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Msb12e ppt ch07
1.
7-1Copyright © 2014,
2011, and 2008 Pearson Education, Inc.
2.
7-2Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Statistics for Business and Economics Chapter 7 Inferences Based on a Single Sample: Tests of Hypotheses
3.
7-3Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Content 1. The Elements of a Test of Hypothesis 2. Formulating Hypotheses and Setting Up the Rejection Region 3. Observed Significance Levels: p-Values 4. Test of Hypothesis about a Population Mean: Normal (z) Statistic 5. Test of Hypothesis about a Population Mean: Student’s t-Statistic
4.
7-4Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Content 6. Large-Sample Test of Hypothesis about a Population Proportion 7. Test of Hypothesis about a Population Variance 8. Calculating Type II Error Probabilities: More about β*
5.
7-5Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Learning Objectives 1. Introduce the concept of a test of hypothesis 2. Provide a measure of reliability for the hypothesis test, called the significance level of the test
6.
7-6Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Learning Objectives 3. Test a specific value of a population parameter (mean, proportion or variance) called a test of hypothesis 4. Show how to estimate the reliability of a test
7.
7-7Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.1 The Elements of a Test of Hypothesis
8.
7-8Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Mean X = 20 Random sample Reject hypothesis! Not close. Reject hypothesis! Not close.
9.
7-9Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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! © 1984-1994 T/Maker Co.
10.
7-10Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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.
11.
7-11Copyright © 2014,
2011, and 2008 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.
12.
7-12Copyright © 2014,
2011, and 2008 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: µ ≠ (some value) Ha: µ < (some value) Ha: µ > (some value)
13.
7-13Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Identifying Hypotheses Example problem: Test that the population mean is not 3 Steps: • State the question statistically (µ ≠ 3) • State the opposite statistically (µ = 3) — Must be mutually exclusive & exhaustive • Select the alternative hypothesis (µ ≠ 3) — Has the ≠, <, or > sign • State the null hypothesis (µ = 3)
14.
7-14Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What Are the Hypotheses? • State the question statistically: µ = 12 • State the opposite statistically: µ ≠ 12 • Select the alternative hypothesis: Ha: µ ≠ 12 • State the null hypothesis: H0: µ = 12 Is the population average amount of TV viewing 12 hours?
15.
7-15Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What Are the Hypotheses? • State the question statistically: µ ≠ 12 • State the opposite statistically: µ = 12 • Select the alternative hypothesis: Ha: µ ≠ 12 • State the null hypothesis: H0: µ = 12 Is the population average amount of TV viewing different from 12 hours?
16.
7-16Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What Are the Hypotheses? • State the question statistically: µ ≤ 20 • State the opposite statistically: µ > 20 • Select the alternative hypothesis: Ha: µ > 20 • State the null hypothesis: H0: µ = 20 Is the average cost per hat less than or equal to $20?
17.
7-17Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What Are the Hypotheses? • State the question statistically: µ > 25 • State the opposite statistically: µ ≤ 25 • Select the alternative hypothesis: Ha: µ > 25 • State the null hypothesis: H0: µ = 25 Is the average amount spent in the bookstore greater than $25?
18.
7-18Copyright © 2014,
2011, and 2008 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.
19.
7-19Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Test Statistic - Example The sampling distribution of assuming µ = 2,400. the chance of observing more than 1.645 standard deviations above 2,400 is only . 05 – if in fact the true mean µ is 2,400. x x
20.
7-20Copyright © 2014,
2011, and 2008 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 α.
21.
7-21Copyright © 2014,
2011, and 2008 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.
22.
7-22Copyright © 2014,
2011, and 2008 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 β.
23.
7-23Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Conclusions and Consequences for a Test of Hypothesis True State of Nature Conclusion H0 True Ha True Accept H0 (Assume H0 True) Correct decision Type II error (probability β) Reject H0 (Assume Ha True) Type I error (probability α) Correct decision
24.
7-24Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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.
25.
7-25Copyright © 2014,
2011, and 2008 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 α that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of α is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test.
26.
7-26Copyright © 2014,
2011, and 2008 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.
27.
7-27Copyright © 2014,
2011, and 2008 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 100α % of the time when H0 is true.
28.
