A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are: Day of the Week Number of Employees Absent Monday 12 Tuesday 9 Wednesday 11 Thursday 10 Friday 9 Saturday 9 What is the critical value of chi-square with α = 0.05? Select one: a. 11.070 b. 12.592 c. 13.388 d. 15.033

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of significance assumed for this case is

The statistic to check the hypothesis is given by:

The table given represent the observed values, we just need to calculate the expected values with the following formula and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

for the free

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