Two independent samples have been selected, 7777 observations from population 1 and 5858 observations from population 2. The sample means have been calculated to be ?⎯⎯⎯1=13.4x¯1=13.4 and ?⎯⎯⎯2=11.3x¯2=11.3. From previous experience with these populations, it is known that the variances are ?21=27σ12=27 and ?22=29σ22=29. For the hypothesis test of ?0:(?1−?2)=1.8H0:(μ1−μ2)=1.8 and ??:(?1−?2)≠1.8Ha:(μ1−μ2)≠1.8 Use ?=0.04α=0.04. (a) Compute the test statistic. ?=z= (b) Find the approximate p-value ?−?????=p−value= The final conclustion is A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8. B. We can reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8 and accept that (?1−?2)≠1.8(μ1−μ2)≠1.8.
Two independent samples have been selected, 7777 observations from population 1 and 5858 observations from population 2. The sample means have been calculated to be ?⎯⎯⎯1=13.4x¯1=13.4 and ?⎯⎯⎯2=11.3x¯2=11.3. From previous experience with these populations, it is known that the variances are ?21=27σ12=27 and ?22=29σ22=29.
For the hypothesis test of ?0:(?1−?2)=1.8H0:(μ1−μ2)=1.8 and ??:(?1−?2)≠1.8Ha:(μ1−μ2)≠1.8 Use ?=0.04α=0.04.
(a) Compute the test statistic.
?=z=
(b) Find the approximate p-value
?−?????=p−value=
The final conclustion is
A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8.
B. We can reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8 and accept that (?1−?2)≠1.8(μ1−μ2)≠1.8.
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