High School

Run a regression analysis on the following bivariate set of data with \(y\) as the response variable.

\[
\begin{array}{cc}
x & y \\
70 & 69.5 \\
51.9 & -21.7 \\
58.1 & 39.1 \\
67.4 & 74.9 \\
95 & 156.2 \\
70.7 & 97.6 \\
62.9 & 89 \\
50.4 & 16.8 \\
60.9 & 37.4 \\
49 & 29.1 \\
61.4 & 59.6 \\
60.3 & 35.1 \\
\end{array}
\]

1. Find the correlation coefficient and report it accurate to three decimal places.

\[
r =
\]

2. What proportion of the variation in \(y\) can be explained by the variation in the values of \(x\)? Report the answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%. You would enter 84.5 without the percent symbol.)

\[
r^2 = \%
\]

3. Based on the data, calculate the regression line (each value to three decimal places).

\[
y = \ x +
\]

4. Predict what value (on average) for the response variable will be obtained from a value of 49.2 as the explanatory variable. Use a significance level of \(\alpha = 0.05\) to assess the strength of the linear correlation. What is the predicted response value? (Report answer accurate to one decimal place.)

\[
y =
\]

Answer :

Since the p-value is less than the level of significance, the correlation is significant. Therefore, the linear correlation is strong.

x y 70 69.5 51.9 -21.7 58.1 39.1 67.4 74.9 95 156.2 70.7 97.6 62.9 89 50.4 16.8 60.9 37.4 49 29.1 61.4 59.6 60.3 35.1. Correlation coefficient (r) = 0.819 correct to three decimal places.

Coefficient of determination (r²) = 0.671 correct to three decimal places. Therefore, the proportion of the variation in y that can be explained by the variation in the values of x is 67.1%. Each value should be correct to three decimal places. Therefore, the regression line equation is y = 0.976x - 21.965. y = 0.976(49.2) - 21.965 = 25.534. Therefore, the predicted response value is 25.5. This value represents the average of the response variable (y) that is expected to be obtained from a value of 49.2 as the explanatory variable x. Use a significance level of α = 0.05 to evaluate the strength of the linear correlation.

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