Setting Up a Two-Mean Test
IntroducedA four unit gap needs an uncertainty model
- Two fictional independent random samples have means 52 and 48 units. Each sample has 25 observations; their standard deviations are 6 and 8 units.
- The observed difference is four units. The inference target is the difference between the two population means, with the subtraction order group 1 minus group 2.
The observed group means are 52 and 48 units. Find the group 1 minus group 2 difference.
52 - 48 = 4 units.
State a population equality claim
- Use $H_0:\mu_1-\mu_2=0$. The preplanned two-sided alternative is $H_a:\mu_1-\mu_2\ne0$.
- Choose a one-sided alternative only if that direction comes from the research question before examining the results.
s1=6,n1=25 and s2=8,n2=25. Compute SE = √(36/25 + 64/25).
√(1.44 + 2.56) = √4 = 2.
Check each group and their relationship
- Check random design, independence between samples, the 10% condition when relevant, and the shape of each sample. Each group has fewer than thirty observations here.
- Matched or repeated measurements require a paired analysis. Separate group labels alone do not prove that their observations are independent.
x̄1=52, x̄2=48, SE=2. Compute t = (x̄1−x̄2)/SE.
(52−48)/2 = 4/2 = 2.
Match each design to the procedure taught here.
The data relationship chooses the procedure; independent and paired formulas answer different designs.
Add the two variance contributions
- For the Welch two-sample method taught here, $SE=\sqrt{s_1^2/n_1+s_2^2/n_2}$. There is no equal-variance pooling assumption.
- Here $SE=\sqrt{36/25+64/25}=2$ units, so $t=(52-48)/2=2$. The variance contributions add even though the means are subtracted.
The variance contributions are 1.44 and 2.56 squared units. Their sum is 4, whose square root is 2 units; do not add the two standard errors directly.
The Welch two-sample t-test taught here pools the sample variances.
False: it keeps s1²/n1 and s2²/n2 separate. A different pooled t procedure requires an equal-variance assumption.
Use degrees of freedom from this procedure
- A two-sample t routine gives Welch degrees of freedom, which can be fractional. For these values, $df\approx44.51$.
- Do not automatically use $25+25-2=48$; that is the pooled equal-variance procedure. A t value of 2 does not alone determine its p-value.
The reference curve, degrees of freedom and tail rule must match the selected procedure.
The null hypothesis for a two-mean test is...
H0 says the two population means are equal.
You know that the two-sample statistic is t = 2. Which additional facts are needed to determine a p-value?
The reference t distribution and tail rule determine the probability.
Carry the statistic to the selected tail rule
- For this two-sided example, the p-value is approximately 0.0516 using $df\approx44.51$. At alpha 0.05, fail to reject population equality.
- The observed four unit difference remains an estimate; non-rejection does not prove equality. Report the uncertainty and use a prechosen direction for any one-sided test.
For this two-sided example, the p-value is approximately 0.0516 using $df\approx44.51$. At alpha 0.05, fail to reject population equality.
Under the null, the numerator of the two-sample t is (x̄1 − x̄2) − 0.
The null value μ1 − μ2 = 0 is subtracted.