OCR MEI S3 (Statistics 3) 2009 January

Question 1
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1
  1. A continuous random variable \(X\) has probability density function $$\mathrm { f } ( x ) = \lambda x ^ { c } , \quad 0 \leqslant x \leqslant 1 ,$$ where \(c\) is a constant and the parameter \(\lambda\) is greater than 1 .
    1. Find \(c\) in terms of \(\lambda\).
    2. Find \(\mathrm { E } ( X )\) in terms of \(\lambda\).
    3. Show that \(\operatorname { Var } ( X ) = \frac { \lambda } { ( \lambda + 2 ) ( \lambda + 1 ) ^ { 2 } }\).
  2. Every day, Godfrey does a puzzle from the newspaper and records the time taken in minutes. Last year, his median time was 32 minutes. His times for a random sample of 12 puzzles this year are as follows. $$\begin{array} { l l l l l l l l l l l l } 40 & 20 & 18 & 11 & 47 & 36 & 38 & 35 & 22 & 14 & 12 & 21 \end{array}$$ Use an appropriate test, with a 5\% significance level, to examine whether Godfrey's times this year have decreased on the whole.
Question 2
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2 A factory manufactures paperweights consisting of glass mounted on a wooden base. The volume of glass, in \(\mathrm { cm } ^ { 3 }\), in a paperweight has a Normal distribution with mean 56.5 and standard deviation 2.9. The volume of wood, in \(\mathrm { cm } ^ { 3 }\), also has a Normal distribution with mean 38.4 and standard deviation 1.1. These volumes are independent of each other. For the purpose of quality control, paperweights for testing are chosen at random from the factory's output.
  1. Find the probability that the volume of glass in a randomly chosen paperweight is less than \(60 \mathrm {~cm} ^ { 3 }\).
  2. Find the probability that the total volume of a randomly chosen paperweight is more than \(100 \mathrm {~cm} ^ { 3 }\). The glass has a mass of 3.1 grams per \(\mathrm { cm } ^ { 3 }\) and the wood has a mass of 0.8 grams per \(\mathrm { cm } ^ { 3 }\).
  3. Find the probability that the total mass of a randomly chosen paperweight is between 200 and 220 grams.
  4. The factory manager introduces some modifications intended to reduce the mean mass of the paperweights to 200 grams or less. The variance is also affected but not the Normality. Subsequently, for a random sample of 10 paperweights, the sample mean mass is 205.6 grams and the sample standard deviation is 8.51 grams. Is there evidence, at the \(5 \%\) level of significance, that the intended reduction of the mean mass has not been achieved?
Question 3
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3 Pathology departments in hospitals routinely analyse blood specimens. Ideally the analysis should be done while the specimens are fresh to avoid any deterioration, but this is not always possible. A researcher decides to study the effect of freezing specimens for later analysis by measuring the concentrations of a particular hormone before and after freezing. He collects and divides a sample of 15 specimens. One half of each specimen is analysed immediately, the other half is frozen and analysed a month later. The concentrations of the particular hormone (in suitable units) are as follows.
Immediately15.2113.3615.9721.0712.8210.8011.5012.05
After freezing15.9610.6513.3815.0012.1112.6512.488.49
Immediately10.9018.4813.4313.1616.6214.9117.08
After freezing9.1315.5311.848.9916.2414.0316.13
A \(t\) test is to be used in order to see if, on average, there is a reduction in hormone concentration as a result of being frozen.
  1. Explain why a paired test is appropriate in this situation.
  2. State the hypotheses that should be used, together with any necessary assumptions.
  3. Carry out the test using a \(1 \%\) significance level.
  4. A \(p \%\) confidence interval for the true mean reduction in hormone concentration is found to be ( \(0.4869,2.8131\) ). Determine the value of \(p\).
Question 4
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4
  1. Explain the meaning of 'opportunity sampling'. Give one reason why it might be used and state one disadvantage of using it. A market researcher is conducting an 'on-street' survey in a busy city centre, for which he needs to stop and interview 100 people. For each interview the researcher counts the number of people he has to ask until one agrees to be interviewed. The data collected are as follows.
    No. of people asked1234567 or more
    Frequency261917131186
    A model for these data is proposed as follows, where \(p\) (assumed constant throughout) is the probability that a person asked agrees to be interviewed, and \(q = 1 - p\).
    No. of people asked1234567 or more
    Probability\(p\)\(p q\)\(p q ^ { 2 }\)\(p q ^ { 3 }\)\(p q ^ { 4 }\)\(p q ^ { 5 }\)\(q ^ { 6 }\)
  2. Verify that these probabilities add to 1 whatever the value of \(p\).
  3. Initially it is thought that on average 1 in 4 people asked agree to be interviewed. Test at the \(10 \%\) level of significance whether it is reasonable to suppose that the model applies with \(p = 0.25\).
  4. Later an estimate of \(p\) obtained from the data is used in the analysis. The value of the test statistic (with no combining of cells) is found to be 9.124 . What is the outcome of this new test? Comment on your answer in relation to the outcome of the test in part (iii).