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Adapted by Peter Au, George Brown College McGraw-Hill Ryerson Copyright © 2011 McGraw-Hill Ryerson Limited.

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Copyright © 2011 McGraw-Hill Ryerson Limited 9.1z Tests about a Difference in Population Means: One-Tailed Alternativez Tests about a Difference in Population Means: One-Tailed Alternative 9.2 z Tests about a Difference in Population Means: Two-Tailed Alternativez Tests about a Difference in Population Means: Two-Tailed Alternative 9.3t Tests about a Difference in Population Means: One-Tailed Alternativet Tests about a Difference in Population Means: One-Tailed Alternative 9.4 t Tests about a Difference in Population Means: Two-Tailed Alternativet Tests about a Difference in Population Means: Two-Tailed Alternative 9.5z Tests about a Difference in Population Proportionsz Tests about a Difference in Population Proportions 9.6 F Tests about a Difference in Population VariancesF Tests about a Difference in Population Variances 9-2

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Copyright © 2011 McGraw-Hill Ryerson Limited Suppose a random sample has been taken from each of two different populations (populations 1 and 2) and suppose that the populations are independent of each other Then the random samples are independent of each other Then the sampling distribution of the difference in sample means is normally distributed or that each of the sample sizes n 1 and n 2 is large ((n 1, n 2 ) is at least 40) is more than sufficient We can easily test a hypothesis about the difference between the means 9-3

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Copyright © 2011 McGraw-Hill Ryerson Limited Suppose we wish to conduct a one-sided hypothesis test about μ 1 - μ 2 The difference between these means can be represented by “D” i.e. μ 1 - μ 2 = D The null hypothesis is: H 0 : μ 1 - μ 2 = D 0 The one-tailed alternative hypothesis is: H a : μ 1 - μ 2 > D 0 or H a : μ 1 - μ 2 < D 0 9-4 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited Often D 0 will be the number 0 In such a case, the null hypothesis H 0 : μ 1 - μ 2 = 0 says there is no difference between the population means μ 1 and μ 2 When D 0 = 0, each alternative hypothesis implies that the population means μ 1 and μ 2 differ Also note the standard deviation of the difference of means is: 9-5 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited The test statistic is: The sampling distribution of this statistic is a standard normal distribution If the populations are normal and the samples are independent... 9-6 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited Reject H 0 : 1 – 2 = D 0 in favor of a particular alternative hypothesis at a level of significance if the appropriate rejection point rule holds or if the corresponding p-value is less than Rules are on the next slide … 9-7 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited Alternative HypothesisReject H 0 if:p-value H a : μ 1 – μ 2 > D 0 z > z α Area under standard normal to the right of z H a : μ 1 – μ 2 < D 0 z < -z α Area under standard normal to the left of –z H a : μ 1 – μ 2 ≠ D 0 *|z| > z α/2 *Twice the area under standard normal to the right of |z| * Note For Two-Tailed Alternative either z > z a/2 or z < –z a/2 9-8 Null Hypothesis: H 0 : 1 – 2 = D 0 L01 L05

