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© 2011 Pearson Education, Inc Statistics for Business and Economics Chapter 6 Inferences Based on a Single Sample: Tests of Hypotheses © 2011 Pearson Education, Inc Content 1. The Elements of a Test of Hypothesis 2. Formulating Hypotheses and Setting Up the Rejection Region 3. Test of Hypothesis about a Population Mean: Normal (z) Statistic 4. Observed Significance Levels: p-Values 5. Test of Hypothesis about a Population Mean: Student’s t-Statistic © 2011 Pearson Education, Inc Content 6. Large-Sample Test of Hypothesis about a Population Proportion 7. Calculating Type II Error Probabilities: More about * 8. Test of Hypothesis about a Population Variance © 2011 Pearson Education, Inc Learning Objectives 1. Test a specific value of a population parameter (mean or proportion), called a test of hypothesis 2. Provide a measure of reliability for the hypothesis test, called the significance level of the test © 2011 Pearson Education, Inc 6.1 The Elements of a Test of Hypothesis © 2011 Pearson Education, Inc Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Random sample Mean X = 20 © 2011 Pearson Education, Inc Reject hypothesis! Not close. What’s a Hypothesis? A statistical hypothesis is a statement about the numerical value of a population parameter. I believe the mean GPA of this class is 3.5! © 2011 Pearson Education, Inc © 1984-1994 T/Maker Co. Null Hypothesis The null hypothesis, denoted H0, represents the hypothesis that will be accepted unless the data provide convincing evidence that it is false. This usually represents the ―status quo‖ or some claim about the population parameter that the researcher wants to test. © 2011 Pearson Education, Inc Alternative Hypothesis The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support. © 2011 Pearson Education, Inc Alternative Hypothesis 1. Opposite of null hypothesis 2. The hypothesis that will be accepted only if the data provide convincing evidence of its truth 3. Designated Ha 4. Stated in one of the following forms Ha: some value) Ha: some value) Ha: some value) © 2011 Pearson Education, Inc Identifying Hypotheses 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) © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc Test Statistic The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses. © 2011 Pearson Education, Inc Test Statistic - Example The sampling distribution of x assuming µ = 2,400. the chance of observing x more than 1.645 standard deviations above 2,400 is only .05 – if in fact the true mean µ is 2,400. © 2011 Pearson Education, Inc Type I Error A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H0 is true. The probability of committing a Type I error is denoted by . © 2011 Pearson Education, Inc Rejection Region The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha. © 2011 Pearson Education, Inc Type II Error A Type II error occurs if the researcher accepts the null hypothesis when, in fact, H0 is false. The probability of committing a Type II error is denoted by . © 2011 Pearson Education, Inc Conclusions and Consequences for a Test of Hypothesis Conclusion True State of Nature H0 True Ha True Accept H0 Correct decision Type II error (Assume H0 True) (probability ) Reject H0 Type I error (Assume Ha True) (probability ) © 2011 Pearson Education, Inc Correct decision Elements of a Test of Hypothesis 1. Null hypothesis (H0): A theory about the specific values of one or more population parameters. The theory generally represents the status quo, which we adopt until it is proven false. 2. Alternative (research) hypothesis (Ha): A theory that contradicts the null hypothesis. The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth. © 2011 Pearson Education, Inc Elements of a Test of Hypothesis 3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis. 4. Rejection region: The numerical values of the test statistic for which the null hypothesis will be rejected. The rejection region is chosen so that the probability is that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test. © 2011 Pearson Education, Inc Elements of a Test of Hypothesis 5. Assumptions: Clear statement(s) of any assumptions made about the population(s) being sampled. 6. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic. © 2011 Pearson Education, Inc Elements of a Test of Hypothesis 7. Conclusion: a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100% of the time when H0 is true. © 2011 Pearson Education, Inc Elements of a Test of Hypothesis 7. Conclusion: b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true. We do not conclude that the null hypothesis is true because we do not (in general) know the probability that our test procedure will lead to an incorrect acceptance of H0 (Type II error). © 2011 Pearson Education, Inc Determining the Target Parameter Parameter Key Words or Phrases Mean; average Type of Data Quantitative p Proportion; percentage; fraction; rate Qualitative 2 Variance; variability; spread Quantitative µ © 2011 Pearson Education, Inc 6.2 Formulating Hypotheses and Setting Up the Rejection Region © 2011 Pearson Education, Inc Steps for Selecting the Null and Alternative Hypotheses 1. Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis will assume one of three forms: a. One-tailed, upper-tailed (e.g., Ha: µ > 2,400) b. One-tailed, lower-tailed (e.g., Ha: µ < 2,400) c. Two-tailed (e.g., Ha: µ ≠ 2,400) © 2011 Pearson Education, Inc Steps for Selecting the Null and Alternative Hypotheses 2. Select the null hypothesis as the status quo, that which will be presumed true unless the sampling experiment conclusively establishes the alternative hypothesis. The null hypothesis will be specified as that parameter value closest to the alternative in onetailed tests and as the complementary (or only unspecified) value in two-tailed tests. (e.g., H0: µ = 2,400) © 2011 Pearson Education, Inc One-Tailed Test A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol ― < ‖ or ― >.‖ © 2011 Pearson Education, Inc Two-Tailed Test A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol ― ≠.‖ © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc Sample Means Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1– Fail to Reject Region Critical Value Ho Value © 2011 Pearson Education, Inc Sample Statistic Rejection Regions (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region Rejection Region 1– 1/2 1/2 Fail to Reject Region Critical Value Ho Value Critical Value © 2011 Pearson Education, Inc Sample Statistic Rejection Regions Alternative Hypotheses LowerTailed = .10 z < –1.28 UpperTailed z > 1.28 Two-Tailed = .05 z < –1.645 z > 1.645 z < –1.96 or z > 1.96 = .01 z < –2.33 z > 2.33 z < –1.645 or z > 1.645 z < –2.575 or z > 2.575 © 2011 Pearson Education, Inc 6.3 Test of Hypotheses about a Population Mean: Normal (z) Statistic © 2011 Pearson Education, Inc Large-Sample Test of Hypothesis about µ One-Tailed Test H 0: µ = µ 0 Ha: µ < µ0 (or Ha: µ > µ0) Two-Tailed Test H0: µ = µ0 Ha: µ ≠ µ0 Test Statistic: Test Statistic: z x µ0 x x µ0 s n z © 2011 Pearson Education, Inc x µ0 x x µ0 s n Large-Sample Test of Hypothesis about µ One-Tailed Test Rejection region: z < –z (or z > z when Ha: µ > µ0) where z is chosen so that P(z > z) = © 2011 Pearson Education, Inc Large-Sample Test of Hypothesis about µ Two-Tailed Test Rejection region: |z| > z where z is chosen so that P(|z| > z) = /2 Note: µ0 is the symbol for the numerical value assigned to µ under the null hypothesis. © 2011 Pearson Education, Inc Conditions Required for a Valid Large-Sample Hypothesis Test for µ 1. A random sample is selected from the target population. 