7-28Copyright © 2014,
2011, and 2008 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 β that our test procedure will lead to an incorrect acceptance of H0 (Type II error).
29.
7-29Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Determining the Target Parameter Parameter Key Words or Phrases Type of Data µ Mean; average Quantitative p Proportion; percentage; fraction; rate Qualitative σ2 Variance; variability; spread Quantitative
30.
7-30Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.2 Formulating Hypotheses and Setting Up the Rejection Region
31.
7-31Copyright © 2014,
2011, and 2008 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)
32.
7-32Copyright © 2014,
2011, and 2008 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 one-tailed tests and as the complementary (or only unspecified) value in two-tailed tests. (e.g., H0: µ = 2,400)
33.
7-33Copyright © 2014,
2011, and 2008 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 “ >.”
34.
7-34Copyright © 2014,
2011, and 2008 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 “ ≠.”
35.
7-35Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Basic Idea Sample Meansµ = 50 H0 Sampling Distribution It is unlikely that we would get a sample mean of this value ... 20 ... if in fact this were the population mean ... therefore, we reject the hypothesis that µ = 50.
36.
7-36Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Rejection Region (One-Tail Test) Ho ValueCritical Value α Sample Statistic Rejection Region Fail to Reject Region Sampling Distribution 1 – α Level of Confidence
37.
7-37Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Rejection Regions (Two-Tailed Test) Ho Value Critical Value Critical Value 1/2 α1/2 α Sample Statistic Rejection Region Rejection Region Fail to Reject Region Sampling Distribution 1 – α Level of Confidence
38.
7-38Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Rejection Regions Alternative Hypotheses Lower- Tailed Upper- Tailed Two-Tailed α = .10 z < –1.282 z > 1.282 z < –1.645 or z > 1.645 α = .05 z < –1.645 z > 1.645 z < –1.96 or z > 1.96 α = .01 z < –2.326 z > 2.326 z < –2.575 or z > 2.575
39.
7-39Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.3 Observed Significance Levels: p-Values
40.
7-40Copyright © 2014,
2011, and 2008 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.
41.
7-41Copyright © 2014,
2011, and 2008 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 α for which H0 can be rejected • Used to make rejection decision • If p-value ≥ α, do not reject H0 • If p-value < α, reject H0
42.
7-42Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating the p- Value for a Test of Hypothesis 1. Determine the value of the test statistic z corresponding to the result of the sampling experiment.
43.
7-43Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating the p- Value 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.
44.
7-44Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating the p- Value 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.
45.
7-45Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Reporting Test Results as p-Values: How to Decide Whether to Reject H0 1. Choose the maximum value of α that you are willing to tolerate. 2. If the observed significance level (p- value) of the test is less than the chosen value of α, reject the null hypothesis. Otherwise, do not reject the null hypothesis.
46.
7-46Copyright © 2014,
2011, and 2008 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 σ to be 15 grams. Find the p-value. How does it compare to α = .05? 368 gm.368 gm.
47.
7-47Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed z Test p-Value Solution z0 1.50 z value of sample statistic (observed) z = x − µ σ n = 372.5− 368 15 25 = +1.50
48.
7-48Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed Z Test p-Value Solution 1/2 p-Value1/2 p-Value z value of sample statistic (observed) p-Value is P(z ≤ –1.50 or z ≥ 1.50) z0 1.50–1.50 From z table: lookup 1.50 .4332 .5000 – .4332 .0668
49.
7-49Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed z Test p-Value Solution 1/2 p-Value .0668 1/2 p-Value .0668 p-Value is P(z ≤ –1.50 or z ≥ 1.50) = .1336 z0 1.50–1.50
50.
7-50Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed z Test p-Value Solution 0 1.50–1.50 z Reject H0Reject H0 1/2 p-Value = .06681/2 p-Value = .0668 1/2 α = .0251/2 α = .025 p-Value = .1336 ≥ α = .05 Do not reject H0. Test statistic is in ‘Do not reject’ region
51.
7-51Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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 σ to be 15 grams. Find the p-value. How does it compare to α = .05? 368 gm.368 gm.
52.
7-52Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed z Test p-Value Solution z0 1.50 z value of sample statistic z = x − µ σ n = 372.5− 368 15 25 = +1.50
53.