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Copyright © 2011 McGraw-Hill Ryerson Limited Test the claim that the new system reduces the mean waiting time Test at the = 0.05 significance level the null H 0 : 1 – 2 = 0 against the alternative H a : 1 – 2 > 0 Use the rejection rule H 0 if z > z At the 5% significance level, z = z 0.05 = 1.645 So reject H 0 if z > 1.645 Use the sample and population data in Example 7.11 to calculate the test statistic 9-9 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited Because z = 14.21 > z 0.05 = 1.645, reject H 0 Conclude that 1 – 2 is greater than 0 and therefore it appears as though the new system does reduce the waiting time Alternatively we can use the p-value The p-value for this test is the area under the standard normal curve to the right of z = 14.21 Since this p value is less than 0.001, we have extremely strong evidence that μ 1 - μ 2 is greater than 0 and, therefore, that the new system reduces the mean customer waiting time 9-10 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited The new system will be implemented only if it reduces mean waiting time by more than 3 minutes Set D 0 = 3, and try to reject the null H 0 : 1 – 2 = 3 in favor of the alternative H a : 1 – 2 > 3 z=2.53 > z 0.05 = 1.645, we reject H 0 in favor of H a There is evidence that the mean waiting time is reduced by more than 3 minutes 9-11 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited The p-value for this test is the area under the standard normal curve to the right of z = 2.53 With Table A.3, the p-value is 0.5 – 0.4943 = 0.0057 There is strong evidence against H 0 Again there is evidence that the mean waiting time is reduced by more than 3 minutes 9-12 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited A 95% confidence interval for the difference in the mean waiting time is: 9-13 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited Alternative HypothesisReject H 0 if:p-value H a : μ 1 – μ 2 ≠ D 0 *|z| > z α/2 *Twice the area under standard normal to the right of |z| * Note For Two-Tailed Alternative either z > z a/2 or z < –z a/2 9-14 Null Hypothesis: H 0 : 1 – 2 = D 0 L01 L05

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Copyright © 2011 McGraw-Hill Ryerson Limited Provide evidence supporting the claim that the new system produces a different mean bank customer waiting time We will test H 0 : μ 1 - μ 2 = 0 versus H a : μ 1 = μ 2 ≠ 0 at the 0.05 level of significance Reject H 0 : μ 1 - μ 2 = 0 if the value of |z| is greater than z α/2 = z 0.025 = 1.96 9-15 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Use the sample and population data in Example 7.11 to calculate the test statistic z = 14.21 is greater than z 0.025 = 1.96 reject H 0 : μ 1 - μ 2 = 0 in favour of H a : μ 1 = μ 2 ≠ 0 Conclude that μ 1 - μ 2 is not equal to 0 There is a difference in the mean customer waiting times 9-16 L05 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Testing the null hypothesis H 0 : μ 1 – μ 2 = D 0 under two conditions 1.When variances are equal, 2.When variances are unequal, 9-17 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited 1.When The test statistic is: 2.When The test statistic is: 9-18 L02 L05

