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Question
The HRD manager of a company wants to find a measure which he can use to fix the monthly income of persons applying for the job in the production department. As an experimental project, he collected data of 7 persons from that department referring to years of service and their monthly incomes.
| Years of service (X) | 11 | 7 | 9 | 5 | 8 | 6 | 10 |
| Monthly Income (₹ 1000's)(Y) | 10 | 8 | 9 | 5 | 9 | 7 | 11 |
- Find the regression equation of income on years of service.
- What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 years?
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Solution
(i) Here, X = Years of service,
Y = Income (₹ in 1000’s)
| X = xi | Y = yi | `"x"_"i"^2` | xi yi |
| 11 | 10 | 121 | 110 |
| 7 | 8 | 49 | 56 |
| 9 | 6 | 81 | 54 |
| 5 | 5 | 25 | 25 |
| 8 | 9 | 64 | 72 |
| 6 | 7 | 36 | 42 |
| 10 | 11 | 100 | 110 |
| 56 | 56 | 476 | 469 |
From the table, we have
n = 7, ∑ xi = 56, ∑ yi = 56, `sum"x"_"i"^2` = 476,
∑ xi yi = 469
`bar"x" = sum "x"_"i"/"n" = 56/7 = 8`
`bar"y" = sum "y"_"i"/"n" = 56/7 = 8`
Now, `"b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar"x"^2)`
∴ `"b"_"YX" = (469 - 7 xx 8 xx 8)/(476 - 7 xx (8)^2) = (469 -448)/(476 - 448) = 21/28 = 3/4`
∴ `"b"_"YX"` = 0.75
Also, a = `bar"y" - "b"_"YX" bar "x" = 8 - 0.75 xx 8 = 8 - 6 = 2`
∴ The regression equation of income (Y) on years of service (X) is
Y = a + bYX X
∴ Y = 2 + 0.75 X
(ii) Given, years of service (X) = 13
∴ Substituting X = 13 in regression equation, we get
Y = 2 + 0.75 (13)
∴ Y = 2 + 9.75
∴ Y = 11.75 (₹ in 1000’s)
∴ I would recommend an initial start of ₹ 11,750 for a person who has served in similar capacity in another company for 13 years.
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