Cramming is a technique often used by students to prepare for exams,in which they study for extended hours only a few days before the exams.Often students drink caffeinated beverages to help them stay awake while cramming for exams.Five students were randomly selected and asked to drink highly caffeinated beverages before their first midterm exam and then drink only water before their second midterm exam in the same subject.The difference of the scores (midterm 2 - midterm 1) is shown below for the five students. + ,the Wilcoxon signed rank test statistic? (Note: Use the positive difference. ) A) 3 B) 12 C) Cannot tell from the given information D) None of the above class=answers-bank-image d-block rel=preload What is the value of,W + ,the Wilcoxon signed rank test statistic? (Note: Use the positive difference. )A) 3B) 12C) Cannot tell from the given informationD) None of the above

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Cramming is a technique often used by students to prepare for exams,in which they study for extended hours only a few days before the exams.Often students drink caffeinated beverages to help them stay awake while cramming for exams.Five students were randomly selected and asked to drink highly caffeinated beverages before their first midterm exam and then drink only water before their second midterm exam in the same subject.The difference of the scores (midterm 2 - midterm 1) is shown below for the five students.  + ,the Wilcoxon signed rank test statistic? (Note: Use the positive difference. )  A) 3 B) 12 C) Cannot tell from the given information D) None of the above class=answers-bank-image d-block rel=preload > What is the value of,W + ,the Wilcoxon signed rank test statistic? (Note: Use the positive difference. )A) 3B) 12C) Cannot tell from the given informationD) None of the above

Ockacia Sterling · Accepted Answer

Final Answer : B, Explanation :  To calculate the Wilcoxon signed rank test statistic, we first need to rank the differences in absolute value from smallest to largest:Student 1: |7|Student 2: |5|Student 3: |3|Student 4: |2|Student 5: |1|Next, we assign a signed rank to each difference, based on the sign of the difference. If the difference is positive, we give it a positive rank; if the difference is negative, we give it a negative rank. If there are ties (i.e., multiple differences with the same absolute value), we give them the average of the ranks that would have been assigned to them if they were distinct.Student 1: -3Student 2: -2Student 3: -1.5Student 4: 1.5Student 5: 3The sum of the signed ranks for the positive differences is W  = 1.5 + 3 = 4.5. The sum of the absolute values of the negative ranks is   = 1 + 2 + 3 = 6.Therefore, W  +   = 4.5 + 6 = 10.5.Since the positive differences were ranked lower than the negative differences, we cannot use the usual distribution to calculate a p-value for the test statistic. Instead, we would need to use a table of critical values or a computer program to determine if the observed test statistic is significant.

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