Mean of two samples when population is normal and Ϭ is known and unequal:
Let's start admit that this is completely unrealistic, that we will find ourselves in the situation to know the population variance, but not the average population. Therefore, to learn method of hypothesis testing, we have limited practical application.. As usual, let's start with an example.
Example:
| A random sample of size 36 from a normal
population with variance 24 gave X̅=15 A second sample of size 28 from another normal population with variance 80 gave X̅2=13.test Ho:μ₁-μ₂=0 against H1: μ₁-μ₂≠0. Let α =0.05 Solution: For Two sided Test:
Z-test = X̅₁ - X̅₂/√(σ₁²/n₁+ σ₂²/n₂)
Z-cal = X̅₁-X̅₂/(sqrt(σ₁ ²/n₁+σ₂ ²/n₂ )
Z-tab = =NORMSINV(1-prob:/2)
P-value = =2*(1-NORMSDIST(Z-cal))
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For One sided Test:
Upper Tail:
Z-test = X̅₁ - X̅₂/√(σ₁²/n₁+ σ₂²/n₂)
Z-cal = X̅₁-X̅₂/(sqrt(σ₁ ²/n₁+σ₂ ²/n₂ )
Z-tab = =NORMSINV(1-prob:)
P-value = NORMSDIST(Z-cal))
Lower Tail:
Z-test = X̅₁ - X̅₂/√(σ₁²/n₁+ σ₂²/n₂)
Z-cal = X̅₁-X̅₂/(sqrt(σ₁ ²/n₁+σ₂ ²/n₂ )
Z-tab = =NORMSINV(1-prob:/2)
P-value = =2*(1-NORMSDIST(Z-cal))
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