2. The sample size n is large (i.e., n ≥ 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.) © 2011 Pearson Education, Inc Possible Conclusions for a Test of Hypothesis 1. If the calculated test statistic falls in the rejection region, reject H0 and conclude that the alternative hypothesis Ha is true. State that you are rejecting H0 at the level of significance. Remember that the confidence is in the testing process, not the particular result of a single test. © 2011 Pearson Education, Inc Possible Conclusions for a Test of Hypothesis 2. If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the level of significance. [Generally, we will not ―accept‖ the null hypothesis unless the probability of a Type II error has been calculated.] © 2011 Pearson Education, Inc Two-Tailed z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified to be 25 grams. Test at the .05 level of significance. 368 gm. © 2011 Pearson Education, Inc Two-Tailed z Test Solution • • • • • H0: = 368 Ha: 368 .05 n 25 Critical Value(s): Reject H 0 .025 –1.96 Reject H 0 .025 0 1.96 z Test Statistic: x 372.5 368 z 0.9 25 n 25 Decision: Do not reject at = .05 Conclusion: No evidence average is not 368 © 2011 Pearson Education, Inc 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? © 2011 Pearson Education, Inc Two-Tailed z Test Solution* • • • • • H0: = 70 Ha: 70 =.05 n = 36 Critical Value(s): Reject H 0 Reject H 0 .025 .025 –1.96 0 1.96 z Test Statistic: x 69.7 70 z .51 3.5 n 36 Decision: Do not reject at = .05 Conclusion: No evidence average is not 70 © 2011 Pearson Education, Inc 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. © 2011 Pearson Education, Inc One-Tailed z Test Solution • • • • • H0: = 368 Test Statistic: Ha: > 368 x 372.5 368 z 1.50 15 = .05 n = 25 n 25 Critical Value(s): Decision: Reject Do not reject at = .05 .05 0 1.645 z Conclusion: No evidence average is more than 368 © 2011 Pearson Education, Inc 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 less than 32? © 2011 Pearson Education, Inc One-Tailed z Test Solution* • • • • • H0: = 32 Ha: < 32 = .01 n = 60 Critical Value(s): Reject .01 -2.33 0 z Test Statistic: x 30.7 32 z 2.65 3.8 n 60 Decision: Reject at = .01 Conclusion: There is evidence average is less than 32 © 2011 Pearson Education, Inc 6.4 Observed Significance Levels: p-Values © 2011 Pearson Education, Inc p-Value The observed significance level, or p-value, for a specific statistical test is the probability (assuming H0 is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis, and supportive of the alternative hypothesis, as the actual one computed from the sample data. © 2011 Pearson Education, Inc p-Value • • Probability of obtaining a test statistic more extreme ( or than actual sample value, given H0 is true Called observed level of significance • • Smallest value of for which H0 can be rejected Used to make rejection decision • • If p-value , do not reject H0 If p-value < , reject H0 © 2011 Pearson Education, Inc Steps for Calculating the pValue for a Test of Hypothesis 1. Determine the value of the test statistic z corresponding to the result of the sampling experiment. © 2011 Pearson Education, Inc Steps for Calculating the pValue for a Test of Hypothesis 2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form > , the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form < , the p-value is the area to the left of, or below, the observed z-value. © 2011 Pearson Education, Inc Steps for Calculating the pValue for a Test of Hypothesis 2b. If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z – that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value. © 2011 Pearson Education, Inc Reporting Test Results as p-Values: How to Decide Whether to Reject H0 1. Choose the maximum value of that you are willing to tolerate. 2. If the observed significance level (p-value) of the test is less than the chosen value of , reject the null hypothesis. Otherwise, do not reject the null hypothesis. © 2011 Pearson Education, Inc 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. © 2011 Pearson Education, Inc Two-Tailed z Test p-Value Solution x 372.5 368 z 1.50 15 n 25 0 1.50 © 2011 Pearson Education, Inc 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 1.50 .5000 – .4332 .0668 z From z table: z value of sample lookup 1.50 © 2011 Pearson Education, Inc 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 © 2011 Pearson Education, Inc z Two-Tailed z Test p-Value Solution p-Value = .1336 = .05 Do not reject H0. 