t-test on independent groups with unequal variance

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Sample average from population which follows normal distribution also follows it. When standard deviation of population is not known, you would have to speculate it with standard deviation of sample. T-statistics follows t-distribution, not normal distribution.

\displaystyle t = \frac{\bar X - \mu}{SD/\sqrt n} = \frac{\bar X - \mu}{SE} \vspace{0.2in}\\

SD: standard deviation; SE: standard error

When you would like to compare average values between separate groups with unequal variance, you could calculate t-statistics with formula below;

\displaystyle t = \frac{(\bar X_2 - \bar X_1)}{SD_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{(\bar X_2 - \bar X_1)}{SE}\vspace{0.2in}\\ SD_p = \sqrt{\frac{(n_1 - 1)SD_1^2 + (n_2 -1)SD_2^2}{n_1 + n_2 - 2}}

SD_p; pooled SD

When t-statistics is greater than a value, null hypothesis is rejected. In one sided test, when it is greater than the value which area under t-distribution curve is smaller than 0.05, it is statistically significant. In two sided test, when it is greater than the value which area under curve is smaller than 0.025, it is statistically significant. T-statistics follows degree of freedom.

Reference:t-distribution

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投稿者: admin

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