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### 11주차

1. 1. Introduction to Probability and Statistics 11th Week (5/24) Hypothesis Testing
2. 2. Hypothesis in statistics, is a claim or statement about a property of a populationHypothesis Testing is to test the claim or statementExample: A conjecture is made that “the average starting salary for computer science gradate is \$30,000 per year”.
3. 3. Nonstatistical Hypothesis Testing… A criminal trial is an example of hypothesis testing without the statistics. In a trial a jury must decide between two hypotheses. The null hypothesis is H0: The defendant is innocent The alternative hypothesis or research hypothesis is H1: The defendant is guilty The jury does not know which hypothesis is true. They must make a decision on the basis of evidence presented.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.3
4. 4. Nonstatistical Hypothesis Testing… In the language of statistics convicting the defendant is called rejecting the null hypothesis in favor of the alternative hypothesis. That is, the jury is saying that there is enough evidence to conclude that the defendant is guilty (i.e., there is enough evidence to support the alternative hypothesis). If the jury acquits it is stating that there is not enough evidence to support the alternative hypothesis. Notice that the jury is not saying that the defendant is innocent, only that there is not enough evidence to support the alternative hypothesis. That is why we never say that we accept the null hypothesis, although most people in industry will say “We accept the null hypothesis”Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.4
5. 5. Question:How can we justify/test this conjecture? A. What do we need to know to justifythis conjecture? B. Based on what we know, how shouldwe justify this conjecture?
6. 6. Answer to A:Randomly select, say 100, computerscience graduates and find out theirannual salaries---- We need to have some sampleobservations, i.e., a sample set!
7. 7. Answer to B:That is what we will learn in thischapter---- Make conclusions based on thesample observations
8. 8. Statistical Reasoning Analyze the sample set in an attempt to distinguish between results that can easily occur and results that are highly unlikely.
9. 9. Statistical DecisionsDecisions about populations on the basis of sample information.Ex) We may wish to decide on the basis of sample data whether a newserum is really effective in curing a disease, or whether one educationalprocedure is better than another
10. 10. Definitions Null Hypothesis (denoted H 0): is the statement being tested in a test of hypothesis. Alternative Hypothesis (H 1): is what is believe to be true if the null hypothesis is false.
11. 11. Null Hypothesis: H0 Must contain condition of equality =, ≥, or ≤ Test the Null Hypothesis directly Reject H 0 or fail to reject H 0
12. 12. Alternative Hypothesis: H1 Must be true if H0 is false ≠, <, > ‘opposite’ of Null Example: H0 : µ = 30 versus H1 : µ > 30
13. 13. Statistical Hypotheses and Null Hypotheses•Statistical hypotheses: Assumptions or guesses about the populations involved. (Such assumptions, which may or may not be true)•Null hypotheses (H0): Hypothesis that there is no difference between the procedures. We formulate it if we want to decide whether one procedure is better than another.•Alternative hypotheses (H1): Any hypothesis that differs from a given null hypothesisExample 1. For example, if the null hypothesis is p = 0.5, possible alternative hypotheses are p =0.7, or p ≠ 0.5.
14. 14. Concepts of Hypothesis Testing (1)…• The two hypotheses are called the null hypothesis and the other the alternative or research hypothesis. The usual notation is: pronounced H “nought”• H0: — the ‘null’ hypothesis• H1: — the ‘alternative’ or ‘research’ hypothesis• The null hypothesis (H0) will always state that the parameter equals the value specified in the alternative 11.14 hypothesis (H1)
15. 15. Stating Your Own HypothesisIf you wish to support your claim, the claim must be stated so that it becomes the alternative hypothesis.
16. 16. Important Notes:H0 must always contain equality; however some claims are not stated using equality. Therefore sometimes the claim and H0 will not be the same.Ideally all claims should be stated that they are Null Hypothesis so that the most serious error would be a Type I error.
