Question

Every morning, Dany jogs the same 2-mile route around her neighborhood. She selects a random sample of days and records the time it takes her to complete the route as well as her pulse rate just after she gets home. She calculated a Pearson correlation coefficient of r = -0.76 and below is a plot of her data:

(Sorry! I can't post the plot)

n= 11

a. Calculate a 95% confidence interval for the population Pearson correlation coefficient between run time and pulse rate. Given: ?∗ = 2.262 (3pt)

Answer 1

We have corelation and also Assuming population of the z given that the calculated a pearson coefficient of r = -0.76 and n= 11 given that, t = 2.262. a brivaniate normal population with correlations , the tranformation sample pearson Correlation from a to is, 2x = { In ( 1+x) = { In (-0-646) 1.76 = = en (0.1363636364) (-1.99 243016) + = -0.99621508 7 2 -0-99 62 1508 This is approximetely normally distributed with variance 6-3) Š. The 95% confidence intercel for the population pearson Correlation coefficient between nun time and pulse rate is, pe (099621508 + 1960) >> € (-0.99621508 $ 1.96 )

> Zg f(-0.99621508 ± 1.96 1.125) => g. € (-0-99621502 + (136 *0.35855339)) 3) 2 € (-0.99621508 -0.69296464,-0.99621508 +0-69296464) - 29 € (-1.689 17972 ,-0.303250 44 ) Now, to obtain 2 = -1.68917972 and zu = -0.30325044. The values of , and zu are then transformed back to the correlation scale using the inverse transformations. e exp (222) - exp ( 22 ) +1 = exp(2x-1.689 17972) - exp(2x-1.68917972) +1 - expl 3.37835944) - exp(-3.37835944) + 1 0.03410335747 -1 0.034103351 47 tl -0.96589664253 1.0341033 5147 = -0.9340426575 3 = -0.934 )

and exp (2Zu)-1 exp (2Zu)+1 - exp (2x +0.30325044)-1 exp (2x - 0.30325044 ) + 1 = exp(-0.60650088) -1 exp(-0.60650088) +1 0.545255 4492 - 1 0.5452554492 ti -0.4547445508 1.545255 4492 = -0.2942843858 Bu = -0.294) The 95% confidence interval for fis, -0.934 < = -0.294