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©2006 Thomson/South-Western 1 Chapter 10 – Hypothesis Testing for the Mean of a Population Slides prepared by Jeff Heyl Lincoln University ©2006 Thomson/South-Western Concise Managerial Statistics Concise Managerial Statistics KVANLI PAVUR KEELING KVANLI PAVUR KEELING
©2006 Thomson/South-Western 2 Hypothesis Testing on the Mean Null hypothesis (H o ): A statement (equality or inequality) concerning a population parameter; the researcher wishes to discredit this statement. Null hypothesis (H o ): A statement (equality or inequality) concerning a population parameter; the researcher wishes to discredit this statement. Alternative hypothesis (H a ): A statement in contradiction to the null hypothesis; the researcher wishes to support this statement. Alternative hypothesis (H a ): A statement in contradiction to the null hypothesis; the researcher wishes to support this statement.
©2006 Thomson/South-Western 3 Type I and Type II Errors =probability of rejecting the H o when H o is true (Type I error) =probability of failing to rejecting the H o when H o is false (Type II error) Actual Situation Conclusion H o TrueH o False Type II error Fail to Reject H o Correct decision Type II error Type I error Reject H o Type I error Correct decision
©2006 Thomson/South-Western 4 Hypothesis Testing Process Determine the H o and H a Determine the H o and H a Determine the significance level Determine the significance level Compare the sample mean (variance) to the hypothesized mean (variance) Compare the sample mean (variance) to the hypothesized mean (variance) Decide whether to fail to reject or reject H o Decide whether to fail to reject or reject H o Determine what the decision means in reference to the problem Determine what the decision means in reference to the problem
©2006 Thomson/South-Western 5 Height Example Height Example H o : = 5.9 H a : ≠ 5.9 =.05 = P(rejecting H o when H o is true) critical value = ± 1.96 Reject H o if > 1.96 X - 5.9 / n X - 5.9 / n
©2006 Thomson/South-Western 6 Height Example Height Example Figure 10.1 Z -k-k 0 k |Z| > k.025 Area =.5 -.025 =.475
©2006 Thomson/South-Western 7 Height Example Height Example Figure 10.2 X 5.79’ µ x = 5.9’ distance =.11’ = 2.38 x.046 x ≈ =.046’.4 75
©2006 Thomson/South-Western 8 Computed Value Computed Value Because - 2.38 < - 1.96, we reject H o thus we conclude that the average population male height is not equal to 5.9 Z = = = -2.38 = Z * X - 5.9 / n 5.79 - 5.9.4 / 75
©2006 Thomson/South-Western 9 Height Example Height Example Figure 10.3 Z -k-k 0 Area =.01 k Area =.5 -.005 =.495.005
©2006 Thomson/South-Western 10 Height Example Height Example Figure 10.4 Z * = -2.38 Z -2.575 2.575 0 Area =.01 Reject H o if Z * falls here
©2006 Thomson/South-Western 11 Hypothesis Testing 5 Step Procedure 3.Define the rejection region 4.Calculate the test statistic 5.Give a conclusion in terms of the problem 1.Set up the null and alternative hypothesis Z =Z =Z =Z = X - µ o / n 2.Define the test statistic
©2006 Thomson/South-Western 12 Everglo Light Example 1.Define the hypotheses H o : µ = 400 H a : µ ≠ 400 3.Define the rejection region reject H o if Z > 1.645 or Z 1.645 or Z < -1.645 2.Define the test statistic Z =Z =Z =Z = X - 400 / n
©2006 Thomson/South-Western 13 Everglo Light Example Figure 10.5 Z -1.645 1.645 0 Area = =.1 Area =.5 -.05 =.45 22 =.05 22 Z * = 2.5
©2006 Thomson/South-Western 14 Everglo Light Example 5.State the conclusion There is sufficient evidence to conclude that the average lifetime of Everglo bulbs is not 400 hours 4.Calculate the value of the test statistic Z * ≈ = = 2.5 208 420 - 400 40 / 25
©2006 Thomson/South-Western 15 Confidence Intervals and Hypothesis Testing When testing H o : µ = µ o versus H a : µ ≠ µ o using the five-step procedure and a significance level, , H o will be rejected if and only if µ o lies outside the (1 - ) 100 confidence interval for µ X - k to X + k n n
©2006 Thomson/South-Western 16 Power of a Statistical Test = P(fail to reject H o when H o is false) 1- = P(rejecting H o when H o is false) 1- = the power of the test
©2006 Thomson/South-Western 17 Probability of Rejecting H o Figure 10.6 X 405413.16B Area = 1 - when µ = 405 Area = =.10 386.84400 A
©2006 Thomson/South-Western 18 Power of Tests 2.Power of test = P(Z > z 1 ) + P(Z z 1 ) + P(Z < z 2 ) 1.Determine z 1 = Z /2 - µ - µ o / n z 1 = -Z /2 - µ - µ o / n
©2006 Thomson/South-Western 19 Power Curve Power curve for test of H o versus H a using five-step procedure Power curve for test of H o versus H a using any other procedure 400 405.1655.10 Figure 10.7 µ 1.0 1 - = P(rejecting H o )
©2006 Thomson/South-Western 20 One Tailed Test for µ H o : µ ≥ 32.5H a : µ < 32.5 1.Define H o and H a prior to observation Z =Z =Z =Z = X - µ o / n 2.Define the test statistic 3. Reject if H o if Z = < -1.645 X - 32.5 / n Example 10.4
©2006 Thomson/South-Western 21 One Tailed Test for µ 5.Study supports the claim that the average mileage for the Bullet is less than 32.5 mpg – supports claim of false advertising Z * = = -2.70 30.4 - 32.5 5.5 / 50 4.The value of the test statistic is Example 10.4