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Copyright © 2011 McGraw-Hill Ryerson Limited If sampled populations are both normal, but sample sizes and variances differ substantially, small-sample estimation and testing can be based on the following “unequal variance” procedure 9-19 Confidence IntervalTest Statistic For both the interval and test, the degrees of freedom are equal to L05 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited AlternativeReject H 0 if:p-value H a : μ 1 – μ 2 > D 0 (One-Tailed)t > t α Area under t distribution to the right of t H a : μ 1 – μ 2 < D 0 (One-Tailed)t < -t α Area under t distribution to the left of t H a : μ 1 – μ 2 ≠ D 0 (Two-Tailed)|t| > t α/2 *Twice the area under t distribution to the right of |t| where t α, t α/2, and p-values are based on (n 1 + n 2 - 2) degrees of freedom * either t > α a/2 or t < –t α/2 9-20 H 0 : μ 1 – μ 2 = D 0 L05 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited If the population of differences is normal, we can reject H 0 : D = D 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than We need a test statistic … 9-21 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited The test statistic is: D 0 = 1 – 2 is the claimed or actual difference between the population means D 0 varies depending on the situation Often D 0 = 0, and the null means that there is no difference between the population means The sampling distribution of this statistic is a t distribution with (n – 1) degrees of freedom Rules are on the next slide … 9-22 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited AlternativeReject H 0 if:p-value H a : μ D > D 0 (One-Tailed)t > t α Area under t distribution to the right of t H a : μ D < D 0 (One-Tailed)t < -t α Area under t distribution to the left of t H a : μ D ≠ D 0 (Two-Tailed)|t| > t α/2 *Twice the area under t distribution to the right of |t| where t α, t α/2, and p-values are based on (n – 1) degrees of freedom * either t > α a/2 or t < –t α/2 9-23 L05 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited Example 9.3 The Coffee Cup Case In order to compare the mean hourly yields obtained by using the Java and Joe production methods, we will test H 0 : μ 1 - μ 2 = 0 versus H a : μ 1 - μ 2 > 0 at the 0.05 level of significance To perform the hypothesis test, we will use the sample information 9-24 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Unequal-variances procedure Consider the bank customer waiting time situation, recall that the bank manager wants to implement the new system only if it reduces the mean waiting time by more than three minutes Therefore, the manager will test the null hypothesis H 0 : μ 1 - μ 2 = 3 versus the alternative hypothesis H a : μ 1 - μ 2 > 3 at α = 0.05 9-25 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Suppose n 1 = 100 and n 2 = 100, computing the sample mean and standard deviation of each sample gives 9-26 L03 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited t = 2.53 is greater than t 0.05 = 1.65 Reject H 0 : μ 1 - μ 2 = 3 in favour of H a :μ 1 2 μ 2 > 3 at α 0.05 The new system reduces the mean customer waiting time by more than three minutes Examine the MegaStat output below t = 2.53, the associated p value is 0.0062, the very small p value tells us that we have very strong evidence against H 0 9-27 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Reject H 0 : μ 1 - μ 2 = 0 if t is greater than t α = t 0.05 = 1.860 Test Statistic: t = 4.6087 > t 0.05 = 1.860 We can reject H 0 Conclude at α = 0.05 the mean hourly yields obtained by using the two production methods differ Note the small p-value in figure 9.1 indicates strong evidence against H 0 9-28 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Example 9.4 The Repair Cost Comparison Case Forest City Casualty currently contracts to have moderately damaged cars repaired at garage 2 However, a local insurance agent suggests that garage 1 provides less expensive repair service that is of equal quality Forest City has decided to give some of its repair business to garage 1 only if it has very strong evidence that μ 1, the mean repair cost estimate at garage 1, is smaller than μ 2, the mean repair cost estimate at garage 2, that is, if μ D = μ 1 - μ 2 is less than zero 9-29 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited We will test H 0 : μ D = 0 (no difference) versus H a : μ D < 0 (difference – garage 1 costs are less than garage 2) at the 0.01 level of significance Reject if t < –t , that is, if t < –t 0.01 With n – 1 = 6 degrees of freedom, t 0.01 = 3.143 So reject H 0 if t < –3.143 9-30 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Calculate the t statistic: Because t = –4.2053 is less than –t 0.01 = – 3.143, reject H 0 Conclude at the = 0.01 significance level that it appears as though the mean repair cost at Garage 1 is less than the mean repair cost of Garage 2 From a computer, for t = -4.2053, the p-value is 0.003 Because this p-value is very small, there is very strong evidence that H 0 should be rejected and that 1 is actually less than 2 9-31 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Example 9.5 Coffee Cup Case (Revisited) In order to compare the mean hourly yields obtained by using the Java and Joe methods Test H 0 : μ 1 - μ 2 = 0 versus H a : μ 1 - μ 2 ≠ 0 at α = 0.05 Reject H 0 : μ 1 - μ 2 = 0 if the absolute value of t is greater than t α/2 = t 0.025 = 2.306 df = n 1 + n 2 - 2 = 5 + 5 - 2 = 8 Test Statistic Because |t| = 4.6087 is greater than t 0.025 = 2.306, reject H 0 in favor of H a Conclude at 5% significance level that the mean hourly yields from the two catalysts do differ 9-32 L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited 9-33 The p-value = 0.0017 The very small p-value indicates that there is very strong evidence against H 0 (that the means are the same). Conclude on basis of p-value the same as before, that the two catalysts differ in their mean hourly yields L02 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited The test statistic is: D 0 = p 1 – p 2 is the claimed or actual difference between the population proportions D 0 is a number whose value varies depending on the situation Often D 0 = 0, and the null means that there is no difference between the population means The sampling distribution of this statistic is a standard normal distribution 9-34 L02

