Question 140714: 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 two tailed 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 387.6959 1 387.6959 8.35 .0068
Residual 1,533.0614 33 46.4564
Total 1,920.7573 34
Regression output confidence interval
variables coefficients std. error t (df = 33) p-value 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!
Probability-and-statistics/140714 (2008-05-08 13:17:51):
In the following regression, X = weekly pay, Y = income tax withheld, and n = 35 McDonald’s employees.
(a) Write the fitted regression equation.
Y = 30.7963 + 0.0343X
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(b) State the degrees of freedom for a two tailed test for zero slope, and use Appendix D to find the critical value at α = .05.
df=33; t=2.035
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(c) What is your conclusion about the slope?
Since p-value is 0.0068 < 5%, reject Ho that claimed the slope was zero.
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(d) Interpret the 95 percent confidence limits for the slope.
With 95% confidence we can say the slope is between the two limits:0.0101.... and 0.0584
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(e) Verify that F = t2 for the slope.
2.889^2 = 8.35
(f) In your own words, describe the fit of this regression.
Since p-value on the regression is < 5% reject Ho which claimed that
X and Y were independent. The correlation is very low as seen in
R2 =0.202. The standard error in predicting Y for any X is 6.816
which is high considering the size of the slope and intercept.
Cheers,
Stan H.
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R2 0.202
Std. Error 6.816
n 35
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ANOVA table
Source........SS....df MS...... F.. p-value
Regression 387.6959 1 387.6959 8.35 .0068
Residual 1,533.0614 33 46.4564
Total 1,920.7573 34
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Regression output confidence interval
variables coefficients std. error t (df = 33) p-value 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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