1/2 p-Value = .0668 1/2 p-Value = .0668 Reject H0 Reject H0 1/2 = .025 1/2 = .025 –1.50 0 1.50 Test statistic is in ‘Do not reject’ region © 2011 Pearson Education, Inc 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 15 grams. Find the pValue. 368 gm. © 2011 Pearson Education, Inc One-Tailed z Test p-Value Solution x 372.5 368 z 1.50 15 n 0 1.50 © 2011 Pearson Education, Inc 25 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 © 2011 Pearson Education, Inc .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 1.50 From z table: lookup 1.50 © 2011 Pearson Education, Inc .5000 – .4332 .0668 z z value of sample statistic One-Tailed z Test p-Value Solution (p-Value = .0668) ( = .05). Do not reject H0. p-Value = .0668 Reject H0 = .05 0 1.50 Test statistic is in ‘Do not reject’ region © 2011 Pearson Education, Inc 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)? © 2011 Pearson Education, Inc p-Value Solution* p-Value is P(z -2.65) = .004. p-Value < ( = .01). Reject H0. Use alternative hypothesis to find direction p-Value .004 .5000 – .4960 .0040 .4960 –2.65 0 z value of sample statistic © 2011 Pearson Education, Inc z From z table: lookup 2.65 6.5 Test of Hypothesis about a Population Mean: Student’s t-Statistic © 2011 Pearson Education, Inc Small-Sample Test of Hypothesis about µ One-Tailed Test H0: µ = µ0 Ha: µ < µ0 (or Ha: µ > µ0) x Test statistic: t s n Rejection region: t < –t (or t > t when Ha: µ > µ0) where t and t are based on (n – 1) degrees of freedom © 2011 Pearson Education, Inc Small-Sample Test of Hypothesis about µ Two-Tailed Test H0: µ = µ0 Ha: µ ≠ µ0 x Test statistic: t s n Rejection region: |t| > t © 2011 Pearson Education, Inc Conditions Required for a Valid Small-Sample Hypothesis Test for µ 1. A random sample is selected from the target population. 2. The population from which the sample is selected has a distribution that is approximately normal. © 2011 Pearson Education, Inc 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. © 2011 Pearson Education, Inc 368 gm. Two-Tailed t Test Solution • • • • • H0: = 368 Ha: 368 = .05 df = 36 – 1 = 35 Critical Value(s): Reject H0 Reject H0 .025 .025 -2.030 0 2.030 t Test Statistic: x 372.5 368 t 2.25 s 12 n 36 Decision: Reject at = .05 Conclusion: There is evidence population average is not 368 © 2011 Pearson Education, Inc 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. © 2011 Pearson Education, Inc Two-Tailed t Test Solution* • • • • • H0: = 3.25 Ha: 3.25 .01 df 64 – 1 = 63 Critical Value(s): Reject H 0 Reject H0 .005 .005 -2.656 0 2.656 t Test Statistic: x 3.238 3.25 t .82 s .117 n 64 Decision: Do not reject at = .01 Conclusion: There is no evidence average is not 3.25 © 2011 Pearson Education, Inc One-Tailed t Test Example Is the average capacity of batteries less than 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. © 2011 Pearson Education, Inc One-Tailed t Test Solution • • • • • H0: = 140 Ha: < 140 = .05 df = 20 – 1 = 19 Critical Value(s): Reject H0 .05 -1.729 0 t Test Statistic: x 138.47 140 t 2.57 s 2.66 n 20 Decision: Reject at = .05 Conclusion: There is evidence population average is less than 140 © 2011 Pearson Education, Inc One-Tailed t Test Thinking Challenge You’re a marketing analyst for WalMart. 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)? © 2011 Pearson Education, Inc One-Tailed t Test Solution* • • • • • H0: = 5 Ha: > 5 = .05 df = 10 – 1 = 9 Critical Value(s): Reject H0 .05 0 1.833 t Test Statistic: x 6.4 5 t 1.31 s 3.373 n 10 Decision: Do not reject at = .05 Conclusion: There is no evidence average is more than 5 © 2011 Pearson Education, Inc 6.6 Large-Sample Test of Hypothesis about a Population Proportion © 2011 Pearson Education, Inc Large-Sample Test of Hypothesis about p One-Tailed Test H0: p = p0 Ha: p < p0 (or Ha: p > p0) Test statistic: z pˆ p0 pˆ where pˆ p0 q0 n q0 1 p0 Rejection region: z < –z (or z > z when Ha: p > p0) Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis © 2011 Pearson Education, Inc Large-Sample Test of Hypothesis about p Two-Tailed Test H0: p = p0 Ha: p ≠ p0 Test statistic: z pˆ p0 pˆ where pˆ p0 q0 n q0 1 p0 Rejection region: |z| < z Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis © 2011 Pearson Education, Inc Conditions Required for a Valid Large-Sample Hypothesis Test for p 1. A random sample is selected from a binomial population. 