252review 11/14/06 (Open this document in 'Outline' view!)The feedback that I get is that there are people who are absolutely sure that there is material we never covered on the final exam. I also remember one very indignant woman who was absolutely sure that you couldn't compare two means unless one was a population mean. This document may help.1. Meansa. Are the means for these three methods the same?b. Is the mean for method 1 above the mean for method 2?c. Is the mean for method 3 above 78?2. Mediansa. Are the medians for these three methods the same?b. Are the medians for method 1 and method 2 the same?c. Is the median for method 3 above 78?3. Variancesa. Are the variances for method 1 and method 2 the same?b. Is the variance for method 3 above 169?4. Distributions - Goodness of Fita. Does method 3 have a Normal distribution?b. Does method 3 have a Normal distribution with a mean of 85 and a standard deviation of 15?5. Proportions1. Are the proportions than need repairs the same for all three groups?2. Is the proportion that needs repairs greater for manufacturer 1 than manufacturer 2?3. Is the proportion that need repairs for manufacturer 1 greater than 1%?6. Regression and Correlation7. Significance© 2004 Roger Even Bove252review 11/14/06 (Open this document in 'Outline' view!)The feedback that I get is that there are people who are absolutely sure that there is material we never covered on the final exam. I also remember one very indignant woman who was absolutely sure that you couldn't compare two means unless one was a population mean. This document may help.M. REVIEW (Assume that 05. in all the following) In each case, be sure to clearly state the null hypothesis and the location of the 'do not reject' or 'reject' region. Note that other methods that we have studied, but which are not mentioned in this review are noted in boldface.1. MeansThe following data is used in the sections on means, medians, variances and distributions. It relates to different methods of training people to pass an exam. Method 1 Method 2 Method 3 74 78 68 88 80 83 82 65 50 93 57 91 55 89 84 70 77 94 81 92 Note: ,36518,462211xx,27879,369222xx,59140,720233xx,6,74.13,00.77111 nsx,5,72.12,80.73222 nsx.9,87.13,00.80333 nsxThese are only given for convenience, you should be able to compute every one of these sums.a. Are the means for these three methods the same?A multiple test of means is usually an Analysis of Variance, For this to be valid, we must assume that the samples come from populations with an underlying Normal distribution and common variance. Note that data is not cross-classified so that a 2-way ANOVA is not applicable.3210:H equalmeansallNotH :1Because there are many examples available, the computation of SSB and SST is not shown here. Youshould be able to do that computation. When you are finished, you should have an ANOVA table with thenumbers below.Source SS DF MS F 05.F 0HBetween 126 2 63 0.34 59.317,2FnsColumn means equalWithin 3131 17 184 Total 3257 19 Explanation: This is one-way ANOVA because the columns are considered three independent randomsamples with no cross classification. Since the Sum of Squares (SS) column must add up, 3131 is found bysubtracting 126 from 3257. Since 20n, the total degrees of freedom are 191 n. Since there are3 random samples or columns, the degrees of freedom for Between is 3 – 1 = 2.Since the Degrees of Freedom (DF) column must add up, 17 = 29 – 2. The Mean Square (MS) column isfound by dividing the SS column by the DF column. 63 is MSB and 17 is MSW. MSWMSBF , andis compared with 05.F from the F table 17,221 dfdf. We accept the null hypothesis if ourcomputed F is less than or equal to 3.59. Because our computed F is less than the table F, do notreject 0H.b. Is the mean for method 1 above the mean for method 2?The data that we can use is repeated here. ,6,79.188,74.13,00.7712111 nssx.5,80.161,72.12,80.7322222 nssxBecause the question does not include an equality, it must be an alternative hypothesis. Neither 77 nor 73.8 is a population mean, so these numbers cannot be in the null hypothesis, since a null hypothesis cannot include sample statistics. We thus conclude that our alternative hypothesis is 211:H , so that our null hypothesis must be the opposite, 210:H. The method taught in class assumes that thedata comes from a Normal Distribution and that these are two samples from populations with equal variances. We could do this problem using the test ratio method, the critical value method or a one-sided confidence interval. Only the test ratio will be done here. For the general formulas used, see the formula table.20.380.7300.7721 xxx 9256221 nnDF ,05. 211ˆ212222112nnsnsnsp= 833.179.176920.64795.943980.161479.1885905.t 2111ˆnnsspx 049.8786.6436667.069.176516169.176 Our hypotheses are 211210::HH or 0:0:10HH Since this is a 1-sided alternate hypothesis, you cannot use 2-sided confidence intervals or tests. Note that 77 and 74.80 cannot appear in the hypotheses because they are sample means, not population means.Test Ratio: 398.0049.8020.30xsxt. The 'do not reject' region is 905.tt . Since our tis below 833.1905.221ttnn, do not reject the null hypothesis.2c. Is the mean for method 3 above 78?The data that we can use is repeated here..9,87.13,00.80333 nsx78:78:3130HH Note that because the question involves an inequality, it is an alternative hypothesis.623333.4987.13nssxWe use only a Test Ratio here, but remember that all such problems can be done using a critical value or a one sided confidence interval. 4326.062333.478800xsxt. The 'do not reject' region is 805.tt . Since our t is below 860.1805.1ttn, do not reject the null hypothesis.Remember, for comparing means or medians of 2 samples, if
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