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Copyright © 2011 McGraw-Hill Ryerson Limited If the population of differences is normal, we can reject H 0 : p 1 – p 2 = D 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than Rules are on the next slide … 9-35 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited For testing the difference of two population proportions 9-36 AlternativeReject H 0 if:p-value H a : p 1 – p 2 > D 0 z > z α Area under the standard normal to the right of z H a : p 1 – p 2 < D 0 z < -z α Area under the standard normal to the left of –z H a : p 1 – p 2 ≠ D 0 |z| > z α/2 *Twice the area under the standard normal to the right of |z| * either t > t a/2 or t < –t a/2 L01 L05

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Copyright © 2011 McGraw-Hill Ryerson Limited If D 0 = 0, estimate by If D 0 ≠ 0, estimate by 9-37 L02 L04

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Copyright © 2011 McGraw-Hill Ryerson Limited Recall from example 7.15 that p 1 is the proportion of all consumers in the Toronto area who are aware of the new product and that p 2 is the proportion of all consumers in the Vancouver area who are aware of the new product To test for the equality of these proportions, we will test H 0 : p 1 - p 2 = 0 versus H a : p 1 - p 2 ≠ 0 at the 0.05 level of significance Samples are large 9-38 L04

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Copyright © 2011 McGraw-Hill Ryerson Limited Since H a : p 1 - p 2 ≠ 0 is of the form H a : p 1 - p 2 ≠ D0 Reject H 0 : p 1 - p 2 = 0 if the absolute value of z is greater than z α/2 = z 0.05/2 = z 0.025 = 1.96 631 out of 1,000 randomly selected Toronto residents were aware of the product and 798 out of 1,000 randomly selected Vancouver residents were aware of the product, the estimate of p = p 1 = p 2 is 9-39 L04

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Copyright © 2011 McGraw-Hill Ryerson Limited Test Statistic Because |z| - 8.2673 is greater than 1.96, we can reject H 0 : p 1 - p 2 = 0 in favour of H a :p 1 - p 2 ≠ 0 The proportions of consumers who are aware of the product in Toronto and Vancouver differ We estimate that the percentage of consumers who are aware of the product in Vancouver is 16.7 percentage points higher than the percentage of consumers who are aware of the product in Toronto 9-40 L02 L04

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Copyright © 2011 McGraw-Hill Ryerson Limited The p value for this test is twice the area under the standard normal curve to the right of |z| = 8.2673 The area under the standard normal curve to the right of 3.29 is 0.0005, the p-value for testing H 0 is less than 2(0.0005) = 0.001 Extremely strong evidence that H 0 : p 1 - p 2 = 0 should be rejected Strong evidence that p 1 and p 2 differ 9-41 L03

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Copyright © 2011 McGraw-Hill Ryerson Limited Population 1 has variance 1 2 and population 2 has variance 2 2 The null hypothesis, H 0, is that the variances are the same H 0 : 1 2 = 2 2 The alternative is that one of them is smaller than the other That population has less variable, more consistent, measurements Suppose 1 2 > 2 2 Let’s look at the ratios of the variances Test H 0 : 1 2 / 2 2 = 1 versus H a : 1 2 / 2 2 > 1 9-42 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited Reject H 0 in favor of H a if s 1 2 /s 2 2 is significantly greater than 1 s 1 2 is the variance of a random sample of size n 1 from a population with variance 1 2 s 2 2 is the variance of a random sample of size n 2 from a population with variance 2 2 To decide how large s 1 2 /s 2 2 must be to reject H 0, describe the sampling distribution of s 1 2 /s 2 2 The sampling distribution of s 1 2 /s 2 2 is described by an F distribution 9-43

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Copyright © 2011 McGraw-Hill Ryerson Limited In order to use the F distribution Employ an F point, which is denoted F a F A is the point on the horizontal axis under the curve of the F distribution that gives a right-hand tail area equal to α Shape depends on two parameters: the numerator number of degrees of freedom (df 1 ) and the denominator number of degrees of freedom (df 2 ) 9-44

