1. 1. Find and Interpret the Regression line 2. Interpret the R-squared
WARM- UP Quiz Review
3. A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq ft) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 0 to 100.
Dependent variable is: Lawn Quality
R-squared = 36.0%
s = with = 6 degrees of freedom
Variable Coefficient SE(Coeff) t-ratio P-value
Constant
Seed Density ? ?
1. Find and Interpret the Regression line
2. Interpret the R-squared
Find and interpret a 95% Confidence Interval for slope.
Is there a significant relationship between seed density and lawn quality?

2. Dependent variable is: Lawn Quality
R-squared = 36.0%
s = with = 6 degrees of freedom
Variable Coefficient SE(Coeff) t-ratio P-value
Constant
Seed Density ? ?
Find and Interpret the Regression line
Predicted Lawn Quality = (Seed Density)
For every additional lbs./500ft2of seeds, Predicted Lawn Quality improves 4.54% on average.
A Lawn with 0 seeds will be rated at 33.1%.
2. Interpret the R-squared
36% of the variation in Predicted Lawn quality is due to seed density.

3. Dependent variable is: Lawn Quality
R-squared = 36.0%
s = with = 6 degrees of freedom
Variable Coefficient SE(Coeff) t-ratio P-value
Constant
Seed Density ? ?
3. Find and interpret a 95% Confidence Interval for slope.
We can be 95% confident that for every increase in seed density the lawn quality is predicted to increase and
We can be 95% confident that the true slope of seed density vs. lawn quality is between and

4. Dependent variable is: Lawn Quality
R-squared = 36.0%
s = with = 6 degrees of freedom
Variable Coefficient SE(Coeff) t-ratio P-value
Constant
Seed Density ? ?
1.8372
0.1158
4. Is there a significant relationship between seed density and lawn quality?
Since >0.05, we fail to reject H0. There is NOT sufficient evidence of a significant relationship b/w S.D. and L.Q.

6. Find and interpret slope and y-int. for the Regression Line.
WARM – UP
Is the height (in inches) of a man related to his I.Q.? The regression analysis from a sample of 26 men is shown. (Assume the assumptions for inference were satisfied.)
Dependent variable is: IQ
R-squared = 47.3%
s =
Variable Coefficient SE(Coeff)
Constant
Height
Find and interpret slope and y-int. for the Regression Line.
For every additional inch of height a mans IQ will increase points. A man 0 inches high will have an IQ of 1.18
2. Is there an association b/t height and IQ in men?
Write the Hypothesis, t & p-values, and conclusion.

7. 2. Is there an association b/t height and IQ in men?
Dependent variable is: IQ
R-squared = 47.3%
s =
Variable Coefficient SE(Coeff)
Constant
Height
n = 26
t-ratio P-Value
1.6418
t-ratio P-Value
1.6418
t-ratio P-Value
1.6418
0.1137
2. Is there an association b/t height and IQ in men?
Write the Hypothesis, t & p-values, and conclusion.
Since the p-value of is above 0.05, we cannot reject H0. In conclusion there is NO Evidence of a significant relationship b/w IQ and height.

8. Regression Inference C.I.
To estimate the rate of change, slope, we use a Confidence Interval. The formula for a confidence interval for 1 is:
Find the t* for a 99% Confidence Interval with n = 20
t* = 2.878

9. WARM – UP What was the Rate of Improvement from the Quiz to the Test .
STUDENT
QUIZ
TEST 1 2 3 4 5 6 7 8 88 84
100 92 70 80 81 95
102 94 90 85 99 97
What was the Rate of Improvement from the Quiz to the Test .
Dependent Variable is: Test
R-squared = 35.3%
s = with 8 – 2 = 6 degrees of freedom
Variable Coefficient SE(Coeff) T-ratio P-Value
Intercept
Quiz
We can be 95% confident that for every additional point increase in quiz grade your test grade is predicted to be and higher.