Question

Call: lm(formula = launch_speed ~ launch_angle, data = muncy) Residuals:     Min      1Q Median      3Q     Max...

Call:

lm(formula = launch_speed ~ launch_angle, data = muncy)

Residuals:

    Min      1Q Median      3Q     Max

-64.802 -9.009   2.401 10.821 20.709

Coefficients:

             Estimate Std. Error t value Pr(>|t|)   

(Intercept) 86.95164    0.78064 111.385 < 2e-16 ***

launch_angle 0.20804    0.02865   7.261 1.77e-12 ***

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 13.74 on 438 degrees of freedom

Multiple R-squared: 0.1074, Adjusted R-squared: 0.1054

F-statistic: 52.72 on 1 and 438 DF, p-value: 1.769e-12

a) Use R the R output to write the equation of the regression line.

b) Use part a to predict the hit distance of a ball with a launch speed of 106 mph, hit at a launch angle of 30+ degrees. Use the standard error to explain why such a ball traveling 350 feet would not be an unreasonable outcome.

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Answer #1

Solutiona:

From output

slope=0.20804

y intercept=86.95164

the equation of the regression line is

launch_speed=86.95164+0.20804*launch_angle

Solutionb:

launch_speed=86.95164+0.20804*launch_angle

Given launch angle =30

launch_speed=86.95164+0.20804*30

=93.19284

predicted launch speed=93.19284

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