In: Statistics and Probability
The data in stat4_prob5 present the performance of a chemical process as a function of sever controllable process variables.
α = 0.05.
Here is the data:
y=c(36.98, 13.74, 10.08, 8.53, 36.42, 26.59, 19.07, 5.96, 15.52, 56.61, 26.72, 20.80, 6.99, 45.93, 43.09, 15.79, 21.60, 35.19, 26.14, 8.60, 11.63, 9.59, 4.42, 38.89, 11.19, 75.62, 36.03)
x1=c(2227.25, 434.90, 481.19, 247.14, 1645.89, 907.59, 608.05, 380.55, 213.40, 2043.36, 761.48, 566.40, 237.08, 1961.49, 1023.89, 411.30, 2244.77, 978.64, 687.62, 468.28, 460.62, 290.42, 233.95, 2088.12, 994.63, 2196.17, 1080.11)
x2=c(2.06, 1.33, 0.97, 0.62, 0.22, 0.76, 1.71, 3.93, 1.97, 5.08, 0.60, 0.90, 0.63, 2.04, 1.57, 2.38, 0.32, 0.44, 8.82, 0.02, 1.72, 1.88, 1.43, 1.35, 1.61, 4.78, 5.88)
excel regression output is
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.836448011 | |||||||
R Square | 0.699645274 | |||||||
Adjusted R Square | 0.674615714 | |||||||
Standard Error | 9.92435002 | |||||||
Observations | 27 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 2 | 5506.286737 | 2753.143368 | 27.9527591 | 5.39031E-07 | |||
Residual | 24 | 2363.82536 | 98.49272332 | |||||
Total | 26 | 7870.112096 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 2.526630302 | 3.610029262 | 0.699891917 | 0.4907239 | -4.924103899 | 9.977364504 | -4.924103899 | 9.977364504 |
X1 | 0.018521965 | 0.002747236 | 6.74203717 | 5.66264E-07 | 0.012851949 | 0.024191981 | 0.012851949 | 0.024191981 |
X2 | 2.185719512 | 0.972694428 | 2.247077243 | 0.034100063 | 0.178176882 | 4.193262142 | 0.178176882 | 4.193262142 |
a)
fitted regression line is
Y hat = 2.5266 + 0.0185 *X1+ 2.1857 *X2
b)
point estimatefor the variance term σ2.=Se^2 = 9.92435002^2 = 98.493
c)
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 2 | 5506.286737 | 2753.143368 | 27.9527591 | 5.39031E-07 |
Residual | 24 | 2363.82536 | 98.49272332 | ||
Total | 26 | 7870.112096 |
d)
for X1
Ho:ß1=0
Ha:ß1╪0
t-value=6.74203717
pvalue=0.000<alpha=0.05,reject Ho
so, slope test is significant
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for X2
Ho:ß2=0
Ha:ß2╪0
t-value=2.2471
p-value=0.0341<alpha=0.05,reject Ho
so, slope test is significant