7-53Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed z Test p-Value Solution Use alternative hypothesis to find direction p-Value is P(z ≥1.50) z value of sample statistic p-Value z0 1.50 From z table: lookup 1.50 .4332 .5000 – .4332 .0668
54.
7-54Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed z Test p-Value Solution p-Value .0668 z value of sample statistic From z table: lookup 1.50 Use alternative hypothesis to find direction .5000 – .4332 .0668 p-Value is P(z ≥ 1.50) = .0668 z0 1.50 .4332
55.
7-55Copyright © 2014,
2011, and 2008 Pearson Education, Inc. α = .05 One-Tailed z Test p-Value Solution 0 1.50 z Reject H0 p-Value = .0668 (p-Value = .0668) ≥ (α = .05). Do not reject H0. Test statistic is in ‘Do not reject’ region
56.
7-56Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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 p-value? How does it compare to α = .01?
57.
7-57Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Use alternative hypothesis to find direction p-Value Solution* z0–2.65 z value of sample statistic From z table: lookup 2.65 .4960 p-Value .004 .5000 – .4960 .0040 p-Value is P(z ≤ -2.65) = .004. p-Value < (α = .01). Reject H0.
58.
7-58Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Converting a Two-Tailed p-Value from a Printout to a One-Tailed p-Value if Ha is of the form > and z is positive or Ha is of the form < and z is negative p = Reported p-value 2 if Ha is of the form > and z is negative Ha is of the form < and z is positive p = 1− Reported p-value 2
59.
7-59Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.4 Test of Hypotheses about a Population Mean: Normal (z) Statistic
60.
7-60Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Large-Sample Test of Hypothesis about µ One-Tailed Test Two-Tailed Test H0: µ = µ0 H0: µ = µ0 Ha: µ < µ0 Ha: µ ≠ µ0 (or Ha: µ > µ0) Test Statistic: Test Statistic: z = x − µ0 σx ≈ x − µ0 s n z = x − µ0 σx ≈ x − µ0 s n
61.
7-61Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Large-Sample Test of Hypothesis about µ One-Tailed Test Rejection region: z < –zα (or z > zα when Ha: µ > µ0) where zα is chosen so that P(z > zα) = α
62.
7-62Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Large-Sample Test of Hypothesis about µ Two-Tailed Test Rejection region: |z| > zα/2 where zα/2 is chosen so that P(|z| > zα/2) = α/2 Note: µ0 is the symbol for the numerical value assigned to µ under the null hypothesis.
63.
7-63Copyright © 2014,
2011, and 2008 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.)
64.
7-64Copyright © 2014,
2011, and 2008 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 α level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.
65.
7-65Copyright © 2014,
2011, and 2008 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 α level of significance. [Generally, we will not “accept” the null hypothesis unless the probability β of a Type II error has been calculated.]
66.
7-66Copyright © 2014,
2011, and 2008 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 σ to be 25 grams. Test at the .05 level of significance. 368 gm.368 gm.
67.
7-67Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed z Test Solution • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 368 µ ≠ 368 .05 25 z0 1.96–1.96 .025 Reject H0 Reject H0 .025 z = x − µ σ n = 372.5− 368 25 25 = +0.9 Do not reject at α = .05 No evidence average is not 368
68.
7-68Copyright © 2014,
2011, and 2008 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 σ = 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?
69.
7-69Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed z Test Solution* • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 70 µ ≠ 70 .05 36 z0 1.96–1.96 .025 Reject H0 Reject H0 .025 z = x − µ σ n = 69.7 − 70 3.5 36 = −.51 Do not reject at α = .05 No evidence average is not 70
70.
7-70Copyright © 2014,
2011, and 2008 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 σ to be 25 grams. Test at the .05 level of significance. 368 gm.368 gm.
71.
7-71Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed z Test Solution • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 368 µ > 368 .05 25 z0 1.645 .05 Reject z = x − µ σ n = 372.5− 368 15 25 = +1.50 Do not reject at α = .05 No evidence average is more than 368
72.
7-72Copyright © 2014,
2011, and 2008 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?
73.
