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

Answer:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

$p_v = P(\chi^2_{4} >13.833)=0.00785$

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 .

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".

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:

$\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{5}$

And the calculations are given by:

$E_{Beverage} =\frac{60}{5}=12$

$E_{Nap} =\frac{60}{5}=12$

$E_{Walk} =\frac{60}{5}=12$

$E_{Snack} =\frac{60}{5}=12$

$E_{Other} =\frac{60}{5}=12$

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:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

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 .