In: Statistics and Probability
Oysters are categorized for retail as small, medium, or large based on their volume. The grading process is slow and expensive when done by hand. A computer reconstruction of oyster volume based on image analysis would be desirable to speed the process. Engineers designed two programs estimating oyster volume, one based on two‑dimensional ( 2D ) image processing and the other based on three‑dimensional ( 3D ) image processing. We want to know if either approach is a good predictor of actual oyster volume. The results of both programs are given in the table for a sample of 30 oysters. Actual volumes are expressed in cubic centimeters (cm3) , 2D reconstructions in thousands of pixels and 3D reconstruction in millions of volume pixels.
To access the complete data set, click the link for your preferred software format: Excel Minitab JMP SPSS TI R Mac-TXT PC-TXT CSV CrunchIt! Consider the program that estimates oyster volume using 3D digital income processing.
(a) Use the software of your choice to create a scatterplot of 3D volume reconstruction and actual volume, using 3D reconstruction for the explanatory variable. Find the equation of the least‑squares regression line. ?̂ = ?
(b) Select the statement that verifies the conditions for inference.
-The relationship is clearly non‑linear, the scatterplot shows some unusual patterns that would indicate Normally distributed residuals or residuals with a constant standard deviation, and the observations are not independent.
-The relationship is clearly linear, the scatterplot shows some unusual patterns that would result in non‑Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.
-The relationship is clearly linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.
-The relationship is non‑linear, the scatterplot shows no unusual pattern that would indicate not Normally distributed residuals or residuals without a constant standard deviation, and the observations are independent.
(c) What are the null and alternative hypotheses to test whether the linear relationship is statistically significant?
-?0:?=0 versus ??:?≠0
-?0:?=0 versus ??:?<0
-?0:?=0 versus ??:?>0
-None of the other choices are correct.
(d) What is the test statistic ? ? (Enter your answer rounded to two decimal places.) ?=
(e) What are the degrees of freedom? (Enter your answer rounded to the nearest whole number.) df=
(f) What is the ? ‑value? You may find Table C helpful.
-?<0.05
-0.05≤?≤0.1
-0.1<?<0.5
-?>0.5
(g) Which is the most appropriate conclusion of your test?
-We have strong evidence of a positive linear relationship between 3D reconstruction and actual oyster volumes.
-None of the other choices are correct.
-We have weak evidence of a positive linear relationship between 3D reconstruction and actual oyster volumes.
-We have no evidence of a positive linear relationship between 3D reconstruction and actual oyster volumes.
The 95% confidence interval for the population slope ? is lower bound to upper bound cm3 per million volume pixels. (Enter your answers rounded to three decimal places.) lower bound:
upper bound:
Data:
Actual | 3D | 2D |
13.04 | 5.136699 | 47.907 |
11.71 | 4.795151 | 41.458 |
17.42 | 6.453115 | 60.891 |
7.23 | 2.895239 | 29.949 |
10.03 | 3.672746 | 41.616 |
15.59 | 5.72888 | 48.07 |
9.94 | 3.987582 | 34.717 |
7.53 | 2.678423 | 27.23 |
12.73 | 5.481545 | 52.712 |
12.66 | 5.016762 | 41.5 |
10.53 | 3.942783 | 31.216 |
10.84 | 4.052638 | 41.852 |
13.12 | 5.334558 | 44.608 |
8.48 | 3.527926 | 35.343 |
14.24 | 5.679636 | 47.481 |
11.11 | 4.013992 | 40.976 |
15.35 | 5.565995 | 65.361 |
15.44 | 6.303198 | 50.91 |
5.67 | 1.928109 | 22.895 |
8.26 | 3.450164 | 34.804 |
10.95 | 4.707532 | 37.156 |
7.97 | 3.019077 | 29.07 |
7.34 | 2.76816 | 24.59 |
13.21 | 4.945743 | 48.082 |
7.83 | 3.138463 | 32.118 |
11.38 | 4.410797 | 45.112 |
11.22 | 4.558251 | 37.02 |
9.25 | 3.449867 | 39.333 |
13.75 | 5.609681 | 51.351 |
14.37 | 5.292105 | 53.281 |
a)
he above scatter plot show's that the actual value is positive correlated with 3D variable and it show any increasing pattern means that they are linearly related each other.
The correlation between between 3D variable and actual value is 0.9765
The equation of the least‑squares regression line.
?̂= ?(b)
?̂=0.41945+2.475*x
Regression Analysis | ||||||
r² | 0.954 | n | 30 | |||
r | 0.977 | k | 1 | |||
Std. Error | 0.649 | Dep. Var. | Actual | |||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 242.8360 | 1 | 242.8360 | 576.93 | 3.17E-20 | |
Residual | 11.7855 | 28 | 0.4209 | |||
Total | 254.6214 | 29 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=28) | p-value | 95% lower | 95% upper |
Intercept | 0.4196 | |||||
3D | 2.4752 | 0.1031 | 24.019 | 3.17E-20 | 2.2641 | 2.6863 |
c) The null and alternative hypotheses to test whether the linear relationship is statistically significant or not
?0:?=0 versus ??:?≠0
The value of F-stat =576.85 and the value of F-tab= 3.17118E-20
The F-stat value is greater than F-tab value means that Reject Ho
i.e. Conclusion: The linear relationship is statistically significant between 3D variable and Response actual value at 5% of level of significance.(?≠0)
The test statistic ? =0.897872771
The p-value is 3.17118E-20
It is appropriate conclusion given below for p-value with 5% l.o.s.
The above p-value for 3D variable is less than alpha value then Reject Ho (null hypothesis)
It is conclude that the variable is significance at 5% of level of significance.
The 95% confidence interval for the population slope ? is lower bound to upper bound cm3 per million
lower bound =2.264
upper bound =2.6863