Analysis of the rate of turnover of employees by a personnel manager produced the following table showing the length of stay of 200 people who left the company for other employment. Use a 1% significance level to test whether length of employment is independent of grade.
Length of employment (years) |
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Grade |
0-2 |
2-5 |
>5 |
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Managerial |
4 |
11 |
6 |
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Skilled |
32 |
28 |
21 |
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Unskilled |
25 |
23 |
50 Analysis of the rate of turnover of employees by a personnel manager produced the following table showing the length of stay of 200 people who left the company for other employment. Use a 1% significance level to test whether length of employment is independent of grade.
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Null hypothesis: Ho : length of employment is independent of grade
Alternate hypothesis:Ha: : length of employment is dependent of grade
degree of freedom(df) =(rows-1)*(columns-1)= | 4 | ||
for 4 df and 0.01 level,critical value χ2= | 13.277 | ||
Decision rule : reject Ho if value of test statistic X2>13.277 |
Applying chi square test of independence: |
Expected | Ei=row total*column total/grand total | 0-2 | 2-5 | >5 | Total |
Managerial | 6.405 | 6.510 | 8.085 | 21 | |
Skilled | 24.705 | 25.110 | 31.185 | 81 | |
Unskilled | 29.890 | 30.380 | 37.730 | 98 | |
total | 61 | 62 | 77 | 200 | |
chi square χ2 | =(Oi-Ei)2/Ei | 0-2 | 2-5 | >5 | Total |
Managerial | 0.903 | 3.097 | 0.538 | 4.5375 | |
Skilled | 2.154 | 0.333 | 3.326 | 5.8131 | |
Unskilled | 0.800 | 1.793 | 3.990 | 6.5830 | |
total | 3.8572 | 5.2222 | 7.8544 | 16.9337 | |
test statistic X2= | 16.934 | ||||
p value = | 0.0020 | from excel: chidist(16.9337,4) |
since test statistic falls in rejection region we reject null hypothesis |
we have sufficient evidence to conclude that length of employment is dependent of grade |
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