If other factors are held constant, how does increasing the sample variance affect the value of the independent-measures t statistic and the likelihood of rejecting the null hypothesis?
To understand how increasing the sample variance affects the independent-measures t statistic and the likelihood of rejecting the null hypothesis, we need to consider the formula for calculating the t statistic in independent-measures t-tests.
The formula for the t statistic in an independent-measures t-test is:
t = (mean1 - mean2) / √[(s^21/n1) + (s^2 2/n2)]
Where:
- mean1 and mean2 are the means of the two independent groups being compared,
- s^2 1 and s^2 2 are the variances of the two independent groups,
- n1 and n2 are the sample sizes of the two independent groups.
When other factors are held constant and the sample variance increases, it leads to an increase in the denominator of the t statistic formula (√[(s^2 1/n1) + (s^2 2/n2)]), which in turn decreases the value of the t statistic.
Now, let's understand the relationship between the t statistic and the likelihood of rejecting the null hypothesis. In hypothesis testing, we compare the calculated t statistic to a critical value to determine if we can reject the null hypothesis. The critical value depends on the significance level and the degrees of freedom.
With a smaller t statistic value resulting from an increased sample variance, it becomes less likely that the calculated t statistic will exceed the critical value, and therefore, less likely that we will reject the null hypothesis. In other words, increasing the sample variance reduces the magnitude of the t statistic and decreases the likelihood of rejecting the null hypothesis.
To summarize, increasing the sample variance decreases the value of the independent-measures t statistic and lowers the likelihood of rejecting the null hypothesis when other factors are held constant.