A random sample of 11 observations was taken from normal population. The sample mean and standard deviation are 74.5 and 9 accordingly. Can we infer at 5% significance level that the population mean is greater than 70?
Let x‾=74.5, σ=9, n=11μ=70, α=0.05Let~\overline{x}=74.5,~\sigma=9,~n=11\\ \mu=70,~\alpha=0.05Let x=74.5, σ=9, n=11μ=70, α=0.05
z=x‾−μσn=74.5−70911≈1.658z=\frac{\overline{x}-\mu}{\sigma}\sqrt{n}=\frac{74.5-70}{9}\sqrt{11}\approx1.658z=σx−μn=974.5−7011≈1.658
P(z>1.658)=1−P(z<1.658)P(z<1.658)>P(z<1.65)P(z>1.658)=1-P(z<1.658)\\ P(z<1.658)>P(z<1.65)P(z>1.658)=1−P(z<1.658)P(z<1.658)>P(z<1.65)
From z-score table: P(z<1.65)=0.9505P(z<1.65)=0.9505P(z<1.65)=0.9505
P(z<1.658)>0.9505P(z>1.658)<1−0.9505=0.0495<0.05=αP(z<1.658)>0.9505\\ P(z>1.658)<1-0.9505=0.0495<0.05=\alphaP(z<1.658)>0.9505P(z>1.658)<1−0.9505=0.0495<0.05=α
Since P(z>1.658)<α,P(z>1.658)<\alpha,P(z>1.658)<α, we can infer that the population mean is greater than 70.
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