The human resource manager at a car dealership wants to know if the ages of its employees are related to the department that they work in. Data was compiled and tabulated in a 2-way contingency table. The employees were classified according to their age and department. Expected counts are printed below observed counts Sales Accounts Marketing Repairs Total 20-29 8 10 27 43 88 17.74 *** 26.20 23.62 30-39 29 26 38 22 * 23.18 26.72 34.24 30.86 40-49 33 32 72 82 219 44.15 50.88 65.20 58.77 50-59 81 106 86 54 327 65.92 75.97 97.36 87.75 Total 151 ** 223 201 749 Chi-Sq = 5.348 + **** + 0.024 + 15.912 + 1.459 + 0.019 + 0.413 + 2.544 + 2.816 + 7.003 + 0.709 + 9.182 + 3.448 + 11.875 + 1.325 + 12.983 = 80.395 DF = *****, P-Value = ****** No cells with expected counts less than 5.
c) Fill in the gaps marked by ‘*’, ‘**’, ‘***’, ‘****’ , ‘*****’ and ‘******’ [8] d) What is the assumption on which these calculations are based? [1]
"\\star" is the total for row 2 and it is given as, "\\star=29+26+38+22=115."
"\\star\\star" is the total for column 2 and it is given as, "\\star\\star=10+26+32+106=174"
"\\star\\star\\star" is the expected count for row 1, column 2. To find its value, we use the formula
"E_{12}=(r_1*c_2)\/n", where "r_1=88" is the total for row 1, "c_2=174" is the total for column 2 and "n=749" is sample size. Therefore,
"E_{12}=(88*174)\/749=20.44(2\\space dp)".
Thus, "\\star\\star\\star=20.44."
"\\star\\star\\star\\star" accounts for the test statistic "(Chi-Sq)" above and it is found using the "(O_{ij}-E_{ij})^2\/E_{ij}". For this case, this value is in row "i=1" and column "j=2" . Therefore ,
"\\star\\star\\star\\star=(O_{12}-E_{12})^2\/E_{12}=(10-20.44)^2\/20.44=109.061638\/20.44=5.335(3dp)"
For this test, the degrees of freedom "DF=(r-1)(c-1)" where "r=4" is the number of rows and "c=4" is the number of columns. Therefore "DF=\\star\\star\\star\\star\\star=(4-1)(4-1)=3*3=9"
To determine the p-value for this test, we find "p(\\chi^2\\gt 80.395)". To find this probability, we consider the degrees of freedom "(DF)" and the test statistic of 80.395 and use the following the command in "R".
> pv=pchisq(80.395,9,lower.tail=FALSE) and the output is,
> pv
[1] 1.348953e-13
The output "pv=1.348953e-13=\\star\\star\\star\\star\\star\\star" is the desired P-Value.
In summary,
"\\star=115, \\space \\star\\star=174,\\star\\star\\star=20.44,\\space \\star\\star\\star\\star=5.335, \\space \\star\\star\\star\\star\\star=9,\\space \\star\\star\\star\\star\\star\\star=1.348953e-13"
The assumption on which these calculations are based is that the expected frequency count for each cell of the table is at least 5.
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