A paper summarized a regression of y = percentage of pups in a capture on x = pe

Question

A paper summarized a regression of y = percentage of pups in a capture on x = percentage of CPV prevalence among adults and pups. The equation of the least-squares line, based on n = 10 observations, was y = 62.9507 - 0.54877x, with r2 = 0.56.

(a) One observation was (25, 70). What is the corresponding residual? (Give the answers to one decimal place.)

What does y =

What does the Residual =

(b) What is the value of the sample correlation coefficient? (Give the answer to three decimal places.)

r =

(c) Suppose that SSTo = 2520 (this value was not given in the paper). What is the value of se? (Give the answer to two decimal places.)

se =

Explanation / Answer

Part a

We are given x = 25, y = 70

We have y = 62.9507 – 0.54877*x

Predicted y when x = 25 is given as below:

Predicted y = 62.9507 – 0.54877*25 = 49.23145

What does y = 70

Residual =| y – predicted y |= 70 - 49.23145 = 20.76855

Residual = 20.8

Part b

Sample correlation coefficient = r = sqrt(r^2)

We are given r^2 = 0.56

So, r = sqrt(0.56) = 0.748331

Sample correlation coefficient = r = 0.748

Part c

We know,

Coefficient of determination = r^2 = SSR/SST

We are given, coefficient of determination = r^2 = 0.56 and SST = 2520

So, 0.56 = SSR/2520

SSR = 0.56*2520 = 1411.2

SSE = SST – SSR

SSE = 2520 – 1411.2 = 1108.8

DF for error = 10 – 1 – 1 = 8

MSE = 1108.8/8 = 138.6

Se = sqrt(MSE) = sqrt(138.6) = 11.77285

Se = 11.77

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