Using the earlier example:

[hoxesi8100@]

Outcome

(O−E)2/E(O - E)^2 / E(O−E)2/E for Subject Line A

(O−E)2/E(O - E)^2 / E(O−E)2/E for Subject Line B

Opened

5

5

Did Not Open

1.25

1.25

χ2=5+5+1.25+1.25=12.5\chi^2 = 5 + 5 + 1.25 + 1.25 = 12.5χ2=5+5+1.25+1.25=12.5

Compare your Chi-Square value to the distribution table.

To determine if the results are statistically denmark phone number material significant, I compare the Chi-Square value (12.5) to a critical value from a Chi-Square distribution table, based on:

Degrees of freedom (df): This is determined by (number of rows −1)×(number of columns −1)(number\ of\ rows\ - 1) \times (number\ of\ columns\ - 1)(number of rows −1)×(number of columns −1). For a 2x2 table, df=1df = 1df=1.
Alpha (α\alphaα): The confidence level of the test. With an alpha of 0.05 (95% confidence), the critical value for df=1df = 1df=1 is 3.84.
In this case:

Chi-Square Value = 12.5
Critical Value = 3.84
Since 12.5>3.8412.5 > 3.8412.5>3.84, the results are statistically significant. This indicates that there is a relationship between the subject line and the open rate.

If the Chi-Square value were lower…