17. 17. Tests of Hypotheses and Significance“Significant”: If on the supposition that a particular hypothesis is true we find that results observed in a random sample differ markedly from those expected under the hypothesis on the basis of pure chance using sampling theory, we would say that the observed differences are significant•We would be inclined to reject the hypothesis if the observed differences are significant.• Tests of hypotheses, tests of significance, or decision rules: Procedures that enable us to decide whether to accept or reject hypotheses or to determine whether observed samples differ significantly from expected results
18. 18. Type I ErrorThe mistake of rejecting the null hypothesis when it is true.The probability of doing this is called the significance level, denoted by α (alpha).Common choices for α: 0.05 and 0.01Example: rejecting a perfectly good parachute and refusing to jump
19. 19. Type II Errorthe mistake of failing to reject the null hypothesis when it is false.denoted by ß (beta)Example: failing to reject a defective parachute and jumping out of a plane with it.
20. 20. Table 7-2 Type I and Type II Errors True State of Nature The null The null hypothesis is hypothesis is true false We decide to Type I error Correct reject the (rejecting a true decision null hypothesis null hypothesis) Decision We fail to Type II error Correct reject the (failing to reject decision null hypothesis a false null hypothesis)
21. 21. Types of Errors… A Type I error occurs when we reject a true null hypothesis (i.e. Reject H0 when it is TRUE) H0 T F Reject I Reject II A Type II error occurs when we don’t reject a false null hypothesis (i.e. Do NOT reject H0 when it is FALSE)Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.21
22. 22. Type of Errors… There are two possible errors. A Type I error occurs when we reject a true null hypothesis. That is, a Type I error occurs when the jury convicts an innocent person. We would want the probability of this type of error [maybe 0.001 – beyond a reasonable doubt] to be very small for a criminal trial where a conviction results in the death penalty, whereas for a civil trial, where conviction might result in someone having to “pay for damages to a wrecked auto”,we would be willing for the probability to be larger [0.49 – preponderance of the evidence ] P(Type I error) = α [usually 0.05 or 0.01]Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.22
23. 23. Type of Errors… A Type II error occurs when we don’t reject a false null hypothesis [accept the null hypothesis]. That occurs when a guilty defendant is acquitted. In practice, this type of error is by far the most serious mistake we normally make. For example, if we test the hypothesis that the amount of medication in a heart pill is equal to a value which will cure your heart problem and “accept the hull hypothesis that the amount is ok”. Later on we find out that the average amount is WAY too large and people die from “too much medication” [I wish we had rejected the hypothesis and threw the pills in the trash can], it’s too late because we shipped the pills to the public.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.23
24. 24. Type of Errors… The probability of a Type I error is denoted as α (Greek letter alpha). The probability of a type II error is β (Greek letter beta). The two probabilities are inversely related. Decreasing one increases the other, for a fixed sample size. In other words, you can’t have α and β both real small for any old sample size. You may have to take a much larger sample size, or in the court example, you need much more evidence.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.24
25. 25. Type of Errors… The critical concepts are theses: 1. There are two hypotheses, the null and the alternative hypotheses. 2. The procedure begins with the assumption that the null hypothesis is true. 3. The goal is to determine whether there is enough evidence to infer that the alternative hypothesis is true, or the null is not likely to be true. 4. There are two possible decisions: Conclude that there is enough evidence to support the alternative hypothesis. Reject the null. Conclude that there is not enough evidence to support the alternative hypothesis. Fail to reject the null.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.25
26. 26. Judging the Test… A statistical test of hypothesis is effectively defined by the significance level ( ) and the sample size (n), both of which are selected by the statistics practitioner. Therefore, if the probability of a Type II error ( ) is too large [we have insufficient power], we can reduce it by increasing , and/or increasing the sample size, n.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.26
27. 27. Judging the Test… The power of a test is defined as 1– . It represents the probability of rejecting the null hypothesis when it is false and the true mean is something other than the null value for the mean. If we are testing the hypothesis that the average amount of medication in blood pressure pills is equal to 6 mg (which is good), and we “fail to reject” the null hypothesis, ship the pills to patients worldwide, only to find out later that the “true” average amount of medication is really 8 mg and people die, we get in trouble. This occurred because the P(reject the null / true mean = 7 mg) = 0.32 which would mean that we have a 68% chance on not rejecting the null for these BAD pills and shipping to patients worldwide.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.27
28. 28. Type I and Type II Errors•Type I error: If we reject a hypothesis when it happens to be true.•Type II error: If we accept a hypothesis when it should be rejected.•In order for any tests of hypotheses or decision rules to be good, they must be designed so as to minimize errors of decision.• An attempt to decrease one type of error is accompanied in general by an increase in the other type of error. The only way to reduce both types of error is to increase the sample size, which may or may not be possible.
