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CT3 April 2012 Exam

T

Turtlelord

Member
Guess I'll get the ball rolling.

What did everyone think of this exam?

I thought it was a pretty hard exam (deceptively hard in fact). At first glance their was nothing that looked horrendous, but I still feel I have lost quite a few marks.
 
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I felt it was a pecuilar paper, although I did appreciate that it made you think about the techniques used rather than repeating methods blindly.

The first question threw me off to begin with (the laptop battery question) and I don't think I got full marks in that question. I used the Normal approximation, but realised after that I foolishly divided by the sample size twice, which clearly would have given a wrong answer.

I had only used the Fisher Transformation to test the correlation coefficient once while doing past paper questions, so I am happy that I spent what I thought would be wasted time knowing when to use it. My answer seemed reasonable, which is always a good thing!

I think I managed to just eek out the marks for the coin toss question in the last 15 minutes.

The one question which really got me was the testing of the binomial parameter, p (one of the last three questions). I used the method of moments to estimate p to be 0.5, and used the chi-squared test to test the distribution. However when I read the next part, which asked us to state the effect of instead being told p=0.5, as opposed to estimating it, I didn't consider that it would mean we could include one more degree of freedom, and we could then easily accept the null Hypothesis again based on the first. And then the next part told us to test p=0.5, which made me think that I'd used the method of moments incorrectly, so I changed the labelling of my initial chi-squared test to part (iv) instead of part (ii) *shakes my head*.
 
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It was a pretty hard paper compared to past years'.

I think the first question was about finding the first quartile, that was all there was to that question.

I messed up the coin tossing question part ii and the question about conditional expectation, and of course the algebra in the CRLB question was beyond me, I was unable to show what was required, wasted quite alotta time on it.

The rest of the paper was quite straight forward though, hope to scrape a pass.
 
For the 1st question what I did was find the mean and then use the following formula:

0.75x + (1-x)*0.25= mean

So the answer was 3.7. What did you guys get
 
Q1

I think 3.7 might not be the correct answer.

If you consider the sample is the true representative of the population, you are trying to estimate x in the following equation:

  • Prob[X > x] = 1 - 0.75 = 0.25

An estimate of x would be the 75th percentile point of the data which is the 0.75 * 24 + 0.5 = 18.5th observation.

Note in the data, the 18th and 19th ranked observation is 2.4.
So, I will go for 2.4 as the answer.
 
shouldnt the abv equation be P[X>x]=0.75 because the question required the time for which the charging will be sufficient 75% of the time, so it would be the 1st quartile rather than the third.
 
@ tirmizi

We are asked to find the longest showing time and 1st quartile would give you a value below the mean so that would be wrong.
 
Wow I really overthought that question! Good job it was only 3 marks!

I think P(X>x) = 0.75 is correct i.e. First quartile because 75% of battery lives are longer than this value, so 0.75 probability the battery will last longer than x
 
Is it incorrect if we use the normal approximation
Mean = 2.23 S.D = .8745
(X-2.23)/.8745 =1 - 0.6745 or X = 2.82
 
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bapan said:
I think 3.7 might not be the correct answer.

If you consider the sample is the true representative of the population, you are trying to estimate x in the following equation:

  • Prob[X > x] = 1 - 0.75 = 0.25

An estimate of x would be the 75th percentile point of the data which is the 0.75 * 24 + 0.5 = 18.5th observation.

Note in the data, the 18th and 19th ranked observation is 2.4.
So, I will go for 2.4 as the answer.
Click to expand...

I did the same.
 
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