to install it. β = P( < 175.34, given that µ = 180), thus… Example 11.1 (revisited) Our original hypothesis… our new assumpation… Effects on β of Changing α Decreasing the significance level α, increases the value of β and vice versa. Change α to .01 in Example 11.1. Stage 1: Rejection region z za z.01 2.33 x x 170 z 2.33 / n 65 / 400 x 177.57 Effects on β of Changing α Stage 2 : Probability of a Type II error b P( x 177 .57 | 180 ) x 177 .57 180 P 65 / 400 / n Pz .75 .2266 Effects on β of Changing α Decreasing the significance level α, increases the value of β and vice versa. Consider this diagram again. Shifting the critical value line to the right (to decrease α) will mean a larger area under the lower curve for β… (and vice versa) Judging the Test A statistical test of hypothesis is effectively defined by the significance level (α) and the sample size (n), both of which are selected by the statistics practitioner. Therefore, if the probability of a Type II error (β) is judged to be too large, we can reduce it by Increasing α, and/or increasing the sample size, n. Judging the Test For example, suppose we increased n from a sample size of 400 account balances to 1,000 in Example 11.1. Stage 1: Rejection region z za z.05 1.645 x x 170 z 1.645 / n 65 / 1,000 x 173.38 Judging the Test Stage 2: Probability of a Type II error b P( x 173.38 | 180) x 173.38 180 P / n 65 / 1 , 000 Pz 3.22 0 (approximat ely ) Judging the Test A statistical test of hypothesis is effectively defined by the significance level (α) and the sample size (n), both of which are selected by the statistics practitioner. Therefore, if the probability of a Type II error (β) is judged to be too large, we can reduce it by Increasing α, and/or increasing the sample size, n. Judging the Test For example, suppose we increased n from a sample size of 400 account balances to 1,000 in Example 11.1. Stage 1: Rejection region z za z.05 1.645 x x 170 z 1.645 / n 65 / 1,000 x 173.38 Judging the Test Stage 2: Probability of a Type II error b P( x 173.38 | 180) x 173.38 180 P 65 / 1,000 / n Pz 3.22 0 (approximat ely ) By increasing the sample size we reduce the probability of a Type II error: Compare β at n=400 and n=1,000… n=400 n=1,000 175.35 173.38 Power Curves Power H0: 0 Power H0: 0 Possible True Values for a Power Possible True Values for a H0: =0 Possible True Values for a = 368 in Example Chi-Square (c2) Test of Variance One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) Chi-Square (c2) Test for Variance 1. Tests one population variance or standard deviation 2. Assumes population is approximately normally distributed 3. Null hypothesis is H0: 2 = 02 4. Test statistic c 2 (n 1) S 2 0 2 Sample variance Hypothesized pop. variance Chi-Square (c2) Distribution Population Select simple random sample, size n. Compute s 2 Sampling Distributions for Different Sample Sizes Compute c2 = (n-1)s 2 /2 0 Astronomical number of c2 values 1 2 3 c2 Finding Critical Value Example What is the critical c2 value given: Ha: 2 > 0.7 Reject n=3 a =.05? a = .05 df = n - 1 = 2 c2 Table (Portion) 0 DF .995 1 ... 2 0.010 5.991 c2 Upper Tail Area … .95 … … 0.004 … … 0.103 … .05 3.841 5.991 Finding Critical Value Example What is the critical c2 value given: Ha: 2 < 0.7 n=3 What do you do a =.05? if the rejection region is on the left? Finding Critical Value Example What is the critical c2 value given: Ha: 2 < 0.7 Upper Tail Area Reject H0 for Lower Critical n=3 Value = 1-.05 = .95 a = .05 a =.05? df = n - 1 = 2 c2 Table (Portion) 0 .103 DF .995 1 ... 2 0.010 c2 Upper Tail Area … .95 … … 0.004 … … 0.103 … .05 3.841 5.991 Chi-Square (c2) 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. Chi-Square (c2) Test Solution • • • • • H0: 2 = 15 Ha: 2 15 a = .05 df = 25 - 1 = 24 Critical Value(s): Test Statistic: Decision: a /2 = .025 Conclusion: 0 12.401 39.364 c2 Chi-Square (c2) Test Solution Test Statistic: c 2 (n 1) S 2 0 2 (25 1) 17.7 2 15 = 33.42 Decision: Do not reject at a = .05 Conclusion: There is no evidence 2 is not 15 2 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

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