3.2. A Trent Honour’s Thesis s
3.2. A Trent Honour’s Thesis student is doing a projectexamining the health of a native brook trout population in ColdCreek. He has collected length, weight and girth measurements onseveral hundred fish but only determined the fat content in 40 ofthe fish. He wants to use the girth and fat content data from the40 fish to be able to predict the fat content of the other fish. Hehas asked you to help him construct this predictive model.
| Girth (mm) | Fat Content (%) |
| 90 | 7.2 |
| 90 | 8.5 |
| 87 | 7.8 |
| 85 | 7.5 |
| 84 | 8.9 |
| 81 | 8.7 |
| 76 | 7.6 |
| 75 | 6.2 |
| 75 | 8.2 |
| 70 | 6.9 |
| 70 | 7.7 |
| 69 | 7.6 |
| 67 | 8.6 |
| 62 | 5.6 |
| 61 | 5.3 |
| 60 | 6.7 |
| 59 | 7.1 |
| 56 | 8.0 |
| 53 | 6.1 |
| 52 | 4.4 |
| 51 | 5.2 |
| 50 | 6.5 |
| 50 | 7.1 |
| 48 | 3.8 |
| 46 | 5.5 |
| 43 | 7.8 |
| 41 | 6.0 |
| 37 | 6.8 |
| 36 | 5.1 |
| 36 | 3.9 |
| 33 | 2.6 |
| 31 | 5.4 |
| 29 | 4.1 |
| 29 | 4.8 |
| 27 | 2.2 |
| 22 | 4.7 |
| 22 | 6.4 |
| 21 | 2.7 |
| 20 | 2.9 |
| 20 | 3.0 |
A). Create a well labelled xy scatter plot on the spreadsheetpage with the data; make note of the general pattern of the plotteddata and whether there are any suspect data points.
B). Use the advanced regression analysis tool to conduct acomplete simple linear regression analysis for the data. You willnot require the options for “residuals” for this analysis. Rememberto complete the five steps of hypothesis testing in the spacesprovided on the worksheet.
C). Use the linear regression equation determined in part “B” tocalculate a set of “predicted y values” for each observed xvalue.
D). Add this set of predicted y values (from part C) to thescatter plot created previously – be sure to show this new set ofvalues as a “line” rather than as a set of “markers”.
**please be sure to show all work and each of the steps for thehypothesis test so that I can learn and understand**
Answer:
a)
….
b)
| Regression Statistics | ||||||
| Multiple R | 0.7782 | |||||
| R Square | 0.6056 | |||||
| Adjusted R Square | 0.5953 | |||||
| Standard Error | 1.1936 | |||||
| Observations | 40 | |||||
| ANOVA | ||||||
| df | SS | MS | F | Significance F | ||
| Regression | 1 | 83.14 | 83.14 | 58.36 | 0.0000 | |
| Residual | 38 | 54.14 | 1.42 | |||
| Total | 39 | 137.28 | ||||
| Coefficients | Standard Error | t Stat | P-value | lower 95% | upper 95% | |
| Intercept | 2.4787 | 0.501 | 4.943 | 0.0000 | 1.4636 | 3.49 |
| X | 0.067 | 0.009 | 7.639 | 0.0000 | 0.0494 | 0.0849 |
Ho: β1= 0H1: β1╪ 0n= 40 alpha = 0.05 estimated std error of slope =Se(ß1) = Se/√Sxx = 1.1936/√18439.1= 0.0088t stat = estimated slope/std error =ß1 /Se(ß1) = (0.0671-0)/0.0088= 7.64Degree of freedom ,df = n-2= 38 p-value = 0.0000 decison : p-value<α , reject Ho Conclusion: Reject Ho and conclude that slope issignificantly different from zero
…………..
c
| x | y | Ŷ | ||||
| 90 | 7.2 | 8.52 | ||||
| 90 | 8.5 | 8.52 | ||||
| 87 | 7.8 | 8.32 | ||||
| 85 | 7.5 | 8.19 | ||||
| 84 | 8.9 | 8.12 | ||||
| 81 | 8.7 | 7.92 | ||||
| 76 | 7.6 | 7.58 | ||||
| 75 | 6.2 | 7.51 | ||||
| 75 | 8.2 | 7.51 | ||||
| 70 | 6.9 | 7.18 | ||||
| 70 | 7.7 | 7.1791 | ||||
| 69 | 7.6 | 7.1120 | ||||
| 67 | 8.6 | 6.977657 | ||||
| 62 | 5.6 | 6.641912 | ||||
| 61 | 5.3 | 6.574763 | ||||
| 60 | 6.7 | 6.507615 | ||||
| 59 | 7.1 | 6.440466 | ||||
| 56 | 8 | 6.239019 | ||||
| 53 | 6.1 | 6.037572 | ||||
| 52 | 4.4 | 5.970423 | ||||
| 51 | 5.2 | 5.903275 | ||||
| 50 | 6.5 | 5.836126 | ||||
| 50 | 7.1 | 5.836126 | ||||
| 48 | 3.8 | 5.701828 | ||||
| 46 | 5.5 | 5.56753 | ||||
| 43 | 7.8 | 5.366083 | ||||
| 41 | 6 | 5.231786 | ||||
| 37 | 6.8 | 4.96319 | ||||
| 36 | 5.1 | 4.896041 | ||||
| 36 | 3.9 | 4.896041 | ||||
| 33 | 2.6 | 4.694595 | ||||
| 31 | 5.4 | 4.560297 | ||||
| 29 | 4.1 | 4.425999 | ||||
| 29 | 4.8 | 4.425999 | ||||
| 27 | 2.2 | 4.291701 | ||||
| 22 | 4.7 | 3.955957 | ||||
| 22 | 6.4 | 3.955957 | ||||
| 21 | 2.7 | 3.888808 | ||||
| 20 | 2.9 | 3.821659 | ||||
| 20 | 3 | 3.821659 |
………………
Please let me know in case of any doubt.
Thanks in advance!
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