WARM – UPIs 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 = 6.3158 Variable Coefficient SE(Coeff) Constant 1.1836 3.2184 Height 1.4492 0.8827

1. Find and interpret slope and y-int. for the Regression Line.

For every additional inch of height a mans IQ will increase For every additional inch of height a mans IQ will increase 1.4492 points. A man 0 inches high will have an IQ of 1.181.4492 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.

ˆ 1.1836 1.4492IQ Height

Dependent variable is: IQ R-squared = 47.3% s = 6.3158 Variable Coefficient SE(Coeff) Constant 1.1836 3.2184 Height 1.4492 0.8827

2. Is there an association b/t height and IQ in men? Write the Hypothesis, t & p-values, and conclusion.

t-ratio P-Value

1.6418

1

1

0

b

bt

SE

1.4492 0

0.8827t

2 1.6418, 99,24p Val tcdf E

0.1137

0 : 0 No Relationship exist b/w IQ & height

: 0 Relationship Existsa

H

H

Since the p-value of .1137 is above 0.05, we cannot reject Since the p-value of .1137 is above 0.05, we cannot reject HH00. In conclusion there is NO Evidence of a significant . In conclusion there is NO Evidence of a significant

relationship b/w IQ and height. relationship b/w IQ and height.

n = 26

t-ratio P-Value

1.6418

t-ratio P-Value

1.6418

Regression InferenceC.I.

• To estimate the rate of change, slope, we use a Confidence Interval. The formula for a confidence interval for 1 is: 1 2 1nb t SE b

Find the t* for a 99% Confidence Interval with n = 20t* = 2.878

WARM – UP1 2 3 4 5 6 7 8

88 84 100 92 70 80 81 92

95 100 102 94 90 85 99 97

STUDENT

QUIZ

TEST

1. What was the Rate of Improvement from the Quiz to the Test . 1 2 1nb t SE b

0.36137 2.447 0.19985

We can be 95% confident that for every additional point increase in quiz grade your test grade is predicted to be - 0.1276 and 0.8504 higher.

Dependent Variable is: TestR-squared = 35.3%s = 4.866 with 8 – 2 = 6 degrees of freedomVariable Coefficient SE(Coeff) T-ratio P-ValueIntercept 64.2170 14.204 1.6249 0.1386Quiz 0.36137 0.19985 1.8082 0.1206

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 = 9.073602 with 8 - 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) t-ratio P-value Constant 33.14815 7.510757 4.413423 0.004503 Seed Density 4.537037 2.469522 ? ?

1. Find and Interpret the Regression line1. Find and Interpret the Regression line2. Interpret the R-squared2. Interpret the R-squared3.3.Find and interpret a 95% Confidence Interval for slope.Find and interpret a 95% Confidence Interval for slope.4.4.Is there a significant relationship between seed density and Is there a significant relationship between seed density and lawn quality?lawn quality?

Dependent variable is: Lawn Quality R-squared = 36.0% s = 9.073602 with 8 - 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) t-ratio P-value Constant 33.14815 7.510757 4.413423 0.004503 Seed Density 4.537037 2.469522 ? ?

1.1. Predicted Lawn Quality = 33.1482 + 4.5370(Seed Density)Predicted Lawn Quality = 33.1482 + 4.5370(Seed Density)For every additional lbs./500ftFor every additional lbs./500ft22of seeds, Lawn Quality of seeds, Lawn Quality improves 4.54%. A Lawn with 0 seeds will be rated at 33.1%.improves 4.54%. A Lawn with 0 seeds will be rated at 33.1%.

2. 36% of the variation in Lawn quality is due to seed density.2. 36% of the variation in Lawn quality is due to seed density.3. 3. 1 2 1nb t SE b

4.5370 2.447 2.4695

We can be 95% confident that for every increase in seed density the lawn quality is predicted to increase - 1.5059 and 10.5799 .

Dependent variable is: Lawn Quality R-squared = 36.0% s = 9.073602 with 8 - 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) t-ratio P-value Constant 33.14815 7.510757 4.413423 0.004503 Seed Density 4.537037 2.469522 ? ?

4. 4.

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

0 : 0 No Relationship

: 0 Relationship Existsa

H

H

22 ( , 99, )nP Val tcdf t E df 1

1

btSE b

1.8372 0.1158