Question

l. and found that participants wearing red outfits won significantly more often than those wearing blue a. In 50 wrestling matches involving red versus blue, suppose that the red outfit won 31 times and lost 19 times. Is this result sufficient to conclude that red wins significantly more than would be expected by chance? Test at the .05 level of significance. b. In 100 matches, suppose red won 62 times and lost 38. Is this sufficient to conclude that red wins sig- nificantly more than would be expected by chance? Again, use a-05 c.Note that the winning percentage for red uniforms in part a is identical to the percentage in part b (31 out of 50 is 62%, and 62 out of 100 is also 62%). Although the two samples have an identical winning percentages, one is significant and the other is not. Explain why the two samples lead to 3.in problem 12, a researcher asked college students to evaluate three new cell phone designs. However, the researcher suspects that college students may have criteria that are different from those used by older adults. To test this hypothesis, the researcher repeats the study using a sample of n- 60 older adults in addition to a sample of n -60 students. The distribution of preference is as follows: Design 1 Design 2 Design3 Stadent Older Adult 21 Do the data indicate that the distribution of preferences for older adults is signif

Answer 1

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The following cross tablulation have been provided. The row and column total have been calculated and they are shown below Column1 27 21 48 Column 2 20 34 54 Column 3 Row 1 Row 2 Total Total 60 60 120 13 18 The expected values are computed in terms of row and column totals. In fact, the formula is Ei C, where Ri corresponds to the total sum of elements in row i, C, corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values Expected Values Row 1 Row 2 Total Column1 48x604 48, 60-24 48 Column 2 54x6027 54x6027 54 Column 3 Total 60 60 120 18x60 120 120 120 18x60 120 120 120 18 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)2/E. The table with squared distances is shown below Squared Distances Column1 (27-240.375 (21-242 0.375 Column 2 (20-27 1.815 34 27 1.815 Column 3 13-92 1.778 (5921.778 Row 1 24 9 Row 2 24 9

The following null and alternative hypotheses need to be tested: HO: The two variables are independent Ha: The two variables are dependent This corresponds to a Chi-Square test of independence. Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df (2 -1) x (3-1) 2, so then the rejection region for this test is R x2:X2 > 5.991 The Chi-Squared statistic is computed as follows: 0.375 +0.375 +1.815+1.8151.778+1.778-7.935 Since it is observed thatx2 -7.9352-5.991, it is then concluded that the null hypothesis is It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level.