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 For every additional inch of height a mans IQ will increase points. A man 0 inches high will have an IQ of Is there an association b/t height and IQ in men? Write the Hypothesis, t & p-values, and conclusion.

Dependent variable is: IQ R-squared = 47.3% s = Variable Coefficient SE(Coeff) Constant Height Is there an association b/t height and IQ in men? Write the Hypothesis, t & p-values, and conclusion. t-ratio P-Value Since the p-value of.1137 is above 0.05, we cannot reject H 0. In conclusion there is NO Evidence of a significant relationship b/w IQ and height. n = 26 t-ratio P-Value t-ratio P-Value

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

WARM – UP STUDENT QUIZ TEST 1.What was the Rate of Improvement from the Quiz to the Test. We can be 95% confident that for every additional point increase in quiz grade your test grade is predicted to be and higher. 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

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 3.Find and interpret a 95% Confidence Interval for slope. 4.Is there a significant relationship between seed density and lawn quality?

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.Predicted Lawn Quality = (Seed Density) For every additional lbs./500ft 2 of seeds, Lawn Quality improves 4.54%. A Lawn with 0 seeds will be rated at 33.1% % of the variation in Lawn quality is due to seed density. 3. We can be 95% confident that for every increase in seed density the lawn quality is predicted to increase and

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 ? ? 4. With a p-value of.1158 we cannot reject H0 so we conclude that there is NOT a significant relationship b/w S.D. and L.Q