7-73Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed z Test Solution* • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 32 µ < 32 .01 60 z0-2.33 .01 Reject z = x − µ σ n = 30.7 − 32 3.8 60 = −2.65 Reject at α = .01 There is evidence average is less than 32
74.
7-74Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.5 Test of Hypothesis about a Population Mean: Student’s t-Statistic
75.
7-75Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Small-Sample Test of Hypothesis about µ One-Tailed Test H0: µ = µ0 Ha: µ < µ0 (or Ha: µ > µ0) Test statistic: Rejection region: t < –tα (or t > tα when Ha: µ > µ0) where tα and tα/2 are based on (n – 1) degrees of freedom t = x − µ s n
76.
7-76Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Small-Sample Test of Hypothesis about µ Two-Tailed Test H0: µ = µ0 Ha: µ ≠ µ0 Test statistic: Rejection region: |t| > tα/2 t = x − µ s n
77.
7-77Copyright © 2014,
2011, and 2008 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.
78.
7-78Copyright © 2014,
2011, and 2008 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. 368 gm.368 gm.
79.
7-79Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed t Test Solution • H0: • Ha: • α = • df = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 368 µ ≠ 368 .05 36 – 1 = 35 t0 2.030-2.030 .025 Reject H0 Reject H0 .025 t = x − µ s n = 372.5− 368 12 36 = +2.25 Reject at α = .05 There is evidence population average is not 368
80.
7-80Copyright © 2014,
2011, and 2008 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.3.25 lb.
81.
7-81Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Two-Tailed t Test Solution* • H0: • Ha: • α = • df = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 3.25 µ ≠ 3.25 .01 64 – 1 = 63 t0 2.656-2.656 .005 Reject H0 Reject H0 .005 t = x − µ s n = 3.238 − 3.25 .117 64 = −.82 Do not reject at α = .01 There is no evidence average is not 3.25
82.
7-82Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed t Test Example Is the average capacity of batteries less than 140 ampere-hours? 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.
83.
7-83Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed t Test Solution • H0: • Ha: • α = • df = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 140 µ < 140 .05 20 – 1 = 19 t0-1.729 .05 Reject H0 t = x − µ s n = 138.47 −140 2.66 20 = −2.57 Reject at α = .05 There is evidence population average is less than 140
84.
7-84Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 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)?
85.
7-85Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Tailed t Test Solution* • H0: • Ha: • α = • df = • Critical Value(s): Test Statistic: Decision: Conclusion: µ = 5 µ > 5 .05 10 – 1 = 9 t0 1.833 .05 Reject H0 t = x − µ s n = 6.4 − 5 3.373 10 = +1.31 Do not reject at α = .05 There is no evidence average is more than 5
86.
7-86Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.6 Large-Sample Test of Hypothesis about a Population Proportion
87.
7-87Copyright © 2014,
2011, and 2008 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: Rejection region: z < –zα (or z > zα when Ha: p > p0) Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis z = ˆp − p0 σ ˆp where σ ˆp = p0q0 n q0 = 1− p0
88.
7-88Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Large-Sample Test of Hypothesis about p Two-Tailed Test H0: p = p0 Ha: p ≠ p0 Test statistic: Rejection region: |z| < zα/2 Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis z = ˆp − p0 σ ˆp where σ ˆp = p0q0 n q0 = 1− p0
89.
7-89Copyright © 2014,
2011, and 2008 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.)
90.
7-90Copyright © 2014,
2011, and 2008 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.
91.
7-91Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Proportion z Test Solution • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: p = .10 p < .10 .05 200 z0-1.645 .05 Reject H0 ( ) 0 0 0 11 .10ˆ 200 2.12 .10 .90 200 p p z p q n − − = = = − Reject at α = .05 There is evidence new system < 10% defective
92.
7-92Copyright © 2014,
2011, and 2008 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?
93.
7-93Copyright © 2014,
2011, and 2008 Pearson Education, Inc. One-Proportion z Test Solution* • H0: • Ha: • α = • n = • Critical Value(s): Test Statistic: Decision: Conclusion: p = .04 p ≠ .04 .05 500 z0 1.96-1.96 .025 Reject H0 Reject H0 .025 ( ) 0 0 0 25 .04ˆ 500 1.14 .04 .96 500 − − = = = p p z p q n Do not reject at α = .05 There is evidence proportion is not 4%
94.
