basic t-test statistics I'm writing an article, and I'm on a deadline so please be kind to me, I'd like to test for statistical significance of the difference before and after an intervention but I'm no real math genious
My sample size is 63 pre and post-intervention interviews, and I have two different mean scores of 134,192 and 178,324
Are these significantly different? How do I do it? In excel or anywhere else is fine. Thanks a whole bunch.
 A: You need to perform the following hypothesis test:
\begin{align}
H_0 &: \mu_1 = \mu_2 \\
H_a &: \mu_1 \neq \mu_2
\end{align}
where $H_0$ and $H_a$ are the null and alternate hypothesis respectively. $\mu_1$ and $\mu_2$ are the means of the two samples. You will have to compute the following test statistic:
\begin{align}
t &= \frac{\hat{\mu}_d}{\sqrt{\hat{s}_d^2/n}}
\end{align}
where $n$ is the number of samples ($63$ in your case), $\hat{\mu}_d$ and $\hat{s}_d^2$ are the sample mean and sample variance of the difference between the paired readings.
The above test statistic has $n-1$ degrees of freedom. So assuming you want to check for significance of the difference between the two means at $5\%$ level, you will have to:


*

*Compute the statistic using your data.

*Compare the resulting number to $2$.

*If your statistic is greater than $2$, the null hypothesis is rejected and there exists a significant difference between the two means at the $5\%$ level. Else, you cannot reject the null hypothesis.


In (2), I have given you the critical value by reading off the t-distribution table here: http://www.math.unb.ca/~knight/utility/t-table.htm. I suggest you to review basic statistical hypothesis testing from a standard text book (e.g. Hogg) in case you want a more detailed understanding. 
A: The paired t test described by Kartik is what you should use when the data is normally distributed or at least approximately so. But if the normality assumption fails a nonparametric paired test such as the Wilcoxon signed-rank test is what you should use.
