Question

An article reported the results of an experiment to determine the effect of load on the drift in signals derived from a piezoelectric force plates. The correlation coefficient y between output and time under a load of 588 N was −0.9515. Measurements were taken 110 times per second for 300 seconds, for a total of 33000 measurements. Find a 95% confidence interval for the population correlation rho. Round the answers to four decimal places.

Answer 1

Answer:

95% confidence interval for the population is; -0.9525 < $p$ < -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 = $\frac{1}{2}$ln($\frac{1 + r}{1 - r}$)

we substitute -0.9515 for r

W = $\frac{1}{2}$ln($\frac{1 + (-0.9515 )}{1 - (-0.9515 )}$)

W = $\frac{1}{2}$ln($\frac{1 - 0.9515 )}{1 + 0.9515 )}$)

W = $\frac{1}{2}$ln($\frac{ 0.0485}{1.9515 }$)

W = $\frac{1}{2}$ln( 0.02485 )

W = $\frac{1}{2}$( -3.6948975 )

W = - 1.8474

So the quantity W is normally distributed with  standard deviation given by;

σ$_w$ = 1 / √(n - 3)

given that n is 33,000, we substitute

σ$_w$ = 1 / √(33,000 - 3)

σ$_w$ = 1 / √32997  

σ$_w$ = 0.0055

Now, at 95% confidence interval, μ$_w$ will be;

⇒ - 1.8474 - 1.96( 0.0055 ) < μ$_w$ < - 1.8474 + 1.96( 0.0055 )

⇒ - 1.8474 - 0.01078 < μ$_w$ < - 1.8474 + 0.01078

⇒ -1.8582 < μ$_w$ < -1.8366

So to obtain 95% confidence interval for p, we use the following equation;

we transform the inequality

p = [( $e^{2u_w} - 1$ ) / ( $e^{2u_w} + 1$ )]

we substitute

$\frac{e^{2(-1.8582 )}-1}{e^{2(-1.8582 )} + 1}$ < $\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}$ < $\frac{e^{2(-1.8366)}-1}{e^{2(-1.8366)} + 1}$

$\frac{e^{-3.7164}-1}{e^{-3.7164} + 1}$ < $\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}$ < $\frac{e^{-3.6732}-1}{e^{-3.6732} + 1}$

$\frac{0.02432-1}{0.02432 + 1}$ < $\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}$ < $\frac{0.025395-1}{0.025395 + 1}$

$\frac{-0.97568}{1.02432}$ < $\frac{e^{2u_w} - 1}{e^{2u_{w}} + 1}$ < $\frac{-0.974605}{1.025395}$

-0.9525 < $p$ < -0.9505

Therefore,  95% confidence interval for the population is; -0.9525 < $p$ < -0.9505