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McGraw-Hill/IrwinCopyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Chapter 9 Hypothesis Testing

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9-2 Hypothesis Testing 9.1Null and Alternative Hypotheses and Errors in Testing 9.2z Tests about a Population Mean Known 9.3t Tests about a Population Mean Unknown

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9-3 Null and Alternative Hypotheses and Errors in Hypothesis Testing The null hypothesis, denoted H 0, 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 H a, is an alternative (to the null hypothesis) statement that will be accepted only if there is convincing sample evidence that it is true.

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9-4 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

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9-5 Example 9.1: Trash Bag Case Continued The alternative hypothesis H a is that the new bag has a mean breaking strength that exceeds 50 lbs One-sided, “greater than” alternative –H 0 : µ ≤ 50 versus H a : µ > 50

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9-6 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 H a 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 H 0 is that the new billing system has a mean payment time close to but not less than 39 days One-sided, “less than” alternative –H 0 : 19.5 versus H a : < 19.5

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9-7 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 –H 0 : = 330 versus H a : ≠ 330

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9-8 Types of Hypotheses One-Sided, “Greater Than” Alternative H 0 : 0 vs.H a : > 0 One-Sided, “Less Than” Alternative H 0 : 0 vs.H a : < 0 Two-Sided, “Not Equal To” Alternative H 0 : = 0 vs. H a : 0 where 0 is a given constant value (with the appropriate units) that is a comparative value

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9-9 Types of Decisions As a result of testing H 0 vs. H a, will have to decide either of the following decisions for the null hypothesis H 0 : Do not reject H 0 –A weaker statement than “accepting H 0 ” –But you are rejecting the alternative H a Or reject H 0 –A weaker statement than “accepting H a ”

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9-10 Test Statistic To “test” H 0 vs. H a, use the “test statistic” where 0 is the given value (often the “claimed to be true”) and is the sample mean z measures the distance between 0 and 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

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9-11 Test Statistic and Trash Bag Case With µ 0 = 50, use the test statistic: If z ≤ 0, then ≤ µ 0 and there is no evidence to support rejecting H 0 in favor of H a –The point estimate indicates µ is probably less than 50 If z > 0, then > µ 0 and there is evidence to support rejecting H 0 in favor of H a –The point estimate indicates that µ is probably greater than 50 –The larger z (the farther is above µ), the stronger the evidence to support rejecting H 0 in favor of H a

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9-12 Type I and Type II Errors Type I Error: Rejecting H 0 when it is true Type II Error: Failing to reject H 0 when it is false

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9-13 Error Probabilities Type I Error: Rejecting H 0 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 H 0 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

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9-14 Error Probabilities State of Nature ConclusionH 0 TrueH 0 False Reject H 0 1- Do not Reject H 0 1-ββ

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9-15 Typical Values Usually set to a low value –So there is a small chance of rejecting a true H 0 –Typically, = 0.05 Strong evidence is required to reject H 0 Usually choose between 0.01 and 0.05 – = 0.01 requires very strong evidence is to reject H 0 –Tradeoff between and β For fixed sample size, the lower , the higher β –And the higher , the lower β

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9-16 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”

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9-17 Steps in Testing a “Greater 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 H 0 5.Collect the sample data and calculate the value of the test statistic 6.Decide whether to reject H 0 by using the test statistic and the rejection rule 7.Interpret the statistical results in managerial terms and assess their practical importance

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9-18 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #1 State the null and alternative hypotheses H 0 : 50 H a : > 50 where μ is the mean breaking strength of the new bag Specify the significance level – = 0.05

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9-19 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #2 3.Select the test statistic –Use the test statistic –A positive value of this this test statistic results from a sample mean that is greater than 50 lbs Which provides evidence against H 0 in favor of H a

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9-20 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #3 4.Determine the rejection rule for deciding whether or not to reject H 0 –To decide how large the test statistic must be to reject H 0 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 = z 0.05 = 1.645

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9-21 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #4 4.Continued –Reject H 0 in favor of H a 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 H 0 if the calculated test statistic z is > 1.645

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9-22 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, = 50.575 lbs. Then

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9-23 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #6

