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Inference for Categorical Data 297 in Visual C#.NET
Inference for Categorical Data 297 PDF 417 Reader In C#.NET Using Barcode recognizer for Visual Studio .NET Control to read, scan PDF 417 image in .NET applications. www.OnBarcode.comPDF417 2d Barcode Decoder In Visual C# Using Barcode reader for .NET framework Control to read, scan read, scan image in VS .NET applications. www.OnBarcode.comThe correct answer is (b) The expected values for the cells are exactly equal to the 25 observed values (eg, for the 1st row, 1st column, Exp = (8) = 5 , so X 2 must 40 equal 0 the variables are independent, and are not related Bar Code Recognizer In C#.NET Using Barcode scanner for .NET Control to read, scan barcode image in .NET applications. www.OnBarcode.comReading Bar Code In C#.NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in .NET applications. www.OnBarcode.comThe correct answer is (e) The expected values for this twoway table are given by the 125 275 matrix: 125 275 Then, X2= (15 12 5)2 ( 25 27 5)2 (10 12 5)2 + + 12 5 27 5 12 5 2 (30 27 5) + = 1 45 27 5 Recognizing PDF 417 In Visual C# Using Barcode decoder for .NET framework Control to read, scan PDF 417 image in .NET framework applications. www.OnBarcode.comPDF417 2d Barcode Reader In .NET Framework Using Barcode recognizer for ASP.NET Control to read, scan PDF 417 image in ASP.NET applications. www.OnBarcode.comThe correct answer is (c) In a 2 test for independence, we are interested in whether or not two categorical variables, measured on a single population, are related In a 2 test for homogeneity of proportions, we are interested in whether two or more populations have the same proportions for each of the values of a single categorical variable The correct answer is (e) Using Ethnicity as the row variable, there are five rows (r = 5) and four columns (c = 4) The number of degrees of freedom for an r c table is (r 1)(c 1) In this question, (5 1)(4 1) = 4 3 = 12 The correct answer is (c) In III, the expected count for each category in a goodnessoffit test is found by multiplying the proportion of the distribution of each category by the sample size The expected count for a test of independence is found by multiplying the row total by the column total and then dividing by n PDF 417 Recognizer In VS .NET Using Barcode recognizer for .NET framework Control to read, scan PDF417 image in VS .NET applications. www.OnBarcode.comPDF417 Scanner In Visual Basic .NET Using Barcode reader for .NET Control to read, scan PDF 417 image in Visual Studio .NET applications. www.OnBarcode.comFree Response
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We note that all expected values are at least 5, so the conditions necessary for the chisquare test are present III X 2 = (O E )2 (110 120)2 = + E 120 + ( 4 5)2 = 679 , df = 4 010 < Pvalue 5 < 015 (from Table C) Using the TI83/84, 2 cdf(679,1000,40)= 0147 IV The Pvalue is greater than any commonly accepted significance level Hence, we do not reject H0 and conclude that we do not have good evidence that the calculator is not correctly generating values from B(4, 03) 2 For a 3 4 twoway table, df = (3 1)(4 1) = 6 0005 < Pvalue < 001 (from Table C) The finding is significant at the 001 level of significance Using the TI83/84, Pvalue = 2 cdf(172,1000,6) = 0009 (a) Is not correct For a given set of observations, they both do produce the same value of chisquare However, they differ in that they are different ways to design a study (b) is correct A test of independence hypothesizes that two categorical variables are independent within a given population A test for homogeneity of proportions hypothesizes that the proportions of the values of a single categorical variable are the same for more than one population (c) Is incorrect It is a reversal of the actual difference between the two designs (d) Is incorrect You always use count data when computing chisquare The expected value of the cell with the frog is 128 (96) = 3977 309 4 5

