a.
Test whether the data suggests a linear relationship between specific gravity and at least one of the predictors at 1% level of significance.
a.
Answer to Problem 56E
There is sufficient evidence to conclude that the there is a use of linear relationship between specific gravity and at least one of the five predictors number of fibers in springwood, number of fibers in summerwood, percentage of springwood, light absorption in springwood and light absorption in summerwood at 1% level of significance.
Explanation of Solution
Given info:
A sample of 20 mature woods were taken and the number of fibers in springwood, number of fibers in summerwood, percentage of springwood, light absorption in springwood and light absorption in summerwood were noted .
The coefficient of determination
Calculation:
The test hypotheses are given below:
Null hypothesis:
That is, there is no use of linear relationship between specific gravity and the five predictors.
Alternative hypothesis:
That is, there is a use of linear relationship between specific gravity and at least one of the five predictors.
Test statistic:
Substitute
P-value:
Software procedure:
- Click on Graph, select View Probability and click OK.
- Select F, enter 5 in numerator df and 14 in denominator df.
- Under Shaded Area Tab select X value under Define Shaded Area By and select Right tail.
- Choose X value as 9.33.
- Click OK.
Output obtained from MINITAB is given below:
Conclusion:
The P-value is 0.000 and the level of significance is 0.01.
The P-value is lesser than the level of significance.
That is
Thus, the null hypothesis is rejected.
Hence, there is sufficient evidence to conclude that there is ause of linear relationship between specific gravity and at least one of the five predictors at 1% level of significance.
b.
Calculate the adjusted
b.
Answer to Problem 56E
The adjusted
The adjusted
Explanation of Solution
Given info:
The
Calculation:
Adjusted
Adjusted
Substitute n as 20,k as 5,
Thus, the adjusted
Adjusted
Substitute n as 20, k as 4,
Thus, the adjusted
c.
Identify whether the data suggests that variables
Test the hypothesis to see whether the variables
c.
Answer to Problem 56E
Yes, the data suggests that variables
There issufficient evidence to conclude the variables
Explanation of Solution
Given info:
The
Calculation:
After dropping the three variables
The test hypotheses are given below:
Null hypothesis:
That is, there is no use of linear relationship betweenspecific gravity and at least one of the predictors, percentage of springwood and light absorption in summerwood.
Alternative hypothesis:
That is, there is use of linear relationship between specific gravity and at least one of the predictors, percentage of springwood and light absorption in summerwood.
From the
Similarly, the sum of squares due to error for the reduced model
Test statistic:
Where,
n represents the total number of observations.
k represents the number of predictors on the full model.
l represents the number of predictors on the reduced model.
Substitute 0.004542for
Critical value:
Software procedure:
- Click on Graph, select View Probability and click OK.
- Select F, enter 3 in numerator df and 14 in denominator df.
- Under Shaded Area Tab select Probability under Define Shaded Area By and select Right tail.
- Choose X Value as 2.32.
- Click OK.
Output obtained from MINITAB is given below:
Conclusion:
The P-value is 0.1197 and the level of significance is 0.05.
The P-value is lesser than the level of significance.
That is,
Thus, the null hypothesis is not rejected.
Hence, there is no sufficient evidence to conclude that there is a use of linear relationship betweenspecific gravity and at least one of the predictor percentage of springwood and light absorption in summerwood at 5% level of significance.
Thus, the variables
d.
Predict the value of specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90.
d.
Answer to Problem 56E
The estimated value for specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90 is 0.5386.
Explanation of Solution
Given info:
The mean and standard deviation for the variable
The estimated regression equation after standardization is
Calculation:
The standardized values
Where,
The standardized value when mean and standard deviation for the variable
Thus, the value of
The standardized value when the mean and standard deviation for the variable
Thus, the value of
The estimated value for specific gravity is,
Thus, the estimated value for specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90 is 0.5386.
e.
Find the 95% confidence interval for the estimated coefficient of
e.
Answer to Problem 56E
The 95% confidence interval for the estimated coefficient of
Explanation of Solution
Calculation:
95% confidence interval:
The confidence interval is calculated using the formula:
Where,
n is the total number of observations.
k is the total number of predictors in the model.
Critical value:
Software procedure:
Step-by-step procedure to find the critical value is given below:
- Click on Graph, select View Probability and click OK.
- Select t, enter 17 as Degrees of freedom, in Shaded Area Tab select Probability under Define Shaded Area By and choose Both tails.
- Enter Probability value as 0.05.
- Click OK.
Output obtained from MINITAB is given below:
The 95% confidence interval is given below:
Thus, the 95% confidence interval for the estimated coefficient of
f.
Find the estimated coefficient and estimated standard deviation of
f.
Answer to Problem 56E
The estimated coefficient of
The estimated standard deviation of
Explanation of Solution
Given info:
Use the information given in part (d) and (e).
Calculation:
The estimated regression equation for standardized model is,
The estimated coefficient of
Thus, the estimated coefficient of
The estimate for
The estimated standard deviation for
Thus, the estimated standard deviation of
g.
Find the 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9.
g.
Answer to Problem 56E
The 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is(0.489, 0.575).
Explanation of Solution
Given info:
The estimated standard deviation for the model with two predictors is 0.02001. The estimated standard deviation for the predicated value when the coefficients
Calculation:
The predicted value for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is calculated as follows:
Thus, the predicted value for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is 0.532.
95% prediction interval:
The confidence interval is calculated using the formula:
Where,
n is the total number of observations.
k is the total number of predictors in the model.
s is the overall standard deviation obtained after fitting the model.
Critical value:
Software procedure:
Step-by-step procedure to find the critical value is given below:
- Click on Graph, select View Probability and click OK.
- Select t, enter 17 as Degrees of freedom, in Shaded Area Tab select Probability under Define Shaded Area By and choose Both tails.
- Enter Probability value as 0.05.
- Click OK.
Output obtained from MINITAB is given below:
The 95% prediction interval is given below:
Thus, the 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is (0.489,0.575).
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Chapter 13 Solutions
Probability and Statistics for Engineering and the Sciences
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