29. 29. Significant Differences Hypothesis testing is designed to detect significant differences: differences that did not occur by random chance. In the “one sample” case: we compare a random sample (from a large group) to a population. We compare a sample statistic to a population parameter to see if there is a significant difference.
30. 30. Level of Significance ( 유의수준 )• Level of significance: In testing a given hypothesis, the maximum probability with which we would be willing to risk a Type I error is called the level of significance
31. 31. Level of Significance•In practice a level of significance of 0.05 or 0.01 is customary, although other values are used.• If for example a 0.05 or 5% level of significance is chosen in designing a test of a hypothesis, then there are about 5 chances in 100 that we would reject the hypothesis when it should be accepted, i.e., whenever the null hypotheses is true, we are about 95% confident that we would make the right decision. In such cases we say that the hypothesis has been rejected at a 0.05 level of significance, which means that we could be wrong with probability 0.05.
32. 32. DefinitionTest Statistic:is a sample statistic or value based on sample dataExample: x – µx z = σ/ n
33. 33. DefinitionCritical Region : is the set of all values of the test statistic that would cause a rejection of the null hypothesis
34. 34. Critical Region• Set of all values of the test statistic that would cause a rejection of the null hypothesis Critical Region
35. 35. Critical Region• Set of all values of the test statistic that would cause a rejection of the • null hypothesis Critical Region
36. 36. Critical Region• Set of all values of the test statistic that would cause a rejection of the null hypothesis Critical Regions
37. 37. DefinitionCritical Value: is the value (s) that separates the critical region from the values that would not lead to a rejection of H 0
38. 38. Critical ValueValue (s) that separates the critical region from the values that would not lead to a rejection of H 0 Critical Value ( z score )
39. 39. Critical ValueValue (s) that separates the critical region from the values that would not lead to a rejection of H 0 Reject H0 Fail to reject H0 Critical Value ( z score )
40. 40. Tests Involving the Normal Distribution- Level of confidence : 0.05•The critical region (or region of rejection of the hypothesis or the region of significance): The set of z scores outside the range -1.96 to 1.96 constitutes• The region of acceptance of the hypothesis (or the region of nonsignificance) : The set of z scores inside the range -1.96 to 1.96 could
41. 41. Tests Involving the Normal Distribution• Decision Rule • When the level of confidence is 0.01, a value 2.58 should be instead of 1.96.
42. 42. Two-tailed,Left-tailed,Right-tailed Tests
43. 43. Left-tailed Test H0: µ ≥ 200 H1: µ < 200
44. 44. Left-tailed Test H0: µ ≥ 200 H1: µ < 200Points Left
45. 45. Left-tailed Test H0: µ ≥ 200 H1: µ < 200Points Left Reject H0 Fail to reject H0 Values thatdiffer significantly from 200 200
46. 46. Right-tailed Test H0: µ ≤ 200 H1: µ > 200
47. 47. Right-tailed Test H0: µ ≤ 200 H1: µ > 200 Points Right
48. 48. Right-tailed Test H0: µ ≤ 200 H1: µ > 200 Points Right Fail to reject H0 Reject H0 Values that differ significantly 200 from 200
49. 49. Two-tailed TestH0: µ = 200H1: µ ≠ 200
50. 50. Two-tailed TestH0: µ = 200 α is divided equally between the two tails of the criticalH1: µ ≠ 200 region
51. 51. Two-tailed Test H0: µ = 200 α is divided equally between the two tails of the critical H1: µ ≠ 200 regionMeans less than or greater than
52. 52. Two-tailed Test H0: µ = 200 α is divided equally between the two tails of the critical H1: µ ≠ 200 regionMeans less than or greater than Reject H0 Fail to reject H0 Reject H0 200 Values that differ significantly from 200
53. 53. Summary of One- and Two-Tail Tests… One-Tail Test Two-Tail Test One-Tail Test (left tail) (right tail)Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.53
54. 54. One-Tailed and Two-Tailed Tests•Two-tailed tests or two-sided tests: When we display interest in extreme values of the statistic S or its corresponding z score on both sides of the mean, i.e., in both tails of the distribution.• One-tailed tests or one-sided tests: When we are interested only in extreme values to one side of the mean, i.e., in one tail of the distribution, as, for example, when we are testing the hypothesis that one process is better than another (which is different from testing whether one process is better or worse than the other).