2. The sample size n is large. (This condition will be satisfied if both np0 ≥ 15 and nq0 ≥ 15.) © 2011 Pearson Education, Inc 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. © 2011 Pearson Education, Inc One-Proportion z Test Solution • • • • • H0: p = .10 Ha: p < .10 = .05 n = 200 Critical Value(s): Reject H0 .05 -1.645 0 Test Statistic: 11 pö p0 200 .10 z 2.12 p0 q0 .10 .90 n 200 Decision: Reject at = .05 Conclusion: There is evidence new z system < 10% defective © 2011 Pearson Education, Inc 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? © 2011 Pearson Education, Inc One-Proportion z Test Solution* • • • • • H0: p = .04 Ha: p .04 = .05 n = 500 Critical Value(s): Reject H 0 Reject H 0 .025 .025 -1.96 0 1.96 z Test Statistic: 25 .04 pö p0 z 500 1.14 p0 q0 .04 .96 n 500 Decision: Do not reject at = .05 Conclusion: There is evidence proportion is not 4% © 2011 Pearson Education, Inc 6.7 Calculating Type II Error Probabilities: More about © 2011 Pearson Education, Inc Type II Error The Type II error probability is calculated assuming that the null hypothesis is false because it is defined as the probability of accepting H0 when it is false.The situation corresponding to accepting the null hypothesis, and thereby risking a Type II error, is not generally as controllable. For that reason, we adopted a policy of nonrejection of H0 when the test statistic does not fall in the rejection region, rather than risking an error of unknown magnitude. © 2011 Pearson Education, Inc Steps for Calculating for a Large-Sample Test about µ 1. Calculate the value(s) of x corresponding to the border(s) of the rejection region. There will be one border value for a one-tailed test and two for a two-tailed test. The formula is one of the following, corresponding to a test with level of significance : Upper-tailed test: s x0 0 z x 0 z n © 2011 Pearson Education, Inc Steps for Calculating for a Large-Sample Test about µ s Lower-tailed test: x0 0 z x 0 z n Two-tailed test: x0, L s 0 z 2 x 0 z 2 n x0, U s 0 z 2 x 0 z 2 n © 2011 Pearson Education, Inc Steps for Calculating for a Large-Sample Test about µ 2. Specify the value of µa in the alternative hypothesis for which the value of is to be calculated. Then convert the border value(s) of x0 to z-value(s) using the alternative distribution with mean µa. The general formula for the z-value is z x0 a x © 2011 Pearson Education, Inc Steps for Calculating for a Large-Sample Test about µ Sketch the alternative distribution (centered at µa) and shade the area in the acceptance (nonrejection) region. Use the z-statistic(s) and Table IV in Appendix B to find the shaded area, which is . © 2011 Pearson Education, Inc Power of Test • Probability of rejecting false H0 • Correct decision • Equal to 1 – • Used in determining test adequacy • Affected by • • • True value of population parameter Significance level Standard deviation & sample size n © 2011 Pearson Education, Inc Two-Tailed z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified to be 15 grams. Test at the .05 level of significance. 368 gm. © 2011 Pearson Education, Inc Finding Power Step 1 Hypothesis: H0: 0 368 Ha: 0 < 368 n 15 25 Reject H0 = .05 Do Not Draw Reject H0 0 = 368 © 2011 Pearson Education, Inc x Finding Power Steps 2 & 3 Hypothesis: H0: 0 368 Ha: 0 < 368 n Reject H0 Do Not Draw Reject H0 15 25 = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify Draw 1– a = 360 © 2011 Pearson Education, Inc x x Finding Power Step 4 Hypothesis: H0: 0 368 Ha: 0 < 368 n Reject H0 Do Not Draw Reject H0 15 25 = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify xL 0 z x 368 1.64 n Draw 363.065 1– a = 360 363.065 x © 2011 Pearson Education, Inc 15 25 Finding