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

95% confidence interval for the population is; -0.9525 <
`p` < -0.9505

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 =
`(1)/(2)`ln(
`(1 + r)/(1 - r)`)

we substitute -0.9515 for r

W =
`(1)/(2)`ln(
`(1 + (-0.9515 ))/(1 - (-0.9515 ))`)

W =
`(1)/(2)`ln(
`(1 - 0.9515 ))/(1 + 0.9515 ))`)

W =
`(1)/(2)`ln(
`( 0.0485)/(1.9515 )`)

W =
`(1)/(2)`ln( 0.02485 )

W =
`(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

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

`(e^(-3.7164)-1)/(e^(-3.7164) + 1)` <
`\frac{e^(2u_w) - 1}{e^{2u_(w)} + 1}` <
`(e^(-3.6732)-1)/(e^(-3.6732) + 1)`

`(0.02432-1)/(0.02432 + 1)` <
`\frac{e^(2u_w) - 1}{e^{2u_(w)} + 1}` <
`(0.025395-1)/(0.025395 + 1)`

`(-0.97568)/(1.02432)` <
`\frac{e^(2u_w) - 1}{e^{2u_(w)} + 1}` <
`(-0.974605)/(1.025395)`

-0.9525 <
`p` < -0.9505

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