Question

A researcher wishes to see if the five ways (drinking decaffeinated beverages, taking a nap, going for a walk, eating a sugary snack, other) people use to combat midday a 0.05 drowsiness are equally distributed among office workers. A sample of 60 office workers is selected, and the following data are obtained. At a 0.10 can it be concluded that there is no preference? Why would the results be of interest to an employer?

Answer 1

`\chi^2 = ((21-12)^2)/(12)+((16-12)^2)/(12)+((10-12)^2)/(12)+((8-12)^2)/(12)+((5-12)^2)/(12)=13.833`

`p_v = P(\chi^2_(4) >13.833)=0.00785`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .

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".

Assume the following dataset:

Method Beverage Nap Walk Snack Other

Number 21 16 10 8 5

The total on this case is 60

We need to conduct a chi square test in order to check the following hypothesis:

H0: Drowsiness are equally distributed among office workers

H1: Drowsiness IS NOT equally distributed among office workers

The level of significance assumed for this case is
`\alpha=0.1`

The statistic to check the hypothesis is given by:

`\chi^2 = \sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

The table given represent the observed values, we just need to calculate the expected values with the following formula
`E_i = (total)/(5)`

And the calculations are given by:

`E_(Beverage) =(60)/(5)=12`

`E_(Nap) =(60)/(5)=12`

`E_(Walk) =(60)/(5)=12`

`E_(Snack) =(60)/(5)=12`

`E_(Other) =(60)/(5)=12`

And the expected values are given by:

Method Beverage Nap Walk Snack Other

Number 12 12 12 12 12

And now we can calculate the statistic:

`\chi^2 = ((21-12)^2)/(12)+((16-12)^2)/(12)+((10-12)^2)/(12)+((8-12)^2)/(12)+((5-12)^2)/(12)=13.833`

Now we can calculate the degrees of freedom for the statistic given by:

`df=(catgories-1)=(5-1)=4`

And we can calculate the p value given by:

`p_v = P(\chi^2_(4) >13.833)=0.00785`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .