Question

_____________________________________________________________________________ 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]

Answer 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.