# Help with chi squared goodness of fit test

#### Sarah20

##### New Member
Volunteers at a teenage distress center have been assigned based on their skills and expertise. It's assumed that 40% are drug related, 25% *** related, 25% stress, and 10% education. For the study sample of 120 calls per month were categoruzed as follows

Observed frequencies: Drug (52) *** (38) Stress (21) Education (9) Total 120.

Apply chi-squared goodness of fit to test the hypothesis of assumed distribution of calls. Interpet results for the purposes of staff allocation at the distress center.

I know how to calculate chi squared but the percentages for each category confuse me! #### Dragan

##### Super Moderator
I know how to calculate chi squared but the percentages for each category confuse me! All you need to do is think of the percentages in terms of proportions (p) to obtain the expected frequencies.

That is, the p's are 0.40, 0.25, 0.25, and 0.10. Thus, with N = 120 your expected frequencies are 48, 30, 30, and 12.

#### Sarah20

##### New Member
Thank you! thats what I figured, but to calculate chi's Xsquared, what should be my denominator for each of the 4 values? i dont' think it would be 30.

#### Dragan

##### Super Moderator
Thank you! thats what I figured, but to calculate chi's Xsquared, what should be my denominator for each of the 4 values? i dont' think it would be 30.

Your chi-square statistic should be computed as follows:

Xsq = (52 - 48)^2 / 48 + (38 - 30)^2 / 30 + (21 - 30)^2 / 30 + (9 - 12)^2 / 12.

Mkay.

#### Sarah20

##### New Member
great thanks! that clears it up!

#### Sarah20

##### New Member
since my x squared is 5.92 and pvalue of 7.82, i guess i can say i do not reject the null since my x squared is smaller than P, but how would i go about interpreting staff allocation at the distress center?

#### Mean Joe

##### TS Contributor
since my x squared is 5.92 and pvalue of 7.82, i guess i can say i do not reject the null since my x squared is smaller than P, but how would i go about interpreting staff allocation at the distress center?
X^2=5.92, but the p-value (chi test with 3 df) = .116

Since the p-value is greater than .05 (the traditional critical value for testing), then you do not reject the null. That is, you accept that the assignments are 40% are drug related, 25% s ex related, 25% stress, and 10% education.