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Independent Two-sample z-test
1. Assumptions
- Experimental Design: The sample forms two independent treatment groups.
- Null Hypothesis: The population means of the two treatment groups are not significantly different from each other.
- Population Distribution: Arbitrary distribution within each treat- ment group.
- Sample Size: Size of each treatment group is equal to or greater than
2. Inputs for independent two-sample z-test
- Sample sizes of treatment groups: n 1 and n 2
- Sample means of treatment groups: ¯x 1 and ¯x 2
- Standard deviations of treatment groups: s 1 and s 2
- Standard errors of treatment group means:
SE 1 =
s 1 √ n 1
and SE 2 =
s 2 √ n 2
- Standard error of the differences: sdiff =
SE^21 + SE^22
- Null hypothesis: μ 1 = μ 2
- The level of the test: α
3. The Five Steps for Performing the Test of Hypothesis
- State null and alternative hypotheses:
H 0 : μ 1 = μ 2 , H 1 : μ 1 6 = μ 2
- Compute test statistic:
z =
(¯x 2 − x¯ 1 ) − (μ 2 − μ 1 ) SEdiff
assuming the null hypothesis.
- Compute 100(1 − α)% confidence interval I for z.
- If z ∈ I, accept H 0 ; if z /∈ I, reject H 1 and accept H 0.
- Compute p-value.
4. Discussion
Since we are assuming that n ≥ 30, the sample standard deviations s 1 and s 1 are close approximations to the population standard deviations σ 1 and σ 2 , so we will assume that the population standard deviations are known and equal to the respective sample standard deviations. Furthermore
E(¯x 2 − x¯ 1 ) = E(¯x 2 ) − E(¯x 1 ) = μ 2 − μ 1.
Also, if the two treatment groups are independent,
Var(¯x 2 − ¯x 1 ) = 1^2 · Var(¯x 2 ) + (−1)^2 Var(¯x 1 ) =
σ^22 n 1
σ^21 n 2
the standard deviation of ¯x 2 − x¯ 1 is
SEdiff =
Var(¯x 2 − x¯ 1 ) =
σ 22 n 1
σ 12 n 2
Finally, because the expected value and variance of
z =
(¯x 2 − x¯ 1 ) − (μ 2 − μ 1 ) SE^2 diff
are μ 2 − μ 1 and SE^2 diff, respectively, E(z) = 0 and σz = 1. By the central limit theorem, z ∼ N (0, 1), so we can use the standard normal table to find confidence intervals and p-values for z.
