In: Statistics and Probability
A horticulturist is studying the relationship between tomato plant height and fertilizer amount. Thirty tomato plants grown in similar conditions were subjected to various amounts of fertilizer over a four-month period, and then their heights were measured. The results are shown in the accompanying table. Fertilizer Amount (ounces) Tomato Plant Height (inches) 1.9 20.4 5.0 29.1 4.2 28.2 1.3 24.3 4.9 29.2 5.4 29.8 3.1 34.6 0.0 25.3 2.3 26.2 2.5 25.7 0.9 26.5 1.0 28.6 4.5 25.7 3.8 36.8 5.3 27.2 2.3 33.4 4.0 25.1 1.4 22.1 3.9 40.6 4.0 24.5 3.6 39.7 0.6 21.1 1.6 25.4 2.5 29.6 4.5 27.6 3.6 32.6 2.0 33.5 0.0 22.5 2.5 27.6 3.1 36.4 Picture Click here for the Excel Data File a. Estimate the linear regression model: Height = β0 + β1Fertilizer + ε. (Round your answers to 3 decimal places.) Heightˆ = 25.000 + 1.290 Fertilizer b-1. Estimate the quadratic regression model: Height = β0 + β1Fertilizer + β2Fertilizer2 + ε. (Round your answers to 3 decimal places.) Heightˆ = 21.100 + 5.460 Fertilizer + 0.759 Fertilizer2 b-2. Using the quadratic results, find the fertilizer amount at which the height reaches a minimum or maximum. (Round your intermediate values and final answer to 3 decimal places.) Fertilizer amount = 3.9 ounces c. Use the best-fitting model to predict, after a four-month period, the height of a tomato plant that received 3.0 ounces of fertilizer. (Round your intermediate values and final answer to 2 decimal places.) Heightˆ = 30.649 inches
(a) Height^=24.956+1.291*Fertilizer
following regression analysis information has been generated using ms-excel
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.386414872 | |||||
R Square | 0.149316453 | |||||
Adjusted R Square | 0.118934898 | |||||
Standard Error | 4.93026189 | |||||
Observations | 30 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 119.4641621 | 119.4642 | 4.914707 | 0.034921943 | |
Residual | 28 | 680.6095046 | 24.30748 | |||
Total | 29 | 800.0736667 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 24.95578271 | 1.891309847 | 13.19497 | 1.54E-13 | 21.08161016 | 28.82995525 |
X Variable 1 | 1.290857863 | 0.582276943 | 2.216914 | 0.034922 | 0.098117631 | 2.483598095 |
(b) Height^=21.0677+5.456*fertilizer-0.759*fertilizer2
following regression analysis information has been generated using ms-excel
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.517861295 | |||||
R Square | 0.268180321 | |||||
Adjusted R Square | 0.213971456 | |||||
Standard Error | 4.656773856 | |||||
Observations | 30 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 2 | 214.5640126 | 107.282 | 4.947167 | 0.014772539 | |
Residual | 27 | 585.5096541 | 21.68554 | |||
Total | 29 | 800.0736667 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 21.06783451 | 2.576457603 | 8.177055 | 8.8E-09 | 15.78138023 | 26.35428878 |
X Variable 1 | 5.455905881 | 2.063550349 | 2.643941 | 0.013482 | 1.221850351 | 9.689961412 |
X Variable 2 | -0.75923721 | 0.362554012 | -2.09414 | 0.045769 | -1.50313659 | -0.01533783 |
(b2) answer is fertilizer=3.594
the quadratice line is given as , Height^=21.0677+5.456*fertilizer-0.759*fertilizer2
using maximization and minimization problem, we differenitate Height with respect ot fertilizer and putting them 0
Height/fertilizer=5.456-2*0.759fertilizer
and putting this equation equal to zero and we get
fertilizer=5.456/(2*0.759)=3.594
since 2Height/fertilizer2 is negative , so at fertilizer=3.594, height will be maximum
(c) for fertilizer=3, the height=21.0677+5.456*3-0.759*3*3=30.60
here R-square for quadratic equation/model is more than linear equation/model , so we used quadratic model for the prediction.