The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a $90 \%$ confidence level with a redshift of 0.0126 .
\begin{tabular}{l|ccccccc}
Redshift & 0.0231 & 0.0535 & 0.0715 & 0.0391 & 0.0439 & 0.0101 \\
\hline Distance & 0.33 & 0.76 & 1.01 & 0.54 & 0.61 & 0.16
\end{tabular}
a. Find the explained variation.
(Round to six decimal places as needed.)
b. Find the unexplained variation.
(Round to six decimal places as needed.)
c. Find the indicated prediction interval.
$\square$ billion light-years $
Solution
Step 1 ：First, we calculate the regression line using the given data. The redshift values are [0.0231, 0.0535, 0.0715, 0.0391, 0.0439, 0.0101] and the corresponding distance values are [0.33, 0.76, 1.01, 0.54, 0.61, 0.16].
Step 2 ：Using the regression line, we can predict the distance values as [0.33076318, 0.75311012, 1.00318396, 0.55305104, 0.6197374, 0.1501543].
Step 3 ：The explained variation is the sum of the squared differences between the predicted distance values and the mean distance value. This is calculated to be approximately 0.457427.
Step 4 ：The unexplained variation is the sum of the squared differences between the actual distance values and the predicted distance values. This is calculated to be approximately 0.000457.
Step 5 ：For a given redshift value of 0.0126, we can predict the distance to be approximately 0.18488677440679632.
Step 6 ：We calculate the t-score for a 90% confidence level to be 2.015048372669157 and the standard error of estimate to be 0.010684052581447383.
Step 7 ：The margin of error is calculated to be 0.015040940214386246.
Step 8 ：Finally, we calculate the prediction interval by adding and subtracting the margin of error from the predicted distance. This gives us a prediction interval of approximately (0.1698458341924101, 0.19992771462118256).
Step 9 ：So, the explained variation is approximately \(\boxed{0.457427}\), the unexplained variation is approximately \(\boxed{0.000457}\), and the prediction interval is approximately \(\boxed{0.170}\) billion light-years \(
From Solvely APP
Solution
Step 1 ：First, we calculate the regression line using the given data. The redshift values are [0.0231, 0.0535, 0.0715, 0.0391, 0.0439, 0.0101] and the corresponding distance values are [0.33, 0.76, 1.01, 0.54, 0.61, 0.16].
Step 2 ：Using the regression line, we can predict the distance values as [0.33076318, 0.75311012, 1.00318396, 0.55305104, 0.6197374, 0.1501543].
Step 3 ：The explained variation is the sum of the squared differences between the predicted distance values and the mean distance value. This is calculated to be approximately 0.457427.
Step 4 ：The unexplained variation is the sum of the squared differences between the actual distance values and the predicted distance values. This is calculated to be approximately 0.000457.
Step 5 ：For a given redshift value of 0.0126, we can predict the distance to be approximately 0.18488677440679632.
Step 6 ：We calculate the t-score for a 90% confidence level to be 2.015048372669157 and the standard error of estimate to be 0.010684052581447383.
Step 7 ：The margin of error is calculated to be 0.015040940214386246.
Step 8 ：Finally, we calculate the prediction interval by adding and subtracting the margin of error from the predicted distance. This gives us a prediction interval of approximately (0.1698458341924101, 0.19992771462118256).
Step 9 ：So, the explained variation is approximately \(\boxed{0.457427}\), the unexplained variation is approximately \(\boxed{0.000457}\), and the prediction interval is approximately \(\boxed{0.170}\) billion light-years \(
