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Chapter 9 Hypothesis Testing McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Hypothesis Testing 9.1 Null and Alternative Hypotheses and Errors in Testing 9.2 z Tests about a Population Mean s Known 9.3 t Tests about a Population Mean s Unknown 9-2 Null and Alternative Hypotheses and Errors in Hypothesis Testing • The null hypothesis, denoted H0, is a statement of the basic proposition being tested. The statement generally represents the status quo and is not rejected unless there is convincing sample evidence that it is false. • The alternative or research hypothesis, denoted Ha, is an alternative (to the null hypothesis) statement that will be accepted only if there is convincing sample evidence that it is true. 9-3 Example 9.1: Trash Bag Case • Tests show the trash bag has a mean breaking strength µ close to but not exceeding 50 lbs – The null hypothesis H0 is that the new bag has a mean breaking strength that is 50 lbs or less • The new bag’s mean breaking strength is not known and is in question, but it is hoped it is stronger than the current one 9-4 Example 9.1: Trash Bag Case Continued • The alternative hypothesis Ha is that the new bag has a mean breaking strength that exceeds 50 lbs • One-sided, “greater than” alternative – H0: µ ≤ 50 Ha: µ > 50 versus 9-5 Example 9.2: Payment Time Case • With new billing system, the mean bill paying time µ is hoped to be less than 19.5 days – Alternative hypothesis Ha is the new billing system has a mean payment time less than 19.5 days • With old billing system, the mean bill paying time µ was close to but not less than 39 days – The null hypothesis H0 is that the new billing system has a mean payment time close to but not less than 39 days • One-sided, “less than” alternative – H0: 19.5 versus Ha: < 19.5 9-6 Example 9.3: The Valentine’s Day Chocolate Case • They have designed a new box • They hope sales increase by 10 percent • Last year’s sales were 330 boxes – The null hypothesis will be that sales this year will not be 330 boxes • Two-sided, “not equal to” alternative – H0: = 330 versus Ha: ≠ 330 9-7 Types of Hypotheses • One-Sided, “Greater Than” Alternative H0: 0 vs. Ha: > 0 • One-Sided, “Less Than” Alternative H0 : 0 vs. Ha : < 0 • Two-Sided, “Not Equal To” Alternative H0 : = 0 vs. Ha : 0 where 0 is a given constant value (with the appropriate units) that is a comparative value 9-8 Types of Decisions • As a result of testing H0 vs. Ha, will have to decide either of the following decisions for the null hypothesis H0: • Do not reject H0 – A weaker statement than “accepting H0” – But you are rejecting the alternative Ha • Or reject H0 – A weaker statement than “accepting Ha” 9-9 Test Statistic • To “test” H0 vs. Ha, use the “test statistic” x 0 x 0 z sx s n where 0 is the given value (often the “claimed to be true”) and x is the sample mean • z measures the distance between 0 and x on the sampling distribution of the sample mean • If the population is normal or n is large*, the test statistic z follows a normal distribution * n ≥ 30, by the Central Limit Theorem 9-10 Test Statistic and Trash Bag Case • With µ0 = 50, use z x 50 x 50 the test statistic: sx s n • If z ≤ 0, then x ≤ µ0 and there is no evidence to support rejecting H0 in favor of Ha – The point estimate x indicates µ is probably less than 50 • If z > 0, then x > µ0 and there is evidence to support rejecting H0 in favor of Ha – The point estimate x indicates that µ is probably greater than 50 – The larger z (the farther x is above µ), the stronger the evidence to support rejecting H0 in favor of Ha 9-11 Type I and Type II Errors Type I Error: Rejecting H0 when it is true Type II Error: Failing to reject H0 when it is false State of Nature Conclusion Reject H0 Do not Reject H0 H0 True Type I Error Correct Decision H0 False Correct Decision Type II Error 9-12 Error Probabilities • Type I Error: Rejecting H0 when it is true – is the probability of making a Type I error – 1 – is the probability of not making a Type I error • Type II Error: Failing to reject H0 when it is false – β is the probability of making a Type II error – 1 – β is the probability of not making a Type II error 9-13 Error Probabilities State of Nature Conclusion Reject H0 Do not Reject H0 H0 True H0 False 1- 1-β β 9-14 Typical Values • Usually set to a low value – So there is a small chance of rejecting a true H0 – Typically, = 0.05 • Strong evidence is required to reject H0 • Usually choose between 0.01 and 0.05 – = 0.01 requires very strong evidence is to reject H0 – Tradeoff between and β • For fixed sample size, the lower , the higher β – And the higher , the lower β 9-15 z Tests about a Population Mean: σ Known • Test hypotheses about a population mean using the normal distribution – Called z tests – Require that the true value of the population standard deviation σ is known • In most real-world situations, σ is not known – But often is estimated from s of a single sample – When σ is unknown, test hypotheses about a population mean using the t distribution • Here, assume that we know σ • Also use a “rejection rule” 9-16 Steps in Testing a “Greater Than” Alternative 1. 