Question

Help! Will fan and medal!

Answer 1

Could mud wrestling be the cause of a rash contracted by Washington State University students in the spring of 1992? Physicians at the WSU student health center wondered this when one male and six female students complained of rashes after participating in a mud wrestling event. Questionnaires were sent to all students in the residence halls that participated in the event. The questionnaire asked about the appearance of a rash and about attendance at the event. The results are summarized below:
Men Women
Developed Rash 12 12
No Rash 38 12
What is the p-value for the chi-square statistic?
Choose one answer.
A. Between 0.05 and 0.01
B. Larger than 0.10
C. Between 0.10 and 0.05
D. Less than 0.01
E. Between 0.001 and 0.01

Answer 2

@agent0smith @kirbykirby

Answer 3

I was thinking C, but I'm not sure.

Answer 4

What kind of null hypothesis are you testing for? Also, is there information about the expected outcomes? Is there a distribution for this? The table only gives observed outcomes.

Answer 5

That is all the info I was given.
but since it is a chi-squared test the null would be observed = expected

Answer 6

Any ideas?

Answer 7

Oh ok . The expected frequencies, you can calculate it for each cell by finding the (row total * column total)/(table total)

Men Women | Row total
Developed Rash 12 12 | 24
No Rash 38 12 | 50
---------------------------------
Col total: 50 24 | 74

So the expected outcomes are:

Men Women
Developed Rash 16.22 7.78
No Rash 33.78 16.22

So the chi-square statistics is \[\sum_{i=1}^4 \frac{(O_i-E_i)^2}{E_i}\\
=\frac{(12-16.22)^2}{16.22}+\frac{(12-7.78)^2}{7.78}+\frac{(38-33.78)^2}{33.78}+\frac{(12-16.22)^2}{16.22}=\chi^2\]
Now the degrees of freedom is
\(df=(\# \text{ rows}-1)(\#\text{ columns }-1)=(2-1)(2-1)=1\)
Now you find in a chi-square table like here http://sites.stat.psu.edu/~mga/401/tables/Chi-square-table.pdf (or a calculator, or using R), you find what your chi-square value is in the table for 1 degree of freedom, and determine which probability this corresponds to, and that's the p-value (With a table you will only the know the interval for the p-value)

Answer 8

The chi-square table I provided might not be adequate for the intervals you need, but I'm sure your prof has provided you with an adequate one.

Answer 9

I believe you should find \(\chi^2=5.002\), so the p-value in what I gave you should lie between 0.05 and 0.025, which would correspond with choice A