55. 55. P Value• The null hypothesis H0 will be an assertion that a populationparameter has a specific value, and the alternative hypothesis H1 will be one of the following assertions:(i) The parameter is greater than the stated value (right-tailed test).(ii) The parameter is less than the stated value (left-tailed test).(iii) The parameter is either greater than or less than the stated value (two- tailed test).• P value of the test: The probability that a value of S in the direction(s) of H1 and as extreme as the one that actually did occur would occur if H0 were true.
56. 56. Interpreting the p-value… The smaller the p-value, the more statistical evidence exists to support the alternative hypothesis. •If the p-value is less than 1%, there is overwhelming evidence that supports the alternative hypothesis. •If the p-value is between 1% and 5%, there is a strong evidence that supports the alternative hypothesis. •If the p-value is between 5% and 10% there is a weak evidence that supports the alternative hypothesis. •If the p-value exceeds 10%, there is no evidence that supports the alternative hypothesis. We observe a p-value of .0069, hence there is overwhelming evidence to support H1: > 170.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.56
57. 57. Interpreting the p-value… Overwhelming Evidence (Highly Significant) Strong Evidence (Significant) Weak Evidence (Not Significant) No Evidence (Not Significant) 0 .01 .05 .10 p=.0069Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.57
58. 58. P Value
59. 59. P Value
60. 60. P Value•Small P values provide evidence for rejecting the null hypothesis in favor of the alternative hypothesis, and large P values provide evidence for not rejecting the null hypothesis in favor of the alternative hypothesis.•The P value and the level of significance do not provide criteria for rejecting or not rejecting the null hypothesis by itself, but for rejecting or not rejecting the null hypothesis in favor of the alternative hypothesis.• When the test statistic S is the standard normal random variable, the table in Appendix C is sufficient to compute the P value, but when S is one of the t, F, or chi-square random variables, all of which have different distributions depending on their degrees of freedom, either computer software or more extensive tables will be needed to compute the P value.
61. 61. Special Tests of Significance for Large Samples: Means
62. 62. Special Tests of Significance for Large Samples: Means
63. 63. Special Tests of Significance for Large Samples: Means
64. 64. Special Tests of Significance for Large Samples: Means
65. 65. Our Problem: The education department at a university has been accused of “grade inflation” so education majors have much higher GPAs than students in general. GPAs of all education majors should be compared with the GPAs of all students.  There are 1000s of education majors, far too many to interview.  How can this be investigated without interviewing all education majors?
66. 66. What we know: The average GPA for all students is 2.70. µ = 2.70 This value is a parameter. To the right is the X = 3.00 statistical information for a random sample s= 0.70 of education majors: N= 117
67. 67. Questions to ask: Is there a difference between the parameter (2.70) and the statistic (3.00)? Could the observed difference have been caused by random chance? Is the difference real (significant)?
68. 68. Two Possibilities:1. The sample mean (3.00) is the same as the pop. mean (2.70).  The difference is trivial and caused by random chance.1. The difference is real (significant).  Education majors are different from all students.