©2006 Thomson/South-Western 22 One Tailed Test for µ Figure 10.8 -1.645 0 Z Area =.5 -.05 =.45 Area = =.05
©2006 Thomson/South-Western 23 Statistical Software Example Z * = 1.531 Figure 10.9 Z 2.33 0 Area =.49 Area = =.01
©2006 Thomson/South-Western 24 Statistical Software Example Figure 10.10
©2006 Thomson/South-Western 25 Statistical Software Example Figure 10.11
©2006 Thomson/South-Western 26 Power of Tests 2.Power of test = P(Z > z 1 ) 1.Determine z 1 = Z /2 - µ - µ o / n H o : µ ≤ µ o verses H a : µ > µ o
©2006 Thomson/South-Western 27 Power of Tests 2.Power of test = P(Z < z 2 ) 1.Determine z 1 = -Z /2 - µ - µ o / n H o : µ ≥ µ o verses H a : µ < µ o
©2006 Thomson/South-Western 28 Power of Tests Tests on a Population Mean ( Known) Two-Tailed Test H o : µ = µ o H a : µ ≠ µ o reject H o if |Z *| > Z /2 One-Tailed Test H o : µ ≤ µ o H a : µ > µ o reject H o if Z * > Z H o : µ ≥ µ o H a : µ < µ o reject H o if Z * < -Z
©2006 Thomson/South-Western 29 Determining the p-Value The p-value is the value of at which the hypothesis test procedure changes conclusions based on a given set of data. It is the largest value of for which you will fail to reject H o
©2006 Thomson/South-Western 30 Determining the p-Value Figure 10.12 -1.96 0 Z -2.575 1.96 2.575 Z * = -2.38 Area =.025 Area =.005 Area =.025 Area =.005
©2006 Thomson/South-Western 31 Determining the p-Value Figure 10.13 0 Z -2.38 2.38 Area = p value Area=.5 -.4913 =.0087 Area =.4913 (Table A.4)
©2006 Thomson/South-Western 32 Procedure for Finding the p-Value For H a : µ ≠ µ o p = 2 (area outside Z *) For H a : µ > µ o p = area to the right of Z * For H a : µ < µ o p = area to the left of Z *
©2006 Thomson/South-Western 33 Determining the p-Value Figure 10.14 0 Z Z * = 1.53 p= area =.5 -.4370 =.0630 Area =.4370
©2006 Thomson/South-Western 34 Interpreting the p-Value Classical Approach reject H o if p-value < fail to reject H o is p-value ≥ General rule of thumb reject H o if p-value is small (p <.01) fail to reject H o is p-value is large (p >.1)
©2006 Thomson/South-Western 35 Interpreting the p-Value Small p Reject H o.01 Large p Fail to reject H o p.1 Inconclusive
©2006 Thomson/South-Western 36 Another Interpretation 1.For a two-tailed test where H o : µ ≠ µ o, the p-value is the probability that the value of the test statistic, Z *, will be at least as large (in absolute value) as the observed Z *, if µ is in fact equal to µ o 2.For a one-tailed test where H a : µ > µ o, the p-value is the probability that the value of the test statistic, Z *, will be at least as large as the observed Z *, if µ is in fact equal to µ o 3.For a one-tailed test where H a : µ < µ o, the p-value is the probability that the value of the test statistic, Z *, will be at least as large as the observed Z *, if µ is in fact equal to µ o
©2006 Thomson/South-Western 37 Statistical Software Example Figure 10.15
©2006 Thomson/South-Western 38 Statistical Software Example Figure 10.16
©2006 Thomson/South-Western 39 Statistical Software Example Figure 10.17
©2006 Thomson/South-Western 40 Practical Versus Statistical Figure 10.18 Z Z * = -2.69 Area= p value =.0036 (from Table A.4)
©2006 Thomson/South-Western 41 Hypothesis Testing on the Mean of a Normal Population ( Unknown) Normal population Normal population Population standard deviation unknown Population standard deviation unknown Small sample Small sample Student t distribution Student t distribution Nonparametric procedure Nonparametric procedure t =t =t =t = X - µ o s / n
©2006 Thomson/South-Western 42 Small Sample Test (a) Normal population Symmetric (nonnormal) population Skewed population (b) Figure 10.19
©2006 Thomson/South-Western 43 Clark Products Example 1. When a question is phrased “Is there evidence to indicate that...,” what follows is the alternative hypothesis H o : µ = 10 and H a : µ ≠ 10 2.The test statistic here is t =t =t =t = X - µ o s / n 3.Using a significance level of.05 and Figure 10.20, the corresponding two-tailed procedure is to reject H o if |t| > t.025,17 = 2.11
©2006 Thomson/South-Western 44 Clark Products Example Figure 10.20 Z -2.11 2.11 Area=.025 Reject H o here
©2006 Thomson/South-Western 45 Clark Products Example Figure 10.21 t * = 1.83 t 2.111.74 0 Area =.05 Area =.025
©2006 Thomson/South-Western 46 Clark Products Example 5. There is insufficient evidence to indicate that the average output voltage is different from 10 volts 4.The value of the test statistic is t * = = 1.83 10.331 - 10.767 / 18
©2006 Thomson/South-Western 47 Clark Products Example Figure 10.22
©2006 Thomson/South-Western 48 Clark Products Example Figure 10.24 Figure 10.23
©2006 Thomson/South-Western 49 Small-Sample Tests on a Normal Population Mean Two-tailed test H o : µ = µ o H a : µ ≠ µ o reject H o if |t *| > t /2, n-1 One tail test H o : µ = µ o H a : µ < µ o reject H o if t * < t , n-1 H o : µ = µ o H a : µ > µ o reject H o if t * > t , n-1
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