7-94Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.7 Test of Hypothesis about a Population Variance
95.
7-95Copyright © 2014,
2011, and 2008 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, σ2 .
96.
7-96Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Test of a Hypothesis about σ 2 One-Tailed Test H0: σ 2 = σ0 2 Ha: σ 2 < σ0 2 (or Ha: σ 2 > σ0 2 ) Test statistic: Rejection region: (or χ 2 > χα 2 when Ha: σ 2 > σ0 2 ) where σ0 2 is the hypothesized variance and the distribution of χ 2 is based on (n – 1) degrees of freedom. χ2 = n −1( )s2 σ0 2 χ2 < χ 1−α( ) 2
97.
7-97Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Test of a Hypothesis about σ 2 Two-Tailed Test H0: σ 2 = σ0 2 Ha: σ 2 ≠ σ0 2 Test statistic: Rejection region: where σ0 2 is the hypothesized variance and the distribution of χ 2 is based on (n – 1) degrees of freedom. χ2 = n −1( )s2 σ0 2 χ2 < χ 1−α 2( ) 2 or χ2 > χ α 2( ) 2
98.
7-98Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Conditions Required for a Valid Hypothesis Test for s2 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.
99.
7-99Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Several χ 2 probability Distributions
100.
7-100Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Critical Values of Chi Square
101.
7-101Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What is the critical χ2 value given: Ha: σ2 > 0.7 n = 3 α =.05? Finding Critical Value Example χ20 Upper Tail Area DF .995 … .95 … .05 1 ... … 0.004 … 3.841 2 0.010 … 0.103 … 5.991 χ2 Table (Portion) df = n - 1 = 2 5.991 Reject α = .05
102.
7-102Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Finding Critical Value Example What is the critical χ2 value given: Ha: σ 2 < 0.7 n = 3 α =.05? What do you do if the rejection region is on the left?
103.
7-103Copyright © 2014,
2011, and 2008 Pearson Education, Inc. What is the critical χ2 value given: Ha: σ 2 < 0.7 n = 3 α =.05? Finding Critical Value Example .103 χ20 Upper Tail Area DF .995 … .95 … .05 1 ... … 0.004 … 3.841 2 0.010 … 0.103 … 5.991 χ2 Table (Portion) Upper Tail Area for Lower Critical Value = 1–.05 = .95α = .05 Reject H0 df = n - 1 = 2
104.
7-104Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Chi-Square (χ2 ) 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.
105.
7-105Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Chi-Square (χ2 ) Test Solution • H0: • Ha: • α = • df = • Critical Value(s): Test Statistic: Decision: Conclusion: σ 2 = 15 σ 2 ≠ 15 .05 25 – 1 = 24 χχ22 00 α/2 = .025 39.36412.401 = 33.42 ( ) 22 2 2 2 0 (25 1) 17.7( 1) 15 n s χ σ −− = = Do not reject at α = .05 There is no evidence σ 2 is not 15
106.
7-106Copyright © 2014,
2011, and 2008 Pearson Education, Inc. 7.8 Calculating Type II Error Probabilities: More about β
107.
7-107Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Type II Error The Type II error probability β 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.
108.
7-108Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating β for a Large-Sample Test about µ 1. Calculate the value(s) of 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 α: Upper-tailed test: x0 = µ0 + zασx ≈ µ0 + zα s n x
109.
7-109Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating β for a Large-Sample Test about µ Lower-tailed test: Two-tailed test: x0 = µ0 − zασx ≈ µ0 − zα s n x0, L = µ0 − zα 2σx ≈ µ0 − zα 2 s n x0, U = µ0 + zα 2σx ≈ µ0 + zα 2 s n
110.
7-110Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating β for a Large-Sample Test about µ 2. Specify the value of µa in the alternative hypothesis for which the value of β is to be calculated. Then convert the border value(s) of to z-value(s) using the alternative distribution with mean µa. The general formula for the z-value is z = x0 − µa σx x0
111.
7-111Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Steps for Calculating β 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 II in Appendix D to find the shaded area, which is β.
112.
7-112Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Power of Test • Probability of rejecting false H0 • Correct decision • Equal to 1 – β • Used in determining test adequacy • Affected by • True value of population parameter • Significance level α • Standard deviation & sample size n
113.
