Question 135274: I saw this question earlier but it wasn't answered and there was no dataset included so I'm posting again.
In the following regression, X = weekly pay, Y = income tax withheld, and n = 35 McDonald’s
employees. (a) Write the fitted regression equation. (b) State the degrees of freedom for a twotailed
test for zero slope, and use Appendix D to find the critical value at α = .05. (c) What is your
conclusion about the slope? (d) Interpret the 95 percent confidence limits for the slope. (e) Verify
that F = t2 for the slope. (f) In your own words, describe the fit of this regression.
R2 0.202
Std. Error 6.816
n 35
ANOVA table
Source---------SS---------df-------MS----------F-----p-value
Regression---387.6956-----1------387.6959----8.35---.0068
Residual-----1,533.0614---33-----46.4564
Total--------1,920.7573---34
Regression output----------------------------------------confidence interval
variables----coeff-----std.error---t(df = 33)----p-val----95%lower----95% upper
Intercept----30.7963---6.4078------4.806---------.0000----17.7595-----43.8331
Slope--------0.0343----0.0119------2.889---------.0068----0.0101------0.0584
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Since nobody else is jumping into reponding on this question,
I'll take a stab at it.
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In the following regression, X = weekly pay, Y = income tax withheld, and n = 35 McDonald’s employees.
(a) Write the fitted regression equation: Y = 0.0343X + 30.7963
--------------------------------
(b) State the degrees of freedom for a two tailed test for zero slope, and use Appendix D to find the critical value at alpha = .05.
df(slope)= 33
------------------------
(c) What is your conclusion about the slope?
Since p-value = 0.0068 reject Ho for alpha = 5%
--------------------------
(d) Interpret the 95 percent confidence limits for the slope.
We are 95% confidence the slope lies between 0.0101 and 0.0584
------------------------------------------------------------
(e) Verify that F = t2 for the slope.
F = 8.35 ; t^2 = 2.889^2 = 8.346
------------------------------------
(f) In your own words, describe the fit of this regression.
It's pretty poor since R = sqrt(0.202)= 0.45
=================================================
Cheers,
Stan H.
=======================
DATA below:
R2 0.202
Std. Error 6.816
n 35
ANOVA table
Source---------SS---------df-------MS----------F-----p-value
Regression---387.6956-----1------387.6959----8.35---.0068
Residual-----1,533.0614---33-----46.4564
Total--------1,920.7573---34
Regression output----------------------------------------confidence interval
variables----coeff-----std.error---t(df = 33)----p-val----95%lower----95% upper
Intercept----30.7963---6.4078------4.806---------.0000----17.7595-----43.8331
Slope--------0.0343----0.0119------2.889---------.0068----0.0101------0.0584
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