Cross Validated
2022-10-26 13:51 UTC
By kurtkim
AI-113-20221026-social-media-4585c8cd
How does Pearson's cumulative test statistic approach Chi-squared distribution?
From Wikipedia, $ \sum_1^k{Z_i^2}$ is Chi-squared distributed( $Z_i$ is a standard normal random variable) Also, it is followed by that Pearson's cumulative test statistic $ \sum_1^n{(O_i-E_i)^2 \over E_i}$ approaches to Chi-squared distribution. ( $O_{i}$ = the number of observations of type $i$ , $E_{i}$ = the expected (theoretical) frequency of type $i$ ) I have been searching for the proof that $ \sum_1^n{(O_i-E_i)^2 \over E_i}$ approaches to $ \sum_1^k{Z_i^2}$ , but I could not find it anywhere. Is there anyone to show the proof?
From Wikipedia, $ \sum_1^k{Z_i^2}$ is Chi-squared distributed( $Z_i$ is a standard normal random variable) Also, it is followed by that Pearson's cumulative test statistic $ \sum_1^n{(O_i-E_i)^2 \over E_i}$ approaches to Chi-squared distribution. ( $O_{i}$ = the number of observations of type $i$ , $E_{i}$ = the expected (theoretical) frequency of type $i$ ) I have been searching for the proof that $ \sum_1^n{(O_i-E_i)^2 \over E_i}$ approaches to $ \sum_1^k{Z_i^2}$ , but I could not find it anywhere. Is there anyone to show the proof?
Full article content could not be extracted automatically. Read the original below.
Source:
Cross Validated
· stats.stackexchange.com