Run a regression analysis on the following bivariate set of data with y as the response variable.
x | y |
---|---|
48 | 41.8 |
39.2 | 67.4 |
34.7 | 68.4 |
42.9 | 50.2 |
49 | 50.6 |
45.6 | 57 |
58.7 | 29.7 |
40.5 | 68.4 |
47.4 | 34.7 |
45.7 | 50.9 |
38.9 | 47.7 |
40.9 | 53.6 |
Find the correlation coefficient and report it accurate to three
decimal places.
r =
What proportion of the variation in y can be explained by
the variation in the values of x? Report 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.)
r² = %
Based on the data, calculate the regression line (each value to
three decimal places)
y = x +
13.Predict what value (on average) for the response variable will
be obtained from a value of 40.5 as the explanatory variable. Use a
significance level of α=0.05α=0.05 to assess the strength of the
linear correlation.
What is the predicted response value? (Report answer accurate to
one decimal place.)
y =
SUMMARY OUTPUT | |||||||||
Regression Statistics | |||||||||
Multiple R | 0.80159 | ||||||||
R Square | 0.642546 | ||||||||
Adjusted R Square | 0.606801 | ||||||||
Standard Error | 7.857172 | ||||||||
Observations | 12 | ||||||||
ANOVA | |||||||||
df | SS | MS | F | Significance F | |||||
Regression | 1 | 1109.729 | 1109.729 | 17.97564 | 0.001717 | ||||
Residual | 10 | 617.3515 | 61.73515 | ||||||
Total | 11 | 1727.08 | |||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | ||
Intercept | 122.7141 | 16.90241 | 7.260157 | 2.73E-05 | 85.05322 | 160.3751 | 85.05322 | 160.3751 | |
x | -1.60333 | 0.378164 | -4.23977 | 0.001717 | -2.44593 | -0.76073 | -2.44593 | -0.76073 |
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