Question

Question 22 1 pts A survey conducted in a small business yielded the result shown in the table. Men Women Total Plan to 41 22 vote Do not plan 34 24 to vote Total Test the independence. Which of the following is not true The degree of freedom is 4 We fail to reject The p-value is 0.4646 The test Statistics is x = 0.5345

Answer 1

The expected values are computed in terms of row and column totals. In fact, the formula is Eij = Rxc, where R; 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 Men women Total 75x63 46x63 Row 1 121 = 39.05 121° = 23.95 63 Row 2 75x58 121 35.95 46x58 121 = 22.05 58 Total 75 46 121 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - 0)2/E. The table with squared distances is shown below: Squared Distances Men women Row 1 (41 39.05) 39.05 = 0.097 = 0.159 (22_23.95) 23.95 (24-22.05) 22.05 Row 2 (34-35.95) 35.95 = 0.106 0.173 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: The two variables are independent He: The two variables are dependent This corresponds to a Chi-Square test of independence. Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = (2-1) (2 - 1) = 1, so then the rejection region for this test is R = {x? : x² > 3.841}. Test Statistics The Chi-Squared statistic is computed as follows: x? (0i; – E;;) Eij = 0.097 + 0.106 + 0.159 + 0.173 = 0.535 Decision about the null hypothesis Since it is observed that x2 = 0.535 < x2 = 3.841, it is then concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level. The corresponding p-value for the test is p = Pr(xỉ > 0.535) = 0.4647.

1st statement is not correct.

Because the degree os freedoms are 1

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