69. 69. The Null and Alternative Hypotheses:1. Null Hypothesis (H0)  The difference is caused by random chance.  The H0 always states there is “no significant difference.” In this case, we mean that there is no significant difference between the population mean and the sample mean.1. Alternative hypothesis (H1)  “The difference is real”.  (H1) always contradicts the H0. One (and only one) of these explanations must be true. Which one?
70. 70. Test the Explanations We always test the Null Hypothesis. Assuming that the H0 is true:  What is the probability of getting the sample mean (3.00) if the H0 is true and all education majors really have a mean of 2.70? In other words, the difference between the means is due to random chance.  If the probability associated with this difference is less than 0.05, reject the null hypothesis.
71. 71. Test the Hypotheses Use the .05 value as a guideline to identify differences that would be rare or extremely unlikely if H0 is true. This “alpha” value delineates the “region of rejection.” Use the Z score formula for single samples and Appendix A to determine the probability of getting the observed difference. If the probability is less than .05, the calculated or “observed” Z score will be beyond ±1.96 (the “critical” Z score).
72. 72. Two-tailed Hypothesis Test: Z= -1.96 Z = +1.96 c cWhen α = .05, then .025 of the area is distributed on either side of the curve in area (C )The .95 in the middle section represents no significant difference between the population and the sample mean.The cut-off between the middle section and +/- .025 is represented by a Z-value of +/- 1.96.
73. 73. Testing Hypotheses:Using The Five Step Model…1. Make Assumptions and meet test requirements.2. State the null hypothesis.3. Select the sampling distribution and establish the critical region.4. Compute the test statistic.5. Make a decision and interpret results.
74. 74. Step 1: Make Assumptions and Meet Test Requirements Random sampling  Hypothesis testing assumes samples were selected using random sampling.  In this case, the sample of 117 cases was randomly selected from all education majors. Level of Measurement is Interval-Ratio  GPA is I-R so the mean is an appropriate statistic. Sampling Distribution is normal in shape  This is a “large” sample (N≥100).
75. 75. Step 2 State the Null Hypothesis H0: μ = 2.7 (in other words, H0: = μ)  You can also state Ho: No difference between the sample mean and the population parameter  (In other words, the sample mean of 3.0 really the same as the population mean of 2.7 – the difference is not real but is due to chance.)  The sample of 117 comes from a population that has a GPA of 2.7.  The difference between 2.7 and 3.0 is trivial and caused by random chance.
76. 76. Step 2 (cont.) State the Alternate Hypothesis H1: μ≠2.7 (or, H0: ≠ μ)  Or H1: There is a difference between the sample mean and the population parameter  The sample of 117 comes a population that does not have a GPA of 2.7. In reality, it comes from a different population.  The difference between 2.7 and 3.0 reflects an actual difference between education majors and other students.  Note that we are testing whether the population the sample comes from is from a different population or is the same as the general student population.
77. 77. Step 3 Select Sampling Distribution and Establish the Critical Region Sampling Distribution= Z  Alpha (α) = .05  α is the indicator of “rare” events.  Any difference with a probability less than α is rare and will cause us to reject the H 0.
78. 78. Step 3 (cont.) Select Sampling Distributionand Establish the Critical Region Critical Region begins at Z= ± 1.96  This is the critical Z score associated with α = .05, two-tailed test.  If the obtained Z score falls in the Critical Region, or “the region of rejection,” then we would reject the H0.
79. 79. Step 4: Use Formula to Compute the TestStatistic (Z for large samples (≥ 100) Χ− µ Z= σ N
80. 80. When the Population σ is not known,use the following formula: Χ−µ Z= s N −1
81. 81. Test the Hypotheses 3.0 − 2.7 Z= = 4.62 .7 117 − 1 We can substitute the sample standard deviation S for σ (pop. s.d.) and correct for bias by substituting N-1 in the denominator. Substituting the values into the formula, we calculate a Z score of 4.62.
82. 82. Step 5 Make a Decision and Interpret Results The obtained Z score fell in the Critical Region, so we reject the H0.  If the H0 were true, a sample outcome of 3.00 would be unlikely.  Therefore, the H0 is false and must be rejected. Education majors have a GPA that is significantly different from the general student body (Z = 4.62, α = .05).* *Note: Always report significant statistics.