Power Step 5 n Hypothesis: H0: 0 368 Ha: 0 < 368 Reject H0 Do Not Draw Reject H0 15 25 = .05 0 = 368 ‘True’ Situation: a = 360 (Ha) Specify z Table Draw xL 0 z = .154 368 1.64 n 363.065 1– =.846 a = 360 363.065 x © 2011 Pearson Education, Inc x 15 25 Properties of and Power 1. For fixed n and , the value of decreases, and the power increases as the distance between the specified null value µ0 and the specified alternative value µa increases. © 2011 Pearson Education, Inc Properties of and Power 2. For fixed n and values of µ0 and µa, the value of increases, and the power decreases as the value of is decreased. © 2011 Pearson Education, Inc Properties of and Power 3. For fixed n and values of µ0 and µa, the value of decreases, and the power increases as the sample size n is increased. © 2011 Pearson Education, Inc 6.8 Test of Hypothesis about a Population Variance © 2011 Pearson Education, Inc Variance Although many practical problems involve inferences about a population mean (or proportion), it is sometimes of interest to make an inference about a population variance, 2. © 2011 Pearson Education, Inc Test of a Hypothesis about 2 One-Tailed Test H0: = 0 Ha: < 0 (or Ha: > 0) Test statistic: 2 2 n 1 s 02 2 2 Rejection region: 1 (or > when Ha: > 0 ) where 0 is the hypothesized variance and the distribution of is based on (n – 1) degrees of © 2011 Pearson Education, Inc freedom. Test of a Hypothesis about 2 Two-Tailed Test H0: = 0 Ha: ≠ 0 Test statistic: 2 2 n 1 s 02 Rejection region: 2 21 2 or 2 2 2 where 0 is the hypothesized variance and the distribution of is based on (n – 1) degrees of © 2011 Pearson Education, Inc freedom. Conditions Required for a Valid Hypothesis Test for s 2 1. A random sample is selected from the target population. 2. The population from which the sample is selected has a distribution that is approximately normal. © 2011 Pearson Education, Inc Several 2 probability Distributions © 2011 Pearson Education, Inc Part of Table VI: Critical Values of Chi Square © 2011 Pearson Education, Inc Finding Critical Value Example What is the critical 2 value given: Ha: 2 > 0.7 Reject n=3 =.05? = .05 df = n - 1 = 2 0 2 Table (Portion) DF .995 1 ... 2 0.010 5.991 2 Upper Tail Area … .95 … … 0.004 … … 0.103 … © 2011 Pearson Education, Inc .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? © 2011 Pearson Education, Inc Finding Critical Value Example What is the critical 2 value given: Ha: 2 < 0.7 Upper Tail Area Reject H0 for Lower Critical n=3 Value = 1–.05 = .95 = .05 =.05? df = n - 1 = 2 0 .103 2 Table (Portion) DF .995 1 ... 2 0.010 2 Upper Tail Area … .95 … … 0.004 … … 0.103 … © 2011 Pearson Education, Inc .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. © 2011 Pearson Education, Inc 2 ( ) Chi-Square Test Solution • • • • • H0: 2 = 15 Ha: 2 15 = .05 df = 25 – 1 = 24 Critical Value(s): /2 = .025 Test Statistic: 2 (n 1)gs2 2 0 (25 1)g17.7 2 2 15 = 33.42 Decision: Do not reject at = .05 Conclusion: There is no evidence 2 2 is not 15 0 12.401 39.364 © 2011 Pearson Education, Inc Key Ideas Key Words for Identifying the Target Parameter – Mean, Average p – Proportion, Fraction, Percentage, Rate, Probability 2 – Variance, Variability, Spread © 2011 Pearson Education, Inc Key Ideas Elements of a Hypothesis Test 1. Null hypothesis (H0) 2. Alternative hypothesis (Ha) 3. Test statistic (z, t, or 2) 4. Significance level () 5. p-value 6. Conclusion © 2011 Pearson Education, Inc Key Ideas Errors in Hypothesis Testing Type I Error = Reject H0 when H0 is true (occurs with probability ) Type II Error = Accept H0 when H0 is false (occurs with probability ) Power of Test = P(Reject H0 when H0 is false) =1– © 2011 Pearson Education, Inc Key Ideas Forms of Alternative Hypothesis Lower-tailed : Ha : < 50 Upper-tailed : Ha : > 50 Two-tailed : Ha : ≠ 50 © 2011 Pearson Education, Inc Key Ideas Using p-values to Decide 1. Choose significance level ( ) 2. Obtain p-value of the test 3. If > p-value, reject H0 © 2011 Pearson Education, Inc
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