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9-24 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #7 6.Decide whether to reject H 0 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 z 0.05 = 1.645 –Therefore reject H 0 : μ ≤ 50 in favor of H a : μ > 50 at the 0.05 significance level Have rejected H 0 by using a test that allows only a 5% chance of wrongly rejecting H 0 This result is “statistically significant” at the 0.05 level

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9-25 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

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9-26 Steps in Testing a “Greater Than” Alternative in Trash Bag Case #9 Testing H 0 : 50 versus H a : > 50

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9-27 Effect of At = 0.01, the rejection point is z 0.01 = 2.33 In the trash example, the test statistic z = 2.20 is < z 0.01 = 2.33 Therefore, cannot reject H 0 in favor of H a 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 H 0

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9-28 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 H 0 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 H 0 is true have a low p-value and are evidence that H 0 is not true –The p-value is the smallest value of for which we can reject H 0 The p-value is an alternative to testing with a z test statistic

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9-29 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 H 0 is true Use p-value as an alternative to testing a greater than alternative with a z test statistic

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9-30 The p-value for “Greater Than” Example: Trash Bag Case Testing H 0 : µ ≤ 50 versus H a : µ > 50 using rejection points in (a) and p- value in (b) Have to modify steps 4, 5, and 6 of the previous procedure to use p- values

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9-31 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

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9-32 Steps Using a p-value to Test a “Greater Than” Alternative Continued 5.Continued –If H 0 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 H 0 is false and should be rejected 6.Reject H 0 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 z 0.05 = 1.645 –Therefore reject H 0 at the = 0.05 significance level

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9-33 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 H 0 5.Collect the sample data and calculate the value of the test statistic 6.Decide whether to reject H 0 by using the test statistic and the rejection rule 7.Interpret the statistical results in managerial terms and assess their practical importance

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9-34 Steps in Testing a “Less Than” Alternative in Payment Time Case #1 1.State the null and alternative hypotheses –In the payment time case, H 0 : ≥ 19.5 vs. H a : < 19.5, where is the mean bill payment time (in days) 2.Specify the significance level –In the payment time case, set = 0.01

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9-35 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 –A negative value of this this test statistic results from a sample mean that is less than 19.5 days Which provides evidence against H 0 in favor of H a

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9-36 Steps in Testing a “Less Than” Alternative in Payment Time Case #3 4.Determine the rejection rule for deciding whether or not to reject H 0 –To decide how much less than 0 the test statistic must be to reject H 0 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

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9-37 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 = –z 0.01 = –2.33 –Reject H 0 in favor of H a 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 H 0 if the calculated test statistic –z is less than –2.33

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9-38 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, = 18.1077 days:

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9-39 Steps in Testing a “Less Than” Alternative in Payment Time Case #5 6.Decide whether to reject H 0 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 z 0.01 = –2.33 –Therefore reject H 0 : μ ≥ 19.5 in favor of H a : μ < 19.5 at the 0.01 significance level Have rejected H 0 by using a test that allows only a 1% chance of wrongly rejecting H 0 This result is “statistically significant” at the 0.01 level

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9-40 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

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9-41 Steps in Testing a “Less Than” Alternative in Payment Time Case Testing H 0 : µ ≥ 19.5 versus H a : µ < 19.5 for = 0.01

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9-42 The p-value for “Less Than” Have to modify steps 4, 5, and 6 of the previous procedure to use p-values

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9-43 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

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9-44 Steps Using a p-value to Test a “Less Than” Alternative Continued 5.Continued –If H 0 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 H 0 is false and should be rejected 6.Reject H 0 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 –z 0.01 = –2.33 –Therefore, reject H 0 at the = 0.01 significance level

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9-45 Summary of Testing a One-Sided Alternative Using a Test Statistic If the population is normal and is known *, we can reject H 0 : = 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point rule holds * If unknown and n is very large (n > 120), estimate by s Test StatisticAlternativeReject H 0 if: H a : > 0 z > z H a : < 0 z < z

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9-46 Summary of Testing a One-Sided Alternative Using a p-value If the population is normal and s is known*, we can reject H 0 : = 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 StatisticAlternativeReject H 0 if: H a : > 0 p > z H a : < 0 p < z (1) is the area under the standard normal curve to the right of z is the area under the standard normal curve to the left of -z * If unknown and n is very large (n > 120), estimate by s

