Aresearcher 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?

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

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:

Method   Beverage   Nap   Walk   Snack   Other

Number       12             12       12         12         12

And now we can calculate the statistic:

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(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 .

33

step-by-step explanation: g=6

step-by-step explanation: