Answer:
95% for the true average increase in total cholesterol under these circumstances
(-2.306 , 9.406)
Step-by-step explanation:
Step(i):-
Given sample size 'n' =41
Mean of the sample(x⁻) = 3.55
The estimated standard error
[tex]S.E = \frac{S.D}{\sqrt{n} }[/tex]
Given estimated standard error ( S.E) = 3.478
Level of significance ∝=0.05
Step(ii):-
95% for the true average increase in total cholesterol under these circumstances
[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]
Degrees of freedom
ν= n-1 = 41-1 =40
t₀.₀₅ = 1.6839
95% for the true average increase in total cholesterol under these circumstances
[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]
( 3.55 - 1.6839 ×3.478 ,3.55 + 1.6839 ×3.478 )
(3.55 - 5.856 , 3.55 + 5.856)
(-2.306 , 9.406)
Conclusion:-
95% for the true average increase in total cholesterol under these circumstances
(-2.306 , 9.406)