A gardener works for 13 hours during a week and charges $169 how much does the gardener charge for each hour?
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Answer:
1) H0: There is significant evidence of independence between marital status and diagnostic
H1: There is significant evidence of an association between marital status and diagnostic
2) The statistic to check the hypothesis is given by:
3)
4)
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.
Step-by-step explanation:
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".
Assume the following dataset:
Alcoholic Undiagnosed Alcoholic Not Alcoholic Total
Married 21 37 58 116
Not Married 59 63 42 164
Total 80 100 100 280
Part 1
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is significant evidence of independence between marital status and diagnostic
H1: There is significant evidence of an association between marital status and diagnostic
The level os significance assumed for this case is
Part 2
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 calculations are given by:
And the expected values are given by:
Alcoholic Undiagnosed Alcoholic Not Alcoholic Total
Married 33.143 41.429 41.429 116
Not Married 46.857 58.571 58.571 164
Total 80 100 100 280
Part 3
And now we can calculate the statistic:
Part 4
Now we can calculate the degrees of freedom for the statistic given by:
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.