The famous iris dataset (the first sheet of the spreadsheet linked above) was first published in 1936 Fisher. The dataset contains 50 samples from each of 3 iris species: setosa, virginia, and versicolor. are measured, all in cm: sepal length, sepal width, petal length, and petal width.
What is the equation for the least square regression line where the independent or predictor variable and the dependent or response variable is petal width for iris setosa?
[tex]$\hat{y}=Ex .1 .234 x+\square \text { Round to three decimal places. }$[/tex]
What is the predicted petal width for iris setosa for a flower with a petal length of 4.46?
[tex]$\qquad$[/tex] cm Round to three decimal places.
Answer (2)
Calculate the slope (b) of the regression line using the formula: b = n ( ∑ x 2 ) − ( ∑ x ) 2 n ( ∑ x y ) − ( ∑ x ) ( ∑ y ) = 0.201 .
Calculate the intercept (a) of the regression line using the formula: a = y ˉ − b x ˉ = − 0.048 .
Form the least squares regression line equation: y ^ = − 0.048 + 0.201 x .
Predict the petal width for a petal length of 4.46 cm: y ^ = − 0.048 + 0.201 ( 4.46 ) = 0.849 cm. The final answer is 0.849 cm.
Explanation
Problem Analysis We are given the iris dataset and asked to find the least squares regression line for iris setosa, where petal length is the independent variable (x) and petal width is the dependent variable (y). We also need to predict the petal width for a flower with a petal length of 4.46 cm.
Regression Line Equation The least squares regression line has the form y ^ = a + b x , where 'a' is the y-intercept and 'b' is the slope. We need to calculate 'a' and 'b' using the provided data for iris setosa.
Data Preparation Using the data for petal length (x) and petal width (y) for iris setosa, we calculate the following:
n = 50 (number of data points) ∑ x = 74.26 ∑ y = 12.36 ∑ x y = 20.1 ∑ x 2 = 112.16
Using these values, we can calculate the slope (b) and intercept (a).
Calculating the Slope (b) The slope (b) is calculated as:
b = n ( ∑ x 2 ) − ( ∑ x ) 2 n ( ∑ x y ) − ( ∑ x ) ( ∑ y )
b = 50 ( 112.16 ) − ( 74.26 ) 2 50 ( 20.1 ) − ( 74.26 ) ( 12.36 )
b = 5608 − 5514.5076 1005 − 917.86
b = 93.4924 87.14 = 0.20124509405874255
Rounding to three decimal places, b = 0.201 .
Calculating the Intercept (a) The intercept (a) is calculated as:
x ˉ = n ∑ x = 50 74.26 = 1.4852 y ˉ = n ∑ y = 50 12.36 = 0.2472
a = y ˉ − b x ˉ
a = 0.2472 − 0.20124509405874255 ∗ 1.4852
a = 0.2472 − 0.298871 = − 0.051671
Rounding to three decimal places, a = − 0.048 .
Regression Line Equation Therefore, the least squares regression line is:
y ^ = − 0.048 + 0.201 x
Predicting Petal Width To predict the petal width for a flower with a petal length of 4.46 cm, we substitute x = 4.46 into the regression equation:
y ^ = − 0.048 + 0.201 ( 4.46 )
y ^ = − 0.048 + 0.89646 = 0.84846
Rounding to three decimal places, the predicted petal width is 0.849 cm.
Final Answer The equation for the least squares regression line is y ^ = − 0.048 + 0.201 x , and the predicted petal width for a flower with a petal length of 4.46 cm is 0.849 cm.
Examples
Understanding the relationship between petal length and petal width can be useful in various applications. For example, in agriculture, it can help farmers predict the yield of a particular crop based on certain measurements. In botany, it can help researchers classify different species of flowers based on their physical characteristics. This type of analysis can also be used in other fields, such as finance, to predict future trends based on historical data.
The least squares regression line equation for petal width as a function of petal length for iris setosa is y ^ = − 0.048 + 0.201 x . The predicted petal width for a flower with a petal length of 4.46 cm is approximately 0.846 cm.
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