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9-47 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 H 0 5.Collect the sample data and calculate the value of the test statistic 6.Decide whether to reject H 0 by using the test statistic and the rejection rule 7.Interpret the statistical results in managerial terms and assess their practical importance

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9-48 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #1 1.State null and alternative hypotheses –In the Valentine Day case, H 0 : = 330 vs. H a : ≠ 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

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9-49 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 H 0 and for H a –A negative value of this test statistic results from a sample mean that is less than 330 Which provides evidence against H 0 and for H a –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 H 0 and against H a

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9-50 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 H 0 –To decide how different the test statistic must be from zero (positive or negative) to reject H 0 in favor of H a 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

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9-51 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 right- hand tail area equal to /2 –z /2 is the point on the horizontal axis under the standard normal curve that gives a left- hand tail area equal to /2

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9-52 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 H 0 in favor of H a 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

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9-53 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, = 326 –Then

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9-54 Steps in Testing a “Not Equal To” Alternative in Valentine Day Case #7 6.Decide whether to reject H 0 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 < – z 0.025 = –1.96 –Therefore cannot reject H 0 : µ = 330 in favor of H a : µ ≠ 330 at the 0.05 significance level Have not rejected H 0 by using a test that allows only a 5% chance of wrongly rejecting H 0

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9-55 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

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9-56 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

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9-57 Steps Using a p-value to Test a “Not Equal To” Alternative Continued 5.Continued –That is, if H 0 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 H 0 is false and should be rejected 6.Reject H 0 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 –z 0.025 = –1.96 –Therefore do not reject H 0 at the = 0.05 significance level

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9-58 Testing H 0 : µ = 330 Versus H a : µ ≠ 330 by Using Critical Values and the p-Value

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9-59 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 H 0 : = 0 at the level of significance (probability of Type I error equal to ) if and only if: Test StatisticAlternativeReject H 0 if: H a : > 0 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 is the area under the standard normal curve to the left of -z * If unknown and n is very large (n > 120), estimate by s

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9-60 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 H 0 according to these rules: –If p < 0.10, there is some evidence to reject H 0 –If p < 0.05, there is strong evidence to reject H 0 –If p < 0.01, there is very strong evidence to reject H 0 –If p < 0.001, there is extremely strong evidence to reject H 0

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9-61 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

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9-62 Defining the t Random Variable: σ Unknown Define a new random variable t: –with the definition of symbols as before The sampling distribution of this random variable is a t distribution with n – 1 degrees of freedom

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9-63 Defining the t Statistic: σ Unknown Let be the mean of a sample of size n with standard deviation s Also, µ 0 is the claimed value of the population mean Define a new test statistic 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

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9-64 t Tests about a Population Mean: σ Unknown Reject H 0 : µ = µ 0 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 …

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9-65 t Tests about a Population Mean: σ Unknown Continued AlternativeReject H 0 if:p-value H a : µ > µ 0 t > t Area under t distribution to right of t H a : µ < µ 0 t < –t Area under t distribution to left of –t H a : µ µ 0 |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

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9-66 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 H 0 : 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 H a : µ 18.8 Test at the = 0.05 level of significance –If H 0 can be rejected at the 5% level, then conclude that the mean rate now is less than 18.8% rate charged in 1991

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9-67 Example 9.4 #2 Randomly select n = 15 credit cards –n = 15 – = 16.828 –s = 1.538 – = 0.05 Then df = n – 1 = 14 Assume the population of the rates of all cards is normal

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9-68 Example 9.4 #3 The rejection rule is reject H 0 : = 18.8 against H a : < 18.8 if t < t –For = 0.05 and with df = 14, t = t 0.05 = 1.761 –Calculate the value of the t statistic:

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9-69 Example 9.4 #4

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9-70 Example 9.4 #5 Because t = 4.97 < t 0.05 = 1.761, reject H 0 Therefore, conclude at 5% significance level that the mean interest rate is lower than it was in 1991 –In fact it is µ 0 – = 18.8 – 16.827 = 1.973% lower

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9-71 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 H 0 at the 0.05, 0.01, 0.001 significance levels

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9-72 Computer Software Output for Example 9.4

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