2. 3. 4. State the null and alternative hypotheses Specify the significance level Select the test statistic Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance 9-17 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #1 • State the null and alternative hypotheses H0: 50 Ha: > 50 where μ is the mean breaking strength of the new bag • Specify the significance level – = 0.05 9-18 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #2 3. Select the test statistic – Use the test statistic z x 50 x 50 sx s n – A positive value of this this test statistic results from a sample mean that is greater than 50 lbs • Which provides evidence against H0 in favor of Ha 9-19 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #3 4. Determine the rejection rule for deciding whether or not to reject H0 – To decide how large the test statistic must be to reject H0 by setting the probability of a Type I error to , do the following: – The probability is the area in the right-hand tail of the standard normal curve – Use the normal table to find the point z (called the rejection or critical point) • • z is the point under the standard normal curve that gives a right-hand tail area equal to Since = 0.05 in the trash bag case, the rejection point is z = z0.05 = 1.645 9-20 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #4 4. Continued – Reject H0 in favor of Ha if the test statistic z is greater than the rejection point z • This is the rejection rule – In the trash bag case, the rejection rule is to reject H0 if the calculated test statistic z is > 1.645 9-21 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #5 5. Collect the sample data and calculate the value of the test statistic – In the trash bag case, assume that σ is known and σ = 1.65 lbs – For a sample of n = 40, x = 50.575 lbs. Then z x 50 s n 50 .575 50 1.65 2.20 40 9-22 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #6 9-23 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #7 6. Decide whether to reject H0 by using the test statistic and the rejection rule – Compare the value of the test statistic to the rejection point according to the rejection rule – In the trash bag case, z = 2.20 is greater than z0.05 = 1.645 – Therefore reject H0: μ ≤ 50 in favor of Ha: μ > 50 at the 0.05 significance level • Have rejected H0 by using a test that allows only a 5% chance of wrongly rejecting H0 • This result is “statistically significant” at the 0.05 level 9-24 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #8 7. Interpret the statistical results in managerial terms and assess their practical importance – Can conclude that the mean breaking strength of the new bag exceeds 50 lbs 9-25 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #9 Testing H0: 50 versus Ha: > 50 9-26 Effect of • At = 0.01, the rejection point is z0.01 = 2.33 • In the trash example, the test statistic z = 2.20 is < z0.01 = 2.33 • Therefore, cannot reject H0 in favor of Ha at the = 0.01 significance level – This is the opposite conclusion reached with =0.05 – So, the smaller we set , the larger is the rejection point, and the stronger is the statistical evidence that is required to reject the null hypothesis H0 9-27 The p-Value • The p-value or the observed level of significance is the probability of the obtaining the sample results if the null hypothesis H0 is true – The p-value is used to measure the weight of the evidence against the null hypothesis • Sample results that are not likely if H0 is true have a low p-value and are evidence that H0 is not true – The p-value is the smallest value of for which we can reject H0 • The p-value is an alternative to testing with a z test statistic 9-28 The p-Value for “Greater Than” Alternative • For a greater than alternative, the p-value is the probability of observing a value of the test statistic greater than or equal to z when H0 is true • Use p-value as an alternative to testing a greater than alternative with a z test statistic 9-29 The p-value for “Greater Than” Example: Trash Bag Case • Testing H0: µ ≤ 50 versus Ha: µ > 50 using rejection points in (a) and pvalue in (b) • Have to modify steps 4, 5, and 6 of the previous procedure to use pvalues 9-30 Steps Using a p-value to Test a “Greater Than” Alternative 4. Collect the sample data and compute the value of the test statistic – In the trash bag case, the value of the test statistic was calculated to be z = 2.20 5. Calculate the p-value by corresponding to the test statistic value – In the trash bag case, the area under the standard normal curve in the right-hand tail to the right of the test statistic value z = 2.20 – The area is 0.5 – 0.4861 = 0.0139 – The p-value is 0.0139 9-31 Steps Using a p-value to Test a “Greater Than” Alternative Continued 5. Continued – If H0 is true, the probability is 0.0139 of obtaining a sample whose mean is 50.575 lbs or higher – This is so low as to be evidence that H0 is false and should be rejected 6. Reject H0 if the p-value is less than – In the trash bag case, a was set to 0.05 – The calculated p-value of 0.0139 is < = 0.05 • This implies that the test statistic z = 2.20 is greater than the rejection point z0.05 = 1.645 – Therefore reject H0 at the = 0.05 significance level 9-32 Steps in Testing a “Less Than” Alternative 1. State the null and alternative hypotheses 2. Specify the significance level 3. Select the test statistic 4. Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance 9-33 Steps in Testing a “Less Than” Alternative in Payment Time Case #1 1. State the null and alternative hypotheses – In the payment time case, H0: ≥ 19.5 vs. Ha: < 19.5, where is the mean bill payment time (in days) 2. Specify the significance level – In the payment time case, set = 0.01 9-34 Steps in Testing a “Less Than” Alternative in Payment Time Case #2 3. Select the test statistic – In the payment time case, use the test statistic x 19 .5 x 19 .5 z sx s n – A negative value of this this test statistic results from a sample mean that is less than 19.5 days • Which provides evidence against H0 in favor of Ha 9-35 Steps in Testing a “Less Than” Alternative in Payment Time Case #3 4. Determine the rejection rule for deciding whether or not to reject H0 – To decide how much less than 0 the test statistic must be to reject H0 by setting the probability of a Type I error to , do the following: – The probability is the area in the left-hand tail of the standard normal curve – Use normal table to find the rejection point –z • –z is the negative of z • –z is the point on the horizontal axis under the standard normal curve that gives a left-hand tail area equal to 9-36 Steps in Testing a “Less Than” Alternative in Payment Time Case #3 4. Continued – Since = 0.01 in the payment time case, the rejection point is –z = –z0.01 = –2.33 – Reject H0 in favor of Ha if the test statistic z is calculated to be less than the rejection point –z • This is the rejection rule – In the payment time case, the rejection rule is to reject H0 if the calculated test statistic –z is less than –2.33 9-37 Steps in Testing a “Less Than” Alternative in Payment Time Case #4 5. Collect the sample data and calculate the value of the test statistic – In the payment time case, assume that σ is known and σ = 4.2 days – For a sample of n=65, x = 18.1077 days: z x 19 .5 s n 18 .1077 19 .5 4.2 2.67 65 9-38 Steps in Testing a “Less Than” Alternative in Payment Time Case #5 6. Decide whether to reject H0 by using the test statistic and the rejection rule – Compare the value of the test statistic to the rejection point according to the rejection rule – In the payment time case, z = –2.67 is less than z0.01 = –2.33 – Therefore reject H0: μ ≥ 19.5 in favor of Ha: μ < 19.5 at the 0.01 significance level • Have rejected H0 by using a test that allows only a 1% chance of wrongly rejecting H0 • This result is “statistically significant” at the 0.01 level 9-39 Steps in Testing a “Less Than” Alternative in Payment Time Case #6 7. Interpret the statistical results in managerial terms and assess their practical importance – Can conclude that the mean bill payment time of the new billing system is less than 19.5 days 9-40 Steps in Testing a “Less Than” Alternative in Payment Time Case Testing H0: µ ≥ 19.5 versus Ha: µ < 19.5 for = 0.01 9-41 The p-value for “Less Than” Have to modify steps 4, 5, and 6 of the previous procedure to use p-values 9-42 Steps Using a p-value to Test a “Less Than” Alternative (Steps 1–3 are the same) 4. Collect the sample data and compute the value of the test statistic – In the payment time case, the value of the test statistic was calculated to be z = –2.67 5. Calculate the p-value by corresponding to the test statistic value – In the payment time case, the area under the standard normal curve in the left-hand tail to the left of the test statistic z = –2.67 – The area is 0.5 – 0.4962 = 0.0038 – The p-value is 0.0038 9-43 Steps Using a p-value to Test a “Less Than” Alternative Continued 5. Continued – If H0 is true, then the probability is 0.0038 of obtaining a sample whose mean is as low as 18.1077 days or lower – This is so low as to be evidence that H0 is false and should be rejected 6. Reject H0 if the p-value is less than – In the payment time case, was 0.01 – The calculated p-value of 0.0038 is < = 0.01 • This implies that the test statistic z = –2.67 is less than the rejection point –z0.01 = –2.33 – Therefore, reject H0 at the = 0.01 significance level 9-44 Summary of Testing a One-Sided Alternative Using a Test Statistic If the population