Question

1. -/1 POINTS ASWSBE14 12.E.009. You may need to use the appropriate technology to answer this question. The following table contains observed frequencies for a sample of 200. Column Variable Row Variable P 16 4046 34 34 Test for independence of the row and column variables using a = 0.05. State the null and alternative hypotheses. O Ho: Variable P is not independent of variable Q. H: Variable P is independent of variable Q. O Ho: The column variable is independent of the row variable. H.: The column variable is not independent of the row variable. OH,: The column variable is not independent of the row variable. H: The column variable is independent of the row variable. O Ho: Variable P is independent of variable Q. H: Variable P is not independent of variable Q. Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) D-value = State your conclusion. Do not reject H. We cannot conclude that there is an association between the column variable and the row variable. Reject H. We conclude that the column and row variables are independent. Reject H. We conclude that there is an association between the column variable and the row variable. Do not reject Ho. We cannot conclude that the column and row variables are independent. Need Help? Read It Watch It Talk to a Tutor

Answer 1

H. The column variable is independent of the row variable. H.: The column variable is not independent of the row variable. EXCEL (PH Stat in Add-Ins) procedure: Step 1: Choose PH Stat > Multiple-Sample Tests > Chi-Square Test. Step 2: In Data, enter the values of Level of Significance as 0.05, Number of Rows as 2. and Number of Columns as 3. Step 3: In Output Options. Select the appropriate Title. Step 4: Click OK. EXCEL (PH Stat in Add-Ins) output:

0.05 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom Results Critical Value 5.991465 Chi-Square Test Statistic 9.632424 p-Value 0.008097 Reject the null hypothesis The value of the test statistic is x2 = 9.632. p-value = 0.0081. Reject H. We conclude that there is an association between the column variable and the row variable.