83. 83. Looking at the curve:(Area C = Critical Region when α=.05) Z= -1.96 Z = +1.96 c c z= +4.62 I
84. 84. Summary: The GPA of education majors is significantly different from the GPA of the general student body. In hypothesis testing, we try to identify statistically significant differences that did not occur by random chance. In this example, the difference between the parameter 2.70 and the statistic 3.00 was large and unlikely (p < .05) to have occurred by random chance.
85. 85. Summary (cont.) We rejected the H0 and concluded that the difference was significant. It is very likely that Education majors have GPAs higher than the general student body
86. 86. Special Tests of Significance for Large Samples: Proportions
87. 87. Special Tests of Significance for Large Samples: Proportions
88. 88. Special Tests of Significance for Large Samples: Difference of Means
89. 89. Special Tests of Significance for Large Samples: Difference of Means
90. 90. Special Tests of Significance for Large Samples: Difference of Means
91. 91. Special Tests of Significance for Large Samples: Difference of Means
92. 92. Special Tests of Significance for Large Samples: Difference of Means
93. 93. Special Tests of Significance for Large Samples: Difference of Means
94. 94. Special Tests of Significance for Large Samples: Difference of Proportions
95. 95. Special Tests of Significance for Large Samples: Difference of Proportions
96. 96. Special Tests of Significance for Large Samples: Difference of Proportions
97. 97. Special Tests of Significance for Small Samples: Means
98. 98. Special Tests of Significance for Small Samples: Means
99. 99. Special Tests of Significance for Small Samples: Means
100. 100. Special Tests of Significance for Small Samples: Means
101. 101. Using the Student’s t Distribution forSmall Samples (One Sample T-Test) When the sample size is small (approximately < 100) then the Student’s t distribution should be used (see Appendix B) The test statistic is known as “t”. The curve of the t distribution is flatter than that of the Z distribution but as the sample size increases, the t-curve starts to resemble the Z-curve (see text p. 230 for illustration)
102. 102. Degrees of Freedom The curve of the t distribution varies with sample size (the smaller the size, the flatter the curve) In using the t-table, we use “degrees of freedom” based on the sample size. For a one-sample test, df = N – 1. When looking at the table, find the t-value for the appropriate df = N-1. This will be the cutoff point for your critical region.
103. 103. Formula for one sample t-test: Χ−µ t= S N −1
104. 104. Example A random sample of 26 sociology graduates scored 458 on the GRE advanced sociology test with a standard deviation of 20. Is this significantly different from the population average (µ = 440)?
105. 105. Solution (using five step model) Step 1: Make Assumptions and Meet Test Requirements: 1. Random sample 2. Level of measurement is interval-ratio 3. The sample is small (<100)
106. 106. Solution (cont.)Step 2: State the null and alternate hypotheses.H0: µ = 440 (or H0: = μ)H1: µ ≠ 440
107. 107. Solution (cont.) Step 3: Select Sampling Distribution and Establish the Critical Region1. Small sample, I-R level, so use t distribution.2. Alpha (α) = .053. Degrees of Freedom = N-1 = 26-1 = 254. Critical t = ±2.060
108. 108. Solution (cont.) Step 4: Use Formula to Compute the Test Statistic Χ−µ 458 − 440t= = = 4.5 S 20 N −1 26 − 1
109. 109. Looking at the curve for the t distributionAlpha (α) = .05 t= -2.060 t = +2.060 c c t= +4.50 I
110. 110. Step 5 Make a Decision and Interpret Results The obtained t score fell in the Critical Region, so we reject the H0 (t (obtained) > t (critical)  If the H0 were true, a sample outcome of 458 would be unlikely.  Therefore, the H0 is false and must be rejected. Sociology graduates have a GRE score that is significantly different from the general student body (t = 4.5, df = 25, α = .05).