is normal and s is known*, we can reject H0: = 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point rule holds Test Statistic z x 0 s Alternative Ha : > 0 Reject H0 if: z > z Ha : < 0 z < z n * If s unknown and n is very large (n > 120), estimate s by s 9-45 Summary of Testing a One-Sided Alternative Using a p-value If the population is normal and s is known*, we can reject H0: = 0 at the level of significance (probability of Type I error equal to ) if and only if the corresponding p-value is less than the specified Test Statistic z x 0 s n Alternative Ha : > 0 Reject H0 if: p > z Ha : < 0 p < z (1) is the area under the standard normal curve to the right of z (2) is the area under the standard normal curve to the left of -z * If s unknown and n is very large (n > 120), estimate s by s 9-46 Steps in Testing a “Not Equal To” Alternative 1. State the null and alternative hypotheses 2. Specify the significance level 3. Select the test statistic 4. Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance 9-47 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #1 1. State null and alternative hypotheses – In the Valentine Day case, H0: = 330 vs. Ha: ≠ 330 2. Specify the significance level – In the Valentine Day case, set = 0.05 3. Select the test statistic – In this case, use the test statistic z x 330 sx x 330 s n 9-48 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #2 3. Continued – A positive value of this test statistic results from a sample mean that is greater than 330 • Which provides evidence against H0 and for Ha – A negative value of this test statistic results from a sample mean that is less than 330 • Which provides evidence against H0 and for Ha – A very small value close to 0 (either slightly positive or slightly negative) of this test statistic results from a sample mean that is nearly 330 • Which provides evidence in favor of H0 and against Ha 9-49 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #3 4. Determine the rejection rule for deciding whether or not to reject H0 – To decide how different the test statistic must be from zero (positive or negative) to reject H0 in favor of Ha by setting the probability of a Type I error to , do the following: • Divide the in half to find /2; /2 is the tail area in both tails of the standard normal curve • Under the standard normal curve, the probability /2 is the area in the right-hand tail and probability /2 is the area in the left-hand tail 9-50 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #4 4. Continued – Use the normal table to find the rejection points z/2 and –z/2 • z/2 is the point on the horizontal axis under the standard normal curve that gives a righthand tail area equal to /2 • –z/2 is the point on the horizontal axis under the standard normal curve that gives a lefthand tail area equal to /2 9-51 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #5 4. Continued – Because = 0.05, /2=0.025 • The area under the standard normal to the right of the mean is 0.5 – 0.025 = 0.475 • From Table A.3, the area is 0.475 for z = 1.96 – Rejection points are z=1.96,–z=– 1.96 – Reject H0 in favor of Ha if the test statistic z satisfies either: • z greater than the rejection point z/2, or • –z less than the rejection point –z/2 • This is the rejection rule 9-52 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #6 5. Collect the sample data and calculate the value of the test statistic – In the Valentine Day case, assume that σ is known and σ = 40 – For a sample of n = 100, x = 326 – Then x 330 326 330 z 1.00 s n 40 100 9-53 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #7 6. Decide whether to reject H0 by using the test statistic and the rejection rule – Compare the value of the test statistic to the rejection point according to the rejection rule – In this case, – z = –1.00 is < – z0.025 = –1.96 – Therefore cannot reject H0: µ = 330 in favor of Ha: µ ≠ 330 at the 0.05 significance level • Have not rejected H0 by using a test that allows only a 5% chance of wrongly rejecting H0 9-54 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #8 7. Interpret the statistical results in managerial terms and assess their practical importance – Cannot conclude that the mean order quantity this year of the Valentine Day box at large retail stores will differ from 330 boxes 9-55 Steps Using a p-value to Test a “Not Equal To” Alternative (Steps 1–3 are the same) 4. Collect the sample data and compute the value of the test statistic – In the Valentine Day case, the value of the test statistic was calculated to be z = –1.00 5. Calculate the p-value by corresponding to the test statistic value – In the Valentine Day case, the area under the standard normal curve in the left-hand tail to the left of the test statistic value z = –1.00 – The area is 0.1587 – The p-value is 0.1587 · 2 = 0.3174 9-56 Steps Using a p-value to Test a “Not Equal To” Alternative Continued 