Note that it is the count that should be inserted in that formula, not the percentage.
I'm having a little trouble with some the stats for a project i'm doing. I haven't taken it in years.
I need to compare the sex ratios in 3 areas and also if they deviate from 50/50. I found a paper that did a similar test and they used a chi squared goodness of fit test.
The percent female for the different areas are:
A: 82%
B: 73%
C: 58%
I tried to do (observed-expected)^2/expected then summed but checked it on an online calculator and it said error.
Any input?
Thanks
Note that it is the count that should be inserted in that formula, not the percentage.
Biology (05-23-2014)
Yea I used the counts. Then for the observed I used the 50% of the total counts for each sex.
My confusion is comparing among areas.
Thanks
For the 'observed', use the observed number of females. For the expected use 0.50*(the number of humans).
You can also use a create a two-way table with female and male as columns and area A, B and C as rows. And do a traditional chi-squared test from that.
Or you can compare the percentage directly: (0.82 - 0.5)/sqrt(0.82*(1-0.82)/n)
That will test if 0.82 is statistically significant from 0.50.
Let me quote from this site [LINK]:
"[Chi-square] Goodness-Of-Fit VS.Test of Independence
Goodness-Of-Fit
A goodness-of-fit test is a one variable Chi-square test. According to Steinberg (2011), “the goal of a Chi-square goodness-of-fit test is to determine whether a set of frequencies or proportions is similar to and therefore “fits” with a hypothesized set of frequencies or proportions” (p. 371). A Chi-square goodness-of-fit test is like to a one-sample t-test. It determines if a sample is similar to, and representative of, a population.
Example of Goodness-Of-Fit:
We might compare the proportion of M&M’s of each color in a given bag of M&M’s to the proportion of M&M’s of each color that Mars (the manufacturer) claims to produce. In this example there is only one variable, M&M’s. M&M’s can be divided into many many categories like Red, Yellow, Green, Blue, and Brown, however there is still only one variable… M&M’s.
Hungry?
Steinberg (2011), notes: “the Chi-square goodness-of-fit test will determine whether or not the relative frequencies in the observed categories are similar to, or statistically different from, the hypothesized relative frequencies within those same categories (p. 371).
Test of Independence
A test of independence is a two variable Chi-square test. Like any Chi-square test the data are frequencies, so there are no scores and no means or standard deviations. Steinberg (2011) points out, “the goal of a two-variable Chi-square is to determine whether or not the first variable is related to—or independent of—the second variable” (p. 382). A two variable Chi-square test or test of independence is similar to the test for an interaction effect in ANOVA, that asks: Is the outcome in one variable related to the outcome in some other variable” (Steinberg, 2011) (p. 382).
Example of Test of Independence
To continue with the M&M’s example, we might investigate whether purchasers of a bag of M&M’s eat certain colors of M&M’s first. Here there are two variables: (1) M&M’s (2) The order based on color that an M&M bag holder/purchaser eats the candies.
Steinberg, W. J. (2011). Statistics alive! (2nd ed.). Thousand Oaks, CA: Sage Publications.
Marczyk, G., DeMatteo, D., & Festinger, D. (2005). Essentials of research design and methodology. Hoboken, NJ: John Wiley & Sons."
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GretaGarbo (06-28-2014)
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