111. 111. Testing Sample Proportions: When your variable is at the nominal (or ordinal) level the one sample z-test for proportions should be used. If the data are in % format, convert to a proportion first. The method is the same as the one sample Z-test for means (see above)
112. 112. Special Tests of Significance for Small Samples: Variance
113. 113. Special Tests of Significance for Small Samples: Variance
114. 114. Special Tests of Significance for Small Samples: Variance
115. 115. Special Tests of Significance for Small Samples: Difference of Means
116. 116. Special Tests of Significance for Small Samples: Difference of Means
117. 117. Special Tests of Significance for Small Samples: Ratios of Variances
118. 118. Special Tests of Significance for Small Samples: Ratios of Variances
119. 119. Special Tests of Significance for Small Samples: Ratios of Variances
120. 120. Concepts of Hypothesis Testing… For example, if we’re trying to decide whether the mean is not equal to 350, a large value of (say, 600) would provide enough evidence. If is close to 350 (say, 355) we could not say that this provides a great deal of evidence to infer that the population mean is different than 350.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.120
121. 121. Concepts of Hypothesis Testing (4)… The two possible decisions that can be made: Conclude that there is enough evidence to support the alternative hypothesis (also stated as: reject the null hypothesis in favor of the alternative) Conclude that there is not enough evidence to support the alternative hypothesis (also stated as: failing to reject the null hypothesis in favor of the alternative) NOTE: we do not say that we accept the null hypothesis if a statistician is around…Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.121
122. 122. Concepts of Hypothesis Testing (2)… The testing procedure begins with the assumption that the null hypothesis is true. Thus, until we have further statistical evidence, we will assume: H0: = 350 (assumed to be TRUE) The next step will be to determine the sampling distribution of the sample mean assuming the true mean is 350. is normal with 350 75/SQRT(25) = 15Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.122
123. 123. Is the Sample Mean in the Guts of the Sampling Distribution??Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.123
124. 124. Three ways to determine this: First way 1. Unstandardized test statistic: Is in the guts of the sampling distribution? Depends on what you define as the “guts” of the sampling distribution. If we define the guts as the center 95% of the distribution [this means α = 0.05], then the critical values that define the guts will be 1.96 standard deviations of X-Bar on either side of the mean of the sampling distribution [350], or UCV = 350 + 1.96*15 = 350 + 29.4 = 379.4 LCV = 350 – 1.96*15 = 350 – 29.4 = 320.6Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.124
125. 125. 1. Unstandardized Test Statistic ApproachCopyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.125
126. 126. Three ways to determine this: Second way 2. Standardized test statistic: Since we defined the “guts” of the sampling distribution to be the center 95% [α = 0.05], If the Z-Score for the sample mean is greater than 1.96, we know that will be in the reject region on the right side or If the Z-Score for the sample mean is less than -1.97, we know that will be in the reject region on the left side. Z=( - )/ = (370.16 – 350)/15 = 1.344 Is this Z-Score in the guts of the sampling distribution???Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.126
127. 127. 2. Standardized Test Statistic ApproachCopyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.127
128. 128. Three ways to determine this: Third way 3. The p-value approach (which is generally used with a computer and statistical software): Increase the “Rejection Region” until it “captures” the sample mean. For this example, since is to the right of the mean, calculate P( > 370.16) = P(Z > 1.344) = 0.0901 Since this is a two tailed test, you must double this area for the p-value. p-value = 2*(0.0901) = 0.1802 Since we defined the guts as the center 95% [α = 0.05], the reject region is the other 5%. Since our sample mean, , is in the 18.02% region, it cannot be in our 5% rejection region [α = 0.05].Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.128
129. 129. 3. p-value approachCopyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.129
130. 130. Statistical Conclusions: Unstandardized Test Statistic: Since LCV (320.6) < (370.16) < UCV (379.4), we reject the null hypothesis at a 5% level of significance. Standardized Test Statistic: Since -Zα/2(-1.96) < Z(1.344) < Zα/2 (1.96), we fail to reject the null hypothesis at a 5% level of significance. P-value: Since p-value (0.1802) > 0.05 [α], we fail to reject the hull hypothesis at a 5% level of significance.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.130
