Assume the gathered data points (`x, y`) from each population respectively are independent and that they are normally distributed.

If the population standard deviation of `x` and `y` is the same and denoted `sigma`, then the sample means `bar x`, `bar y` are such that

`bar x` ~ N(`alpha,sigma/sqrt(13))` `bar y`...

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Assume the gathered data points (`x, y`) from each population respectively are independent and that they are normally distributed.

If the population standard deviation of `x` and `y` is the same and denoted `sigma`, then the sample means `bar x`, `bar y` are such that

`bar x` ~ N(`alpha,sigma/sqrt(13))` `bar y` ~ N(`beta, sigma/sqrt(13))` where `bar x` and `bar y ` are independent.

To test `alpha > beta` (or `alpha < beta`, but not both of these as it is a one-sided test) against the null `alpha=beta`

the test statistic that we calculate is of the form

`(bar x - bar y)/(SE(bar x - bar y))`

where `SE` is the standard error.

Given that `SE(bar x - bar y) = sqrt(V(bar x) + V(bar y)) = sqrt((2sigma^2)/13) = sqrt(2/13) sigma`

The test statistic is of the form

`sqrt(13/2)((bar x - bar y)/(sigma))`

Now, the 95th percentile of the standard normal distribution is 1.645. The null hypothesis is that `bar x = bar y`. So, if we are testing the alternative `bar x > bar y` at the 5% level, then we check whether our computed test statistic is greater than 1.645. This is the case if

`sqrt(13/2)((bar x - bar y)/sigma) > 1.645` implies `bar x - bar y > 1.645 sqrt(2/13)sigma`

implies `bar x - bar y > 0.645` population standard deviations.

**Thus, the two means need to be at least 0.645 population standard deviations apart to conclude at the 5% level of significance that one is larger than the other (pre-specifying beforehand which one we expect to be larger, since the test is one-sided) and reject the null that they are the same.**