Question

You perform a X2 goodness-of-fit test to see if the number of birthdays occurring each month matches the expected number (assuming each month is equally likely to be the birth month for any given individual). You get 20.5 as your X2 value. What is the P-value for this test?

Answer 1

Answer:

$p_v = P(\chi^2_{11} >20.5)=0.0389$

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

"=1-CHISQ.DIST(20.5,11,TRUE)"

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".

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

H0: Each month is equally likely to be the birth month for any given individual

H1: Each month is NOT equally likely to be the birth month for any given individual

The statistic to check the hypothesis is given by:

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

After calculate the statistic we got $\chi^2 = 20.5$

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

$df=categories-1=12-1=11$

And we have categories =12  since we have 12 months in a year

And we can calculate the p value given by:

$p_v = P(\chi^2_{11} >20.5)=0.0389$

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

"=1-CHISQ.DIST(20.5,11,TRUE)"