The 99% of a confidence interval for the average maximum HP for the experimental engine. (843.0906, 916.9094)
Given,
In the question:
Given that the mean of the Population = 830HP
Given that the size of the sample  'n' = 25
Given that the mean of the sample = 880HP
Given that the sample standard deviation = 55HP
To find the confidence interval for the average maximum HP for the experimental engine.
Use a significance level of α=0.01.
Now, According to the question:
Degrees of freedom = n-1 =25 - 1 = 24
t₀.₀₀₅ = 3.3554
The 99% of a confidence interval for the average maximum HP for the experimental engine.
[tex](x -\frac{t_0_._0_0_1}{2},24\frac{S.D.}{\sqrt{n} } , x^2 +\frac{t_0_._0_0_1}{2},24\frac{S.D.}{\sqrt{n} })[/tex]
(880 - 3.3554 [tex]\frac{55}{\sqrt{25} }[/tex] , 880 + 3.3554 [tex]\frac{55}{\sqrt{25} }[/tex])
(843.0906, 916.9094)
Hence, The 99% of a confidence interval for the average maximum HP for the experimental engine. (843.0906, 916.9094)
Learn more about Significance level at:
https://brainly.com/question/13909985
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