Question 3 (b)
Number of pairs from the probable error, and the limits of the correlation coefficient
4 marks
- Time to write
- About 8 minutes
- Topic
- Correlation
Question
Question 3 (b) · 4 marksIn a factory, the correlation coefficient worked out for a sample containing a certain number of observations on two items produced is 0.85 while the probable error is 0.047. Find out the number of pairs of observations (N) taken as the sample and also find out the limit of the correlation coefficient for another sample of the same size from the same population.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- Find…Two asks in one question: answer both.
- Find outFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Formula of the probable error 1 mark 2 min
- 2 Number of pairs N 2 marks 3 min
- 3 Limits of r for another sample 1 mark 2 min
Short on time? Step 2 carries 2 of the 4 marks: guard its minutes.
Try it first. About 8 minutes for 4 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Probable error. The syllabus places the probable and standard errors of the correlation coefficient in Unit 3. The probable error of r, for N pairs of observations, is
P.E.(r) = 0.6745 × (1 − r²) / √N,
and the correlation coefficient of another sample of the same size from the same population is expected to lie within r ± P.E.(r).
Number of pairs (N). With r = 0.85 and P.E. = 0.047:
- 1 − r² = 1 − 0.7225 = 0.2775.
- 0.047 = 0.6745 × 0.2775 / √N, so √N = 0.6745 × 0.2775 / 0.047 = 0.18717 / 0.047 = 3.9824.
- N = (3.9824)² = 15.86, so N = 16 pairs (a sample size must be a whole number; with N = 16, P.E. = 0.18717 / 4 = 0.0468, which rounds to the given 0.047).
Limits of the correlation coefficient.
r ± P.E. = 0.85 ± 0.047, so for another sample of 16 pairs from the same population the correlation coefficient is expected to lie between 0.803 and 0.897.
Comment. r = 0.85 is more than six times its probable error (6 × 0.047 = 0.282), so the correlation between the two items is significant: it is not a chance result of the sample.
Where the 4 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Formula of the probable error | 1 |
| 2 | Number of pairs N | 2 |
| 3 | Limits of r for another sample | 1 |
| Total | 4 |
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