A recent report from Pythagoras College determined that although only 4 percent of students wore baseball hats to school regularly, 60 percent of students caught cheating were wearing baseball hats at the time. Pythagoras College concluded that students who wear baseball hats are more likely to cheat than students who do not.
Which of the following does most to strengthen the conclusion made by Pythagoras College?
Choices
A. No cheating incidents have been reported in classrooms where baseball hats are prohibited.
B. Baseball hats make students easier to identify.
C. The number of students caught cheating exceeds the number of students wearing baseball hats on any given day.
D. Many of the students caught cheating have been caught more than once.
E. Students at Pythagoras College cheat more often than students at other colleges do.
OA is A.
I agree with the OA. However, I don't understand why C is incorrect.
Let's suposse that the 100% of the students with hats (4% of all students) were caught cheating. In this sense, and making some calculations, 6 2/3% of all students were caught cheating. Therefore, 2 2/3% or 2.6666% of all students were caucght cheating and not were wearing a hat.
Now, let's compare:
- 100% of students with hats were caught cheating.
- [(2.666 /96)*100]% of the students without hats were caught cheating. (96%: students without hats).
It's clear that students with hats are more likely to cheat. This option would strengthen the conclusion.
Please, explain.
Which of the following does most to strengthen the conclusion made by Pythagoras College?
Choices
A. No cheating incidents have been reported in classrooms where baseball hats are prohibited.
B. Baseball hats make students easier to identify.
C. The number of students caught cheating exceeds the number of students wearing baseball hats on any given day.
D. Many of the students caught cheating have been caught more than once.
E. Students at Pythagoras College cheat more often than students at other colleges do.
OA is A.
I agree with the OA. However, I don't understand why C is incorrect.
Let's suposse that the 100% of the students with hats (4% of all students) were caught cheating. In this sense, and making some calculations, 6 2/3% of all students were caught cheating. Therefore, 2 2/3% or 2.6666% of all students were caucght cheating and not were wearing a hat.
Now, let's compare:
- 100% of students with hats were caught cheating.
- [(2.666 /96)*100]% of the students without hats were caught cheating. (96%: students without hats).
It's clear that students with hats are more likely to cheat. This option would strengthen the conclusion.
Please, explain.
