Question

We assume that our wages will increase as we gain experience and become more valuable to our employers. Wages also increase because of inflation. By examining a sample of employees at a given point in time, we can look at part of the picture. How does length of service (LOS) relate to wages? The data here (data303.dat) is the LOS in months and wages for 60 women who work in Indiana banks. Wages are yearly total income divided by the number of weeks worked. We have multiplied wages by a constant for reasons of confidentiality.

(a) Plot wages versus LOS. Consider the relationship and whether or not linear regression might be appropriate. (Do this on paper. Your instructor may ask you to turn in this graph.)

(b) Find the least-squares line. Summarize the significance test for the slope. What do you conclude?

Wages = + LOS

t =

P =

(c) State carefully what the slope tells you about the relationship between wages and length of service.

(d) Give a 95% confidence interval for the slope.

( , )

worker  wages   los     size
1       50.6041 18      Large
2       37.983  71      Small
3       69.2473 101     Small
4       60.7764 63      Small
5       43.1003 73      Large
6       37.7743 57      Small
7       49.7362 97      Large
8       38.0134 51      Large
9       51.4777 101     Large
10      40.4452 51      Small
11      62.5658 55      Large
12      39.8094 84      Small
13      56.6751 17      Small
14      39.3967 27      Large
15      43.31   68      Large
16      46.515  35      Large
17      49.5578 70      Large
18      60.087  121     Small
19      75.6425 127     Large
20      57.1271 79      Large
21      95.585  64      Large
22      60.9494 73      Small
23      49.6756 56      Large
24      53.9179 52      Small
25      64.7448 58      Large
26      70.3789 34      Small
27      48.8014 44      Small
28      44.2582 25      Large
29      47.0324 53      Large
30      39.0001 98      Large
31      50.5997 52      Small
32      52.3797 96      Large
33      59.7787 82      Large
34      37.7668 46      Small
35      54.7275 61      Large
36      67.0155 36      Large
37      73.678  79      Large
38      53.5782 55      Small
39      40.2197 60      Large
40      48.6178 19      Small
41      54.2275 47      Small
42      41.1371 36      Small
43      58.5678 73      Large
44      44.6214 24      Small
45      84.7947 91      Large
46      46.7653 83      Small
47      76.4572 131     Large
48      42.9859 167     Large
49      40.5102 43      Small
50      40.579  25      Large
51      50.8892 53      Large
52      70.2534 63      Large
53      49.0235 98      Large
54      73.2916 48      Small
55      46.6731 33      Small
56      39.3245 52      Large
57      50.4908 250     Small
58      45.0845 55      Large
59      54.8646 25      Small
60      47.41   160     Large

Answer 1

a) The scatter plot of wages versus LOS is

Scatterplot of wages vs los 100 90 80 70 60 40 30 150 50 100 200 250 los O 50 wages

b) Using Minitab software, (Stat -> Regression -> Regression -> Fit Regression Model), we get the following output :

Minitab Untitled File Edit Data Calc Stat Graph Editor Tools Window Help Assistant X Session Regression Analysis: wages versus los Model Summary R-sq (adj) 0.52% R-sq (pred) 0.00% R-sq 2.21% 12.6794 Coefficients Coef SE Coef T-Value P-Value Term VIF 49.84 3.22 15.50 0.000 Constant los 0.0468 0.0409 1.14 0.257 1.00 Regression Equation wages 49.84 0.0468 los Worksheet1 ** C4-T C1 CZ C3 C5 C6 worker los size wages 50.6041 18 Large 1 1 71 Small 37.9830 2 2 101 Small 3 3 69.2473 4 60.7764 63 Small 4 73 Large 5 43.1003 5 57 Small 6 37.7743 97 Large 7 49.7362 7 51 Large 38.0134 51.4777 101 Large 9 9 40.4452 10 51 Small 10 55 Large 11 62.5658 11 Current Worksheet: Worksheet 1 Type here to search n

The least square line is :

$\hat{wages} = 49.84 + 0.0468*los$

Testing the significance for the slope,

The value of the test statistic t = 1.14

and P-value = 0.257

since p-value >0.05 , so at 5% level of significance, we can conclude that slope is not statistically significant.

d) a 95% confidence interval for the slope

$= \hat{\beta }\pm t_{0.025,n-2}*SE(\hat{\beta })$

$= 0.0468 \pm t_{0.025,n-2}*0.0409$

$= 0.0468 \pm (2.00172*0.0409 )$

$=(-0.0350703, 0.128670)$