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Statistics for Management Fundamentals of Hypothesis Testing
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Lesson Topics 1. What is a Hypothesis? Hypothesis Testing Methodology Hypothesis Testing Process Level of Significance, a Errors in Making Decisions 2. Hypothesis Testing: Steps Z Test for the Mean (s Known) Connection to Confidence Interval Estimation Hypothesis Testing Methodology
State its opposite, the Alternative Hypothesis ( H 1 : < 3)
Hypotheses are mutually exclusive & exhaustive
Sometimes it is easier to form the alternative hypothesis first.
Identify the Problem
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Population Assume the population mean age is 50. (Null Hypothesis) REJECT The Sample Mean Is 20 Sample Null Hypothesis Hypothesis Testing Process No, not likely!
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Sample Mean = 50 Sampling Distribution It is unlikely that we would get a sample mean of this value ... ... if in fact this were the population mean. ... Therefore, we reject the null hypothesis that = 50. 20 H 0 Reason for Rejecting H 0
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H 0 : Innocent Jury Trial Hypothesis Test Actual Situation Actual Situation Verdict Innocent Guilty Decision H 0 True H 0 False Innocent Correct Error Do Not Reject H 0 1 - Type II Error ( ) Guilty Error Correct Reject H 0 Type I Error ( ) Power (1 - ) Result Possibilities
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Reduce probability of one error and the other one goes up. & Have an Inverse Relationship
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Z 0 Reject H 0 Z 0 Reject H 0 H 0 : H 1 : < 0 H 0 : 0 H 1 : > 0 Must Be Significantly Below = 0 Small values don’t contradict H 0 Don’t Reject H 0 ! Rejection Region
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. Test at the 0.05 level.
368 gm. Example: One Tail Test H 0 : 368 H 1 : > 368 _
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Z .04 .06 1.6 . 5495 . 5505 .5515 1.7 .5591 .5599 .5608 1.8 .5671 .5678 .5686 .5738 .5750 Z 0 Z = 1 1.645 .50 -. 05 .45 . 05 1.9 .5744 Standardized Normal Probability Table (Portion) What Is Z Given = 0.05 ? = .05 Finding Critical Values: One Tail Critical Value = 1.645
Test Statistic: Decision: Conclusion: Do Not Reject at = .05 No Evidence True Mean Is More than 368 Z 0 1.645 .05 Reject Example Solution: One Tail H 0 : 368 H 1 : > 368
Does an average box of cereal contains 368 grams of cereal? A random sample of 25 boxes showed X = 372.5 . The company has specified to be 15 grams. Test at the 0.05 level.
368 gm. Example: Two Tail Test H 0 : 368 H 1 : 368
Test Statistic: Decision: Conclusion: Do Not Reject at = .05 No Evidence that True Mean Is Not 368 Z 0 1.96 .025 Reject Example Solution: Two Tail -1.96 .025 H 0 : 386 H 1 : 386
If not normal, only slightly skewed & a large sample taken
Parametric test procedure
t test statistic
t-Test: Unknown
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Example: One Tail t-Test Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5 , and S= 15 . Test at the 0.01 level. 368 gm. H 0 : 368 H 1 : 368 is not given,
Test Statistic: Decision: Conclusion: Do Not Reject at = .01 No Evidence that True Mean Is More than 368 Z 0 2.4377 .01 Reject Example Solution: One Tail H 0 : 368 H 1 : 368
Do not reject at = .05 Z Test for Proportion: Solution H 0 : p .04 H 1 : p .04 Critical Values: 1.96 Test Statistic: Decision: Conclusion: We do not have sufficient evidence to reject the company’s claim of 4% response rate. Z p - p p (1 - p) n s = .04 -.05 .04 (1 - .04) 500 = -1.14 Z 0 Reject Reject .025 .025
t Test for Differences in Two Means (Variances Unknown)
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Developing the Pooled-Variance t Test (Part 1)
Setting Up the Hypothesis:
H 0 : 1 2 H 1 : 1 > 2 H 0 : 1 - 2 = 0 H 1 : 1 - 2 0 H 0 : 1 = 2 H 1 : 1 2 H 0 : 1 2 H 0 : 1 - 2 0 H 1 : 1 - 2 > 0 H 0 : 1 - 2 H 1 : 1 - 2 < 0 OR OR OR Left Tail Right Tail Two Tail H 1 : 1 < 2
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Developing the Pooled-Variance t Test (Part 2)
Calculate the Pooled Sample Variances as an Estimate of the Common Populations Variance:
= Pooled-Variance = Variance of Sample 1 = Variance of sample 2 = Size of Sample 1 = Size of Sample 2
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t X X S n S n S n n df n n P 1 2 1 2 2 1 1 2 2 2 2 1 2 1 2 1 1 1 1 2 Hypothesized Difference Developing the Pooled-Variance t Test (Part 3)
You’re a financial analyst for Charles Schwab. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:
NYSE NASDAQ Number 21 25
Mean 3.27 2.53
Std Dev 1.30 1.16
Assuming equal variances, is there a difference in average yield ( = 0.05 )?
Test Statistic: Decision: Conclusion: Reject at = 0.05 There is evidence of a difference in means. t 0 2.0154 -2.0154 .025 Reject H 0 Reject H 0 .025 t 3 27 2 53 1 510 21 25 2 03 . . . . Solution
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