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Sektor404 Sektor404
wrote...
Posts: 125
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11 years ago
One year, at England University, there were these FOUR students who were taking Chemistry and who did pretty well on all of the quizzes and labs. These students were so confident that they decided to party the weekend before their final exams. They partied and drank alcohol, which resulted in them sleeping all day Sunday and waking up late on the Monday morning of their final examination. Rather than attend the exam, they arrived after the exam and explained to their professor why they were late and had missed the exam. They told him they went away and on the way back home they had a flat tyre with no spare tyre to fix it. The professor agreed and let them sit their examination the very next day, but they were shocked upon finding their very first question in the exam. It said: Which tyre?

1) There are 44 different ways in which the four students could have answered this question.
 What is the probability that the four students passed the chemistry examination?

2) A statistician and a mathematician were arguing over the validity of the
above probability calculation. The statistician felt the calculation was
valid because she felt each tyre was equally likely to be picked. The
mathematician felt students were twice as likely to pick a front tyre over
a rear tyre going flat, because many would think that the front tyres are
more likely to run over something first that could damage a tyre. The
statistician decided to settle the argument by surveying 40 students from
her class. The following results were obtained (REFER TO ATTACHMENT)

Perform a chi-squared goodness of fit test, using a 0.05 significance level,
to see if the data supports the statistician’s hypothesis.

3) The mathematician claimed the data clearly confirmed his argument as
almost twice an many students picked the front tyre. Conduct another
chi-squared goodness of fit test with a 0.05 significance level to see if the
data supports the mathematician’s point of view.

4) Given your results from above, what conclusion can be made?

Please show all working, thank you Slight Smile
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wrote...
Valued Member
On Hiatus
11 years ago
1) As the exercise says, there are 44 different combinations of answers that the four students can give. To pass the test all four students must choose the same tire (any of them). Since the car has 4 tires, there are 4 correct combinations: one for each tire. This means that the probability of answering the question correctly is 4/44 = 1/43 (If we assume that each student will randomly choose one tire)

2) 40 students took part to this test.
At first, we assume that every tire has the same probability to be selected by students. According to that, we would expect that each tire was selected by 10 students (null hypothesis).

We apply the x2 test: x2=(11-10)2/10  + (15-10)2/10 + (8-10)2/10 + (6-10)2/10 = 1/10 + 25/10 + 4/10 + 16/10 = 4,6
Degrees of freedom are: k-1=4-1=3.

For p=0,05 and dregrees of freedom=3, we have x2=7,815 (You have the diagrams that shows the x2 values, right?)
Because 4,6<7,815 , we can't reject the Null Hypothesis. (However, we also can't accept it.)

3)
The mathematician believes that it is twice as likely for a frond tire to be selected. Since he didn't care if the students choose the right or the left tire, in that case wee just assume that about 27 students selected frond tires and about 13 students selected back tires. (null hypothesis) (you could use the numbers 26,66 and 13,33 for higher accuracy of the test, but for simplicity I rounded these numbers)
The observed numbers are 11+15=26 and 8+6=14 respectively.
So, we do the x2 test:
x2= (27-26)2/27 + (13-14)2/14 =0,1
In this case, the degrees of freedom are 2-1=1.
0,1<3,841 so, the null hypothesis can't be rejected.

4) Both the hypothesis of the statistician and the mathematician can't be rejected, so with our data, we can't tell if the mathematician of the statistician is right.

Please also check the exercise yourself because I am not sure if I did that correctly (it's been a while since I studied statistics)
bio_man,  duddy,  megamind442
Sektor404 Author
wrote...
11 years ago
Thanks Alexx!
Sektor404 Author
wrote...
11 years ago
Alexx,

For 1) A friend of mine had 0.017 probability of passing the exam (4/256)...is this correct?

Another note, the final values given to me by my friend were 4.6 < 7.815 and 0.94 < 7.815.  I know you had the 4.6 one but what about the other 0.94 result? Could she have calculated it wrong? Thanks!
wrote...
Valued Member
On Hiatus
11 years ago
1) 1/43=4/256
But my calculator says that 4/256=0,015625

2) Your friend probably though that since we assume that it's twice as likely to choose the frond tires, we will also assume that the students choose the tires as following:
frond left:13,33 front right13,33 rear left:6,66 rear right:6,66
However, the mathematician did NOT mention if the students would prefer the left or right tires.
So, it would be unfair for him to assume that the students who selected front tires, evenly selected the left and right tires (and respectively for rear tires too).

That's why, as a null hypothesis I just assumed that 26,66 students (rounded: 27) selected a front tire, and 13,33 (rounded 13) selected a rear tire.

I think that many people will think like your friend did, and I think both answers would be considered correct.

(However, the two answers are different. If you check the two x2 values, they are not even close to each other.)
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