Answer:
95% confidence interval for the population is; -0.9525 < [tex]p[/tex] < -0.9505
Step-by-step explanation:
Given the data in the question;
correlation coefficient y between output and time under a load of 588 N was −0.9515
so first we determine the quantity W.
we know that; W = [tex]\frac{1}{2}[/tex]ln([tex]\frac{1 + r}{1 - r}[/tex])
we substitute -0.9515 for r
W = [tex]\frac{1}{2}[/tex]ln([tex]\frac{1 + (-0.9515 )}{1 - (-0.9515 )}[/tex])
W = [tex]\frac{1}{2}[/tex]ln([tex]\frac{1 - 0.9515 )}{1 + 0.9515 )}[/tex])
W = [tex]\frac{1}{2}[/tex]ln([tex]\frac{ 0.0485}{1.9515 }[/tex])
W = [tex]\frac{1}{2}[/tex]ln( 0.02485 )
W = [tex]\frac{1}{2}[/tex]( -3.6948975 )
W = - 1.8474
So the quantity W is normally distributed with standard deviation given by;
σ[tex]_w[/tex] = 1 / √(n - 3)
given that n is 33,000, we substitute
σ[tex]_w[/tex] = 1 / √(33,000 - 3)
σ[tex]_w[/tex] = 1 / √32997
σ[tex]_w[/tex] = 0.0055
Now, at 95% confidence interval, μ[tex]_w[/tex] will be;
⇒ - 1.8474 - 1.96( 0.0055 ) < μ[tex]_w[/tex] < - 1.8474 + 1.96( 0.0055 )
⇒ - 1.8474 - 0.01078 < μ[tex]_w[/tex] < - 1.8474 + 0.01078
⇒ -1.8582 < μ[tex]_w[/tex] < -1.8366
So to obtain 95% confidence interval for p, we use the following equation;
we transform the inequality
p = [( [tex]e^{2u_w} - 1[/tex] ) / ( [tex]e^{2u_w} + 1[/tex] )]
we substitute
[tex]\frac{e^{2(-1.8582 )}-1}{e^{2(-1.8582 )} + 1}[/tex] < [tex]\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}[/tex] < [tex]\frac{e^{2(-1.8366)}-1}{e^{2(-1.8366)} + 1}[/tex]
[tex]\frac{e^{-3.7164}-1}{e^{-3.7164} + 1}[/tex] < [tex]\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}[/tex] < [tex]\frac{e^{-3.6732}-1}{e^{-3.6732} + 1}[/tex]
[tex]\frac{0.02432-1}{0.02432 + 1}[/tex] < [tex]\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}[/tex] < [tex]\frac{0.025395-1}{0.025395 + 1}[/tex]
[tex]\frac{-0.97568}{1.02432}[/tex] < [tex]\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}[/tex] < [tex]\frac{-0.974605}{1.025395}[/tex]
-0.9525 < [tex]p[/tex] < -0.9505
Therefore, 95% confidence interval for the population is; -0.9525 < [tex]p[/tex] < -0.9505
