Question 3 OR (e)
Means, correlation and standard error from two regression lines
9 marks
- Time to write
- About 18 minutes
- Topic
- Regression
Question
Question 3 OR (e) · 9 marksA company wishes to analyse the link between Employee Training Hours (X) and Employee Productivity Score (Y). The two regression lines below summarize their findings:
5Y-4X-280=0 and 6Y-5X-330=0
Determine the average Employee Training Hours and the average Employee Productivity Score.
Find the correlation coefficient between the training hours and the productivity score.
With the variance of Employee Productivity Score as 36, calculate the standard error of estimate of productivity based on training hours.
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
- Determine…Find…Two asks in one question: answer both.
- DetermineFormula 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 Means from the two lines 3 marks 5 min
- 2 Identifying the lines and the correlation coefficient 4 marks 7 min
- 3 Standard error of estimate 2 marks 4 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 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.
The two regression lines cut each other at the point of the two means (X̄, Ȳ), so the means are the solution of
5Y − 4X = 280 … (1) and 6Y − 5X = 330 … (2).
Multiplying (1) by 6 and (2) by 5: 30Y − 24X = 1,680 and 30Y − 25X = 1,650. Subtracting, X = 30. Putting X = 30 in (1): 5Y = 280 + 120 = 400, so Y = 80.
Average training hours X̄ = 30; average productivity score Ȳ = 80.
First decide which line is which; r² = byx × bxy can never exceed 1.
- Taking (1) as Y on X: Y = 56 + 0.8X, so byx = 0.8; and (2) as X on Y: X = 1.2Y − 66, so bxy = 1.2. byx × bxy = 0.8 × 1.2 = 0.96, which is admissible.
- The other way round: (2) as Y on X gives byx = 5/6 and (1) as X on Y gives bxy = 5/4, a product of 1.0417 > 1, which is impossible.
So (1) is the regression line of Y on X and (2) that of X on Y. The correlation coefficient is the geometric mean of the regression coefficients, with their common (positive) sign:
r = +√(0.8 × 1.2) = √0.96 = 0.9798.
Training hours and productivity are very highly, positively correlated.
(iii) Standard error of estimate
The variance of the productivity score is 36, so σy = 6. The standard error of estimate of productivity (Y) from training hours (X) is
Syx = σy √(1 − r²) = 6 × √(1 − 0.96) = 6 × 0.2 = 1.2 (score points).
Estimates of productivity from the line of Y on X are therefore expected to be within about 1.2 points of the actual score on average: a very close fit.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Means from the two lines | 3 |
| 2 | Identifying the lines and the correlation coefficient | 4 |
| 3 | Standard error of estimate | 2 |
| Total | 9 |
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