131. 131. Example 11.1… A department store manager determines that a new billing system will be cost-effective only if the mean monthly account is more than \$170. A random sample of 400 monthly accounts is drawn, for which the sample mean is \$178. The accounts are approximately normally distributed with a standard deviation of \$65. Can we conclude that the new system will be cost-effective?Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.131
132. 132. Example 11.1… The system will be cost effective if the mean account balance for all customers is greater than \$170. We express this belief as a our research hypothesis, that is: H1: > 170 (this is what we want to determine) Thus, our null hypothesis becomes: H0: = 170 (this specifies a single value for the parameter of interest) – Actually H0: μ < 170Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.132
133. 133. Example 11.1… What we want to show: H1: > 170 H0: < 170 (we’ll assume this is true) Normally we put Ho first. We know: n = 400, = 178, and = 65 = 65/SQRT(400) = 3.25 α = 0.05Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.133
134. 134. Example 11.1… Rejection Region… The rejection region is a range of values such that if the test statistic falls into that range, we decide to reject the null hypothesis in favor of the alternative hypothesis. is the critical value of to reject H0.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.134
135. 135. Example 11.1… At a 5% significance level (i.e. =0.05), we get [all α in one tail] Zα = Z0.05 = 1.645 Therefore, UCV = 170 + 1.645*3.25 = 175.35 Since our sample mean (178) is greater than the critical value we calculated (175.35), we reject the null hypothesis in favor of H1 OR (>1.645) Reject null OR p-value = P( > 178) = P(Z > 2.46) = 0.0069 < 0.05 Reject nullCopyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.135
136. 136. Example 11.1… The Big Picture… H1: > 170 =175.34 H0: = 170 =178 Reject H0 in favor ofCopyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.136
137. 137. Conclusions of a Test of Hypothesis… If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative hypothesis is true. If we fail to reject the null hypothesis, we conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true. This does not mean that we have proven that the null hypothesis is true! Keep in mind that committing a Type I error OR a Type II error can be VERY bad depending on the problem.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.137
138. 138. One tail test with rejection region on right The last example was a one tail test, because the rejection region is located in only one tail of the sampling distribution: More correctly, this was an example of a right tail test. H1: μ > 170 H0: μ < 170Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.138
139. 139. One tail test with rejection region on left The rejection region will be in the left tail.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.139
140. 140. Two tail test with rejection region in both tails The rejection region is split equally between the two tails.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.140
141. 141. Example 11.2… Students work AT&T’s argues that its rates are such that customers won’t see a difference in their phone bills between them and their competitors. They calculate the mean and standard deviation for all their customers at \$17.09 and \$3.87 (respectively). Note: Don’t know the true value for σ, so we estimate σ from the data [σ ~ s = 3.87] – large sample so don’t worry. They then sample 100 customers at random and recalculate a monthly phone bill based on competitor’s rates. Our null and alternative hypotheses are H1: ≠ 17.09. We do this by assuming that: H0: = 17.09Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.141
142. 142. Example 11.2… The rejection region is set up so we can reject the null hypothesis when the test statistic is large or when it is small. stat is “small” stat is “large” That is, we set up a two-tail rejection region. The total area in the rejection region must sum to , so we divide α by 2.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.142
143. 143. Example 11.2… At a 5% significance level (i.e. = .05), we have /2 = .025. Thus, z.025 = 1.96 and our rejection region is: z < –1.96 -or- z > 1.96 -z.025 +z.025 z 0Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.143
144. 144. Example 11.2… From the data, we calculate = 17.55 Using our standardized test statistic: We find that: Since z = 1.19 is not greater than 1.96, nor less than –1.96 we cannot reject the null hypothesis in favor of H1. That is “there is insufficient evidence to infer that there is a difference between the bills of AT&T and the competitor.”Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.144
145. 145. Probability of a Type II Error – A Type II error occurs when a false null hypothesis is not rejected or “you accept the null when it is not true” but don’t say it this way if a statistician is around. In practice, this is by far the most serious error you can make in most cases, especially in the “quality field”.Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.145
146. 146. Probability you ship pills whose mean amount of medication is 7 mg approximately 67%Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 11.146