5. Continued – That is, if H0 is true, then the probability is 0.3174 of obtaining a sample whose mean is at least as extreme as 326 – This probability is not so low as to be evidence that H0 is false and should be rejected 6. Reject H0 if the p-value is less than a – In the Valentine Day case, was 0.05 – Calculated p-value of 0.3174 is greater than • This implies that the test statistic z = –1.00 is greater than the rejection point –z0.025 = –1.96 – Therefore do not reject H0 at the = 0.05 significance level 9-57 Testing H0: µ = 330 Versus Ha: µ ≠ 330 by Using Critical Values and the p-Value 9-58 Summary of Testing a Two-Sided Not Equal Alternative Using a p-value If the population is normal and s is known*, we can reject H0: = 0 at the level of significance (probability of Type I error equal to ) if and only if: Test Statistic z x 0 s n Alternative Ha : > 0 Reject H0 if: z > z/2 or p < (1) z > –z/2 or p < (2) (1) is the area under the standard normal curve to the right of z (2) is the area under the standard normal curve to the left of -z * If s unknown and n is very large (n > 120), estimate s by s 9-59 Weight of Evidence Against the Null • Calculate the test statistic and the corresponding p-value • Rate the strength of the conclusion about the null hypothesis H0 according to these rules: – If p < 0.10, there is some evidence to reject H0 – If p < 0.05, there is strong evidence to reject H0 – If p < 0.01, there is very strong evidence to reject H0 – If p < 0.001, there is extremely strong evidence to reject H0 9-60 t Tests about a Population Mean: σ Unknown • Suppose the population being sampled is normally distributed • The population standard deviation σ is unknown, as is the usual situation – If the population standard deviation σ is unknown, then it will have to estimated from a sample standard deviation s • Under these two conditions, have to use the t distribution to test hypotheses 9-61 Defining the t Random Variable: σ Unknown • Define a new random variable t: t x s n – with the definition of symbols as before • The sampling distribution of this random variable is a t distribution with n – 1 degrees of freedom 9-62 Defining the t Statistic: σ Unknown • Let x be the mean of a sample of size n with standard deviation s • Also, µ0 is the claimed value of the population mean • Define a new test statistic x 0 t s n • If the population being sampled is normal, and s is used to estimate σ, then … • The sampling distribution of the t statistic is a t distribution with n – 1 degrees of freedom 9-63 t Tests about a Population Mean: σ Unknown • Reject H0: µ = µ0 in favor of a particular alternative hypothesis Ha at the a level of significance if and only if the appropriate rejection point rule or, equivalently, the corresponding p-value is less than • We have the following rules … 9-64 t Tests about a Population Mean: σ Unknown Continued Alternative Reject H0 if: p-value Ha: µ > µ0 t > t Area under t distribution to right of t Ha: µ < µ0 t < –t Area under t distribution to left of –t Ha: µ µ0 |t| > t /2 * Twice area under t distribution to right of |t| t, t/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n) * either t > t/2 or t < –t/2 9-65 Example 9.4 #1 • The current mean credit card interest rate is as it was in 1991, at the rate of 18.8% – This is the null hypothesis H0: µ = 18.8 • The alternative to be tested is that the current mean interest rate is not as it was back in 1991, but in fact has decreased – This is the alternative hypothesis Ha: µ 18.8 • Test at the = 0.05 level of significance – If H0 can be rejected at the 5% level, then conclude that the mean rate now is less than 18.8% rate charged in 1991 9-66 Example 9.4 #2 • Randomly select n = 15 credit cards – n = 15 – x = 16.828 – s = 1.538 – = 0.05 • Then df = n – 1 = 14 • Assume the population of the rates of all cards is normal 9-67 Example 9.4 #3 • The rejection rule is reject H0: = 18.8 against Ha: < 18.8 if t < t – For = 0.05 and with df = 14, t = t0.05 =1.761 – Calculate the value of the t statistic: t x 18.8 s n 16.827 18.8 1.538 15 4.97 9-68 Example 9.4 #4 9-69 Example 9.4 #5 • Because t = 4.97 < t0.05 = 1.761, reject H0 • Therefore, conclude at 5% significance level that the mean interest rate is lower than it was in 1991 – In fact it is µ0 – x = 18.8 – 16.827 = 1.973% lower 9-70 p-value for Example 9.4 • The p-value is the left-hand tail area under the t curve with df = 14 to the left of t = 4.97 – This t value is off the t table, so use statistics software on a computer to calculate the p-value • p = 0.000103 – This says that if the claimed mean of 18.8 % is true, then there is only a 0.01% chance of randomly selecting a sample of 15 credit whose mean rate would be as low as 16.827% or lower – This p-value is less than 0.05, 0.01, and 0.001 • Reject H0 at the 0.05, 0.01, 0.001 significance levels 9-71 Computer Software Output for Example 9.4 9-72