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Chapter 5: Problem 49

The paper "Effects of Canine Parvovirus (CPV) on Gray Wolves in Minnesota" (Journal of Wildlife Management \([1995]: 565-570\) ) summarized a regression of \(y=\) percentage of pups in a capture on \(x=\) percentage of \(\mathrm{CPV}\) prevalence among adults and pups. The equation of the least-squares line, based on \(n=10\) observations, was \(\hat{y}=62.9476-0.54975 x\), with \(r^{2}=.57\) a. One observation was \((25,70)\). What is the corresponding residual? b. What is the value of the sample correlation coefficient? c. Suppose that \(\mathrm{SSTo}=2520.0\) (this value was not given in the paper). What is the value of \(s_{e} ?\)

Short Answer

Expert verified
a. The corresponding residual for the observation (25,70) is 20.29. b. The value of the sample correlation coefficient is -0.75. c. The value of \(s_{e}\) is 11.674.

Step by step solution

01

Calculate the residual for the observation (25,70)

Residual is calculated by subtracting the predicted value from the actual value. The equation of the line is given by \( \hat{y} = 62.9476 - 0.54975x \) from the exercise. Let's substitute \(x = 25\) into the equation to find the predicted value (\(\hat{y}\)): \(\hat{y} = 62.9476 - 0.54975*25 = 49.71 \). The actual value (\(y\)) for \(x = 25\) is 70. Hence, the residual is \(y - \hat{y} = 70 - 49.71 = 20.29 \)
02

Determine the value for correlation coefficient

The value of the square of the correlation coefficient (\(r^2\)) is given as 0.57 in the exercise. So, the correlation coefficient (\(r\)) is \(\sqrt{0.57}\) or - \(\sqrt{0.57}\). As we have a negative slope in the regression equation, we will take the negative root which gives \(r = -0.75\).
03

Compute the value for standard error (\(s_{e}\))

Standard error (\(s_{e}\)) is calculated by taking the square root of the residual sum of squares (SSR) divided by the degrees of freedom. The value of SSR can be calculated using the given value of total sum of squares (SSTo) and \(r^{2}\). SSR = SSTo*(1-\(r^{2}\)). plugging in the values, we get, SSR = 2520.0*(1-0.57) = 1084.40. As there are 10 observations, the degrees of freedom would be n-2=10-2=8. Let's substitute these values in the \(s_{e}\) equation: \(s_{e} = \sqrt{SSR/degree\ of\ freedom} = \sqrt{1084.40/8} = 11.674.

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