14.3p – 32.24 = 127.92 14.3p – 32.24 + 32.24 = 127.92 + 32.24 14.3p = 160.16 StartFraction 14.3 p Over 14.3 EndFraction equals S
1 answer:
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Answer:
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