Certainly! Let's break down the solution step-by-step.
### Part c: Determine the y-intercept [tex]\( a \)[/tex]
We are given the following information:
- Mean height ([tex]\( \bar{x} \)[/tex]): 63.86
- Mean armspan ([tex]\( \bar{y} \)[/tex]): 159.77
- Slope ([tex]\( b \)[/tex]): 1.90
To find the y-intercept ([tex]\( a \)[/tex]), we use the formula for the y-intercept in a linear regression line:
[tex]\[ a = \bar{y} - b \cdot \bar{x}\][/tex]
Substituting the given values:
[tex]\[ a = 159.77 - 1.90 \cdot 63.86 \][/tex]
Performing the calculations, we get:
[tex]\[ a = 159.77 - 121.334 \][/tex]
[tex]\[ a = 38.436 \][/tex]
Rounding to two decimal places, we obtain:
[tex]\[ a = 38.44 \][/tex]
So, the verified y-intercept is:
[tex]\[ a = 38.44 \][/tex]
### Part d: Predict the armspan for someone 61 inches tall
Now that we have our linear regression equation [tex]\( y = a + bx \)[/tex], where:
- [tex]\( a = 38.44 \)[/tex] (y-intercept)
- [tex]\( b = 1.90 \)[/tex] (slope)
- [tex]\( x \)[/tex] is the height for prediction (61 inches)
We substitute [tex]\( x = 61 \)[/tex] into the regression equation to predict the armspan:
[tex]\[ y = 38.44 + 1.90 \cdot 61 \][/tex]
Performing the calculations, we get:
[tex]\[ y = 38.44 + 115.9 \][/tex]
[tex]\[ y = 154.34 \][/tex]
Rounding to one decimal place, we obtain:
[tex]\[ y = 154.3 \][/tex]
Therefore, the predicted armspan for someone 61 inches tall is:
[tex]\[ \text{Predicted Armspan} = 154.3 \, \text{cm} \][/tex]
