Segment S1S2S3Total
Region133
Region2895122162
Region320428145530
Region478587143
Region51228112215
All496471861,053
For the categories above, when i calculate the expected frequency, the count is less than 5 for S2-Region1 & S3-Region1.
If Chi-Square test is not valid, what other test that I can perform to see whether the two variables (Region & Segment) is statistically significant?
2006-12-23 01:21:50 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
How about Fisher exact test? I think it's better to be used according you still found the count of less than 5% in 2 regions.
2006-12-23 03:23:22 · answer #1 · answered by eddy 3 · 0⤊ 0⤋
A chi-sq. attempt is any statistical speculation attempt wherein the attempt statistic has a chi-sq. distribution while the null speculation is genuine, or any wherein the threat distribution of the attempt statistic (assuming the null speculation is genuine) could be made to approximate a chi-sq. distribution as heavily as needed by making the pattern length great sufficient. extremely, a chi-sq. attempt for independence evaluates statistically significant variations between proportions for 2 or greater communities in a documents set. an beautiful characteristic of the chi-sq. goodness-of-greater healthful attempt is that that is utilized to any univariate distribution for which you will calculate the cumulative distribution function. The chi-sq. goodness-of-greater healthful attempt is utilized to binned documents (i.e., documents positioned into training). that is unquestionably no longer a restrict for the reason that for non-binned documents you could purely calculate a histogram or frequency table in the previous producing the chi-sq. attempt. in spite of the indisputable fact that, the fee of the chi-sq. attempt statistic are based on how the information is binned. yet another draw back of the chi-sq. attempt is that it calls for a sufficient pattern length to confirm that the chi-sq. approximation to be legitimate
2016-12-18 18:07:32 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Dear CK, it would help if you put spaces between the numbers to enable us to double check the calculations.
Also what is the nature of the data? Is it height, deaths, bacterial counts etc? The nature of the data influences the choice of statistical test.
2006-12-23 02:03:31 · answer #3 · answered by claudeaf 3 · 0⤊ 0⤋
The t-test offers much more reliable results than the chi test.
2006-12-23 02:11:23 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
Your numbers are gobbledygook.
Please check what you are writing before you press the last button, to avoid people thinking you are as slovenly as you seem to be.
2006-12-23 03:08:32 · answer #5 · answered by Anonymous · 1⤊ 0⤋