7-113Copyright © 2014,
2011, and 2008 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 σ to be 15 grams. Test at the .05 level of significance. 368 gm.368 gm.
114.
7-114Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Finding Power Step 1 xµ0 = 368 Reject H0 Do Not Reject H0 Hypothesis: H0: µ0 ≥ 368 Ha: µ0 < 368 α = .05 Draw15 25 n σ =
115.
7-115Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Finding Power Steps 2 & 3 xµa = 360 ‘True’ Situation: µa = 360 (Ha) Draw Specify β 1–β xµ0 = 368 Reject H0 Do Not Reject H0 Hypothesis: H0: µ0 ≥ 368 Ha: µ0 < 368 α = .05 Draw15 25 n σ =
116.
7-116Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Finding Power Step 4 363.065363.065 xxµµaa = 360= 360 ‘True’ Situation: µa = 360 (Ha) Draw Specify 1–1–ββ xµ0 = 368 Reject H0 Do Not Reject H0 Hypothesis: H0: µ0 ≥ 368 Ha: ι0 < 368 α = .05 Draw15 25 n σ = xL = µ0 − z σ n = 368 − 1.64 15 25 = 363.065
117.
7-117Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Finding Power Step 5 363.065363.065 xxµµaa = 360= 360 ‘True’ Situation: µa = 360 (Ha) Draw Specify xµ0 = 368 Reject H0 Do Not Reject H0 Hypothesis: H0: µ0 ≥ 368 Ha: µ0 < 368 α = .05 Draw15 25 n σ = β = .154 1–β =.846 z Table xL = µ0 − z σ n = 368 − 1.64 15 25 = 363.065
118.
7-118Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Properties of β and Power 1. For fixed n and α, the value of β decreases, and the power increases as the distance between the specified null value µ0 and the specified alternative value µa increases.
119.
7-119Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Properties of β and Power 2. For fixed n and values of µ0 and µa, the value of β increases, and the power decreases as the value of α is decreased.
120.
7-120Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Properties of β and Power 3. For fixed n and values of µ0 and µa, the value of β decreases, and the power increases as the sample size n is increased.
121.
7-121Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Key Ideas Key Words for Identifying the Target Parameter µ – Mean, Average p – Proportion, Fraction, Percentage, Rate, Probability σ 2 – Variance, Variability, Spread
122.
7-122Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Key Ideas Elements of a Hypothesis Test 1. Null hypothesis (H0) 2. Alternative hypothesis (Ha) 3. Test statistic (z, t, or χ2 ) 4. Significance level (α) 5. p-value 6. Conclusion
123.
7-123Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Key Ideas Errors in Hypothesis Testing Type I Error = Reject H0 when H0 is true (occurs with probability α) Type II Error = Accept H0 when H0 is false (occurs with probability β ) Power of Test = P(Reject H0 when H0 is false) = 1 – β
124.
7-124Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Key Ideas Forms of Alternative Hypothesis Lower-tailed : Ha : µ < 50 Upper-tailed : Ha : µ > 50 Two-tailed : Ha : µ ≠ 50
125.
7-125Copyright © 2014,
2011, and 2008 Pearson Education, Inc. Key Ideas Using p-values to Decide 1. Choose significance level (α ) 2. Obtain p-value of the test 3. If α > p-value, reject H0
Editor's Notes
As a result of this class, you will be able to ...
As a result of this class, you will be able to ...
As a result of this class, you will be able to ...
As a result of this class, you will be able to ...
:1, 1, 3
:1, 1, 3
Rejection region does NOT include critical value.
Rejection region does NOT include critical value.
Rejection region does NOT include critical value.
:1, 1, 3
.
:1, 1, 3
:1, 1, 3
Allow students about 10 minutes to finish this.
Assume that the population is normally distributed. Allow students about 10 minutes to solve this.
Note: More than 5 have been sold (6.4), but not enough to be significant.
:1, 1, 3
:1, 1, 3
:1, 1, 3
As a result of this class, you will be able to...
As a result of this class, you will be able to...
As a result of this class, you will be able to...
As a result of this class, you will be able to...
As a result of this class, you will be able to...
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