Click to select options are both yes/no A Cleveland-based company is interested

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Click to select options are both yes/no

A Cleveland-based company is interested in possible differences in days worked by salaried employees in four departments in the financial area. A survey of salaried employees was randomly chosen across the four departments. Descriptive statistics, by department, are provided below. Note that because of the casual sampling methodology in this survey, the sample sizes are unequal A one-way ANOVA was conducted to determine whether the mean annual attendance rates are the same for employees in these four departments. The Excel partial ANOVA output is provided below Mean 261.2 238.0 244.4 258.6 Std. Dev Department 4.95 Budgets 2.24 Payables 2.70 Pricing 4 6 10.11 Collections ANOVA table Source df MS Treatment (Between Groups) Error (Within groups Total ? 905.000 5,656.00

Explanation / Answer

Follow the steps to complete the ANOVA table.

1. Given, MSTr=905.00, df(Between)=k-1=4-1=3. Therefore, SSTr=MSTr*df(Between)=905*3=2715

2. Given, SSE=5656, df(Within)=N-k=(5+9+4+6)-4=20. Therefore, MSE=SSE/df(Within)=5656/20=282.8

3. F=MSTr/MSE=905/282.8=3.20

4. SS(Total)=SSTr+SSE=2715+5656=8371

5. df(Total)=df(between)+df(within)=3+20=23

Using technology, the critical F value at alpha=0.01 and for (3,20) degress of freedom is: 4.94.

Using sample data, the calculated value of test statistic, Fobserved is:3.20.

The null hypothesis should not be rejected. [Per rejection rule based on critical value, reject null hypothesis if observed F >critical F, here, observed test statistic does not fall in critical region, thus, fail to reject null hypothesis].

The compnay cannot conclude. Ans>No.

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