Solution Found!
Bone LengthUse the results of Section to answer the
Chapter 5, Problem 10AYU(choose chapter or problem)
Problem 10AYU
Bone LengthUse the results of Problem in Section to answer the following questions:
(a)Predict the mean length of the right tibia of all rats whose right humerus is 25.83 mm.
(b) Construct a 95% confidence interval for the mean length found in part (a).
(c) Predict the length of the right tibia of a randomly selected rat whose right humerus is 25.83 mm.
(d) Construct a 95% prediction interval for the length found in part (c).
(e) Explain why the predicted lengths found in parts (a) and (c) are the same, yet the intervals constructed in parts (b) and (d) are different.
Problem
Bone Length Research performed at NASA and led by Dr. Emily R. Morey-Hollon measured the lengths of the right humerus and right tibia in 11 rats that were sent into space on Spacelab Life Sciences 2. The following data were collected:
Use the results from Problem in Section to answer the following questions:
(a) Treating the length of the right humerus as the explanatory variable,x, determine the estimates of β0 and β1.
(b) Compute the standard error of the estimate.
(c) Determine whether the residuals are normally distributed.
(d) If the residuals are normally distributed, determine Sb1.
(e) If the residuals are normally distributed, test whether a linear relation exists between the length of the right humerus, x, and the length of the right tibia, y, at the α = 0.01 level of significance.
(f) If the residuals are normally distributed, construct a 99% confidence interval for the slope of the true least-squares regression line.
(g) What is the mean length of the right tibia on a rat whose right humerus is 25.93 mm?
problem
Use the results from Problems
BoneLength Research performed at NASA and led by Emily R. Morey-Holton measured the lengths of the right humerus and right tibia in 11 rats who were sent to space on Spacelab Life Sciences 2. The following data were collected.
(a)Find the least-squares regression line treating the length of the right humerus,x,as the explanatory variable and the length of the right tibia,y,as the response variable.
(b)Interpret the slope and y-intercept, if appropriate.
(c)Determine the residual if the length of the right humerus is 26.11 mm and the actual length of the right tibia is 37.96 mm. Is the length of this tibia above or below average?
(d) Draw the least-squares regression line on the scatter diagram.
(e)Suppose one of the rats sent to space experienced a broken right tibia due to a severe landing. The length of the right humerus is determined to be 25.31 mm. Use the least-squares regression line to estimate the length of the right tibia.
problem
Bone LengthResearch performed at NASA and led by Emily R. Morey-Holton measured the lengths of the right humerus and right tibia in 11 rats that were sent to space on SpaceIhb Life Sciences 2. The following data were collected.
(a) Draw a scatter diagram treating the length of the right humerus as the explanatory variable and the length of the right tibia as the response variable.
(b) Compute the linear correlation coefficient between the length of the right humerus and the length of the right tibia.
(c) Does a linear relation exist between the length of the right humerus and the length of the right tibia?
(d) Convert the data to inches (1 mm = 0.03937 inch), and recompute the linear correlation coefficient. What effect did the conversion from millimeters to inches have on the linear correlation coefficient?
Questions & Answers
QUESTION:
Problem 10AYU
Bone LengthUse the results of Problem in Section to answer the following questions:
(a)Predict the mean length of the right tibia of all rats whose right humerus is 25.83 mm.
(b) Construct a 95% confidence interval for the mean length found in part (a).
(c) Predict the length of the right tibia of a randomly selected rat whose right humerus is 25.83 mm.
(d) Construct a 95% prediction interval for the length found in part (c).
(e) Explain why the predicted lengths found in parts (a) and (c) are the same, yet the intervals constructed in parts (b) and (d) are different.
Problem
Bone Length Research performed at NASA and led by Dr. Emily R. Morey-Hollon measured the lengths of the right humerus and right tibia in 11 rats that were sent into space on Spacelab Life Sciences 2. The following data were collected:
Use the results from Problem in Section to answer the following questions:
(a) Treating the length of the right humerus as the explanatory variable,x, determine the estimates of β0 and β1.
(b) Compute the standard error of the estimate.
(c) Determine whether the residuals are normally distributed.
(d) If the residuals are normally distributed, determine Sb1.
(e) If the residuals are normally distributed, test whether a linear relation exists between the length of the right humerus, x, and the length of the right tibia, y, at the α = 0.01 level of significance.
(f) If the residuals are normally distributed, construct a 99% confidence interval for the slope of the true least-squares regression line.
(g) What is the mean length of the right tibia on a rat whose right humerus is 25.93 mm?
problem
Use the results from Problems
BoneLength Research performed at NASA and led by Emily R. Morey-Holton measured the lengths of the right humerus and right tibia in 11 rats who were sent to space on Spacelab Life Sciences 2. The following data were collected.
(a)Find the least-squares regression line treating the length of the right humerus,x,as the explanatory variable and the length of the right tibia,y,as the response variable.
(b)Interpret the slope and y-intercept, if appropriate.
(c)Determine the residual if the length of the right humerus is 26.11 mm and the actual length of the right tibia is 37.96 mm. Is the length of this tibia above or below average?
(d) Draw the least-squares regression line on the scatter diagram.
(e)Suppose one of the rats sent to space experienced a broken right tibia due to a severe landing. The length of the right humerus is determined to be 25.31 mm. Use the least-squares regression line to estimate the length of the right tibia.
problem
Bone LengthResearch performed at NASA and led by Emily R. Morey-Holton measured the lengths of the right humerus and right tibia in 11 rats that were sent to space on SpaceIhb Life Sciences 2. The following data were collected.
(a) Draw a scatter diagram treating the length of the right humerus as the explanatory variable and the length of the right tibia as the response variable.
(b) Compute the linear correlation coefficient between the length of the right humerus and the length of the right tibia.
(c) Does a linear relation exist between the length of the right humerus and the length of the right tibia?
(d) Convert the data to inches (1 mm = 0.03937 inch), and recompute the linear correlation coefficient. What effect did the conversion from millimeters to inches have on the linear correlation coefficient?
ANSWER:
Answer :
Step 1 of 4 :
Given, Research performed at NASA and led by Dr. Emily R. Morey-Hollon measured the lengths of the right humerus and right tibia in 11 rats that were sent into space on Spacelab Life Sciences 2
Right humerus |
Right tibia |
24.8 |
36.05 |
24.59 |
35.57 |
24.59 |
35.57 |
24.29 |
34.58 |
23.81 |
34.2 |
24.87 |
34.73 |
25.9 |
37.38 |
26.11 |
37.96 |
26.63 |
37.46 |
26.31 |
37.75 |
26.84 |
38.5 |
- From Excel
>enter the data in excel sheet
> select Data - Data analysis - regression
> select the y and x range
> select the output range then ok.
The least square regression line be
y = 1.3901*x + 1.113953
We have to predict mean length of the right tibia of all rats whose right humerus is 25.83 mm
Let x = 25.83
Therefore, y = 1.3901*25.83 + 1.113953
y = 37.02mm
b) we have to Construct a 95% confidence interval for the mean length found in part (a)
Hence, 95% confidence interval is
36.34
36.34 2.228 (0.4585)
(35.31, 37.361)
c)
We have to predict mean length of the right tibia of a randomly selected rat whose right humerus is 25.83 mm
Let x = 25.83
Therefore, y = 1.3901*25.83 + 1.113953
y = 37.02mm
d) we have to Construct a 95% prediction interval for the length found in part (c).
Hence, 95% confidence interval is
37.02
37.02 2.228 (0.4585)
(36, 38.041)
d)
are the same, the intervals are different because the distribution of the means, part (a), has less variability than the distribution of the individuals, part (c).