Minitab was used to fit the model E[y] = 0 + 1X1 + 2X2 to n=20 data points and t
ID: 3177030 • Letter: M
Question
Minitab was used to fit the model E[y] = 0 + 1X1 + 2X2 to n=20 data points and the following is the output.
Predictor
Coef
SE Coef
T
P
Constant
506.346
45.17
11.21
0.000
X1
-941.9
275.08
***
0.003
X2
-429.060
379.83
-1.13
0.274
Predictor
Coef
SE Coef
T
P
Constant
506.346
45.17
11.21
0.000
X1
-941.9
275.08
***
0.003
X2
-429.060
379.83
-1.13
0.274
Minitab was used to fit the model E[y] = beta_0 + beta_1X_1 + beta_2X_2 to n=20 data points and the following is the output. What is the least squares prediction equation? Interpret the values of beta_1 and beta_2. Test for a significant linear relationship between y and X_1. Find the missing t-statistic in the coefficients table. Create a confidence interval for the estimate of beta_2. Based on this interval, is the relationship between y and X_2 significant? Interpret R^2 and R_a^2. Conduct the global utility test for the model. In a backwards elimination procedure, which variable would be removed first?Explanation / Answer
Part a:
Least square prediction equation:
y^ = 506.346 – 941.9x1 – 429.060x2
Part b:
Here we have to slopes. The slope b1 says that if there is one unit increment on the variable X1 keeping the variable X2 as constant then the variable y is going to be decreased by 941.9 units.
Here we have to slopes. The slope b2 says that if there is one unit increment on the variable X2 keeping the variable X1 as constant then the variable y is going to be decreased by429.06 units.
Part c:
t = -941.9/275.08 = -3.424
The p-value is .003. This is less than .05 (level of significance); the null hypothesis can be rejected. There is sufficient evidence to conclude that there is a linear relationship between y and X1.
Part d:
Here degrees of freedom is 18 and we get the critical t for 5% level of significance from t-table as 2.101.
Confidence Interval:
-429.06 (-/+) ( 2.101 * 379.83)
= -429.06 (-/+)798.0228
= -1227.08 and 368.96
(-1227.08 and 368.96)