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Copyright © 2011 McGraw-Hill Ryerson Limited Suppose we randomly select independent samples from two normally distributed populations with variances 1 2 and 2 2 If the null hypothesis H 0 : 1 2 / 2 2 = 1 is true, then the population of all possible values of s 1 2 /s 2 2 has an F distribution with df 1 = (n 1 – 1) numerator degrees of freedom and with df 2 = (n 2 – 1) denominator degrees of freedom 9-45 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Recall that the F point F is the point on the horizontal axis under the curve of the F distribution that gives a right-hand tail area equal to The value of F depends on (the size of the right-hand tail area) and df 1 and df 2 Different F tables for different values of See: Table A.6 for = 0.10Table A.6 for = 0.10 Table A.7 for = 0.05Table A.7 for = 0.05 Table A.8 for = 0.025Table A.8 for = 0.025 Table A.9 for = 0.01Table A.9 for = 0.01 9-46 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Independent samples from two normal populations Test H 0 : 1 2 = 2 2 versus H a : 1 2 > 2 2 Use the test statistic F = s 1 2 /s 2 2 The p-value is the area to the right of this value of F under the F curve having df 1 = (n 1 – 1) numerator degrees of freedom and df 2 = (n 2 – 1) denominator degrees of freedom Reject H 0 at the significance level if: F > F , or p-value < 9-47 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Independent samples from two normal populations Test H 0 : 1 2 = 2 2 versus H a : 1 2 < 2 2 Use the test statistic F = s 2 2 /s 1 2 The p-value is the area to the right of this value of F under the F curve having df 1 = (n 1 – 1) numerator degrees of freedom and df 2 = (n 2 – 1) denominator degrees of freedom Reject H 0 at the significance level if: F > F , or p-value < 9-48 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Independent samples from two normal populations Test H 0 : 1 2 = 2 2 versus H a : 1 2 ≠ 2 2 Use the test statistic The p-value is twice the area to the right of this value of F under the F curve having df 1 = (n 1 – 1) numerator degrees of freedom and df 2 = (n 2 – 1) denominator degrees of freedom Reject H 0 at the significance level if: F > F , or p-value < 9-49 L01

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Copyright © 2011 McGraw-Hill Ryerson Limited The production supervisor wishes to use Figure 9.13 to determine whether σ 1 2, the variance of the average production yields obtained by using the Java method, is smaller than σ 2 2, the variance of the yields obtained by using the Joe method Test the hypotheses H 0 : σ 1 2 = σ 2 2 versus H a : σ 1 2 σ 2 2 9-50 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Using the Excel output we can compute the test statistic 9-51 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited Compare this value with F a based on df 1 = n 2 - 1 = 5 - 1 = 4 numerator degrees of freedom and df 2 = n 1 - 1 = 5 - 1 = 4 denominator degrees of freedom at the 0.05 level of significance F 0.05 = 6.39 F = 1.2544 is not greater than F 0.05 = 6.39 we cannot reject H 0 at α = 0.05 We cannot conclude that σ 1 2 is less than σ 2 2 9-52 L06

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Copyright © 2011 McGraw-Hill Ryerson Limited It is possible to compare two populations using a one-tail or a two-tailed test Hypothesis tests can be conducted on such populations (using CI’s, rejection points, or p-values) Populations may be independent or dependent (paired difference experiments) The value of σ may be known or unknown. This affects the type of test statistic we use (i.e. t or z) Independent tests can involve an equal variances assumption or an unequal variances assumption Two population variances can be compared using the F distribution 9-53

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Copyright © 2011 McGraw-Hill Ryerson Limited 9-54

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Copyright © 2011 McGraw-Hill Ryerson Limited 9-55

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Copyright © 2011 McGraw-Hill Ryerson Limited 9-56

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Copyright © 2011 McGraw-Hill Ryerson Limited 9-57

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Lecture 8: Hypothesis Testing

Lecture 8: Hypothesis Testing

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