WARM – UP Can you accurately predict your TEST grade based on your QUIZ grade? 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 Describe the association between Quiz Grades and Test Grades. Write the equation of the line of regression. Use this model to predict a Test Grade from a Quiz grade of 50. Is this a good model?

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 Describe the association between Quiz Grades and Test Grades Write the equation of the line of regression. Use this model to predict a Test Grade from a Quiz grade of 50. Is this a good model? TEST Grade Residuals QUIZ Grade QUIZ Grade TEST = 64.217 + 0.361 (QUIZ) 82.286 = = 64.217 + 0.361(50)

WARM – UP # Hours of Study Test Grade .5 3 1.5 2 1 72 98 82 89 76 73 Does a significant relationship exist between number of hours studying and test grades? H0: β1 = 0 Ha: β1 ≠ 0 β=True Slope of the linear relationship. Lin Reg t-Test Since the p-value is less than 0.05, we REJECT H0. There is sufficient evidence to conclude that # hours of study and Test grades are related.

Is there a relationship between the amount of lead verses the amount of Zinc present in fish? State Hypothesis and Conclusion. Dependent Variable is: Zinc R-squared = 85.7% s = 15.3330 with 8 – 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept 23.0796 14.204 Lead 27.5335 4.5837 Because the p-val < 0.05 Reject Ho. There is evidence to conclude that a relationship exists b/t lead and zinc content in fish.

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:

WARM – UP 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 Find the 95 % Conf. Interval for the Rate of Improvement from the Quiz to the Test . Dependent Variable is: Test R-squared = 35.3% s = 4.866 with 8 – 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept 64.2170 14.204 1.6249 0.1386 Quiz 0.36137 0.19985 1.8082 0.1206 We can be 95% confident that for every additional point increase in quiz grade your test grade is predicted to be - 0.1276 to 0.8504 points higher.

Powerboat Registrations In Thousands EXAMPLE: Manatees living off the coast of Florida are often killed by powerboats. Here are data from 1984 – 1990 Powerboat Registrations In Thousands Manatees Killed 459 24 585 33 614 645 39 675 43 711 52 719 51 Dependent Variable is: Manatees R-squared = 98.2% s = 1.9391 with 7 – 2 = 5 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept -29.0001 14.204 1.6249 0.1386 Boats 0.1084 0.0155 6.9932 <0.0001 Find the 95% confidence interval for the rate at which the Manatee deaths change with regards to powerboat registration. We can be 95% confident that for every additional 1000 powerboat registered, between 0.0686 and 0.1483more Manatees are killed.

Estimate the Rate of change for the relationship between lead and zinc using a 99% Confidence Interval Metal Concentrations in 1999 Spokane River Fish Samples (concentrations in parts per million (ppm) Source: WA State Dept. of Ecology report) Lead Zinc     0.73 45.3   1.14 50.8   0.60 40.2   1.59 64.0 4.34 150.   1.98 106.   3.12 90.8   1.80 58.8 Dependent Variable is: Zinc R-squared = 85.7% s = 15.3330 with 8 – 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept 23.0796 14.204 1.6249 0.1386 Lead 27.5335 4.5837 6.0068 0.0010 We can be 99% confident that the true slope of the relationship between lead and zinc is between .

Show that the p-value is 0 Show that the p-value is 0.0010 by using the t formula and calculating the p-value. Lead Zinc     0.73 45.3   1.14 50.8   0.60 40.2   1.59 64.0 4.34 150.   1.98 106.   3.12 90.8   1.80 58.8 Dependent Variable is: Zinc R-squared = 85.7% s = 15.3330 with 8 – 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept 23.0796 14.204 1.6249 0.1386 Lead 27.5335 4.5837 6.0068 0.0010 SE(b1)

HW Page 661: 11, 12, 16-18

Is there a relationship between the amount of lead verses the amount of Zinc present in fish? State Hypothesis and Conclusion . Metal Concentrations in 1999 Spokane River Fish Samples (concentrations in parts per million (ppm) Source: WA State Dept. of Ecology report) Lead Zinc     0.73 45.3   1.14 50.8   0.60 40.2   1.59 64.0 4.34 150.   1.98 106.   3.12 90.8   1.80 58.8 Dependent Variable is: Zinc R-squared = 85.7% s = 15.3330 with 8 – 2 = 6 degrees of freedom Variable Coefficient SE(Coeff) T-ratio P-Value Intercept 23.0796 14.204 1.6249 0.1386 Lead 27.5335 4.5837 6.0068 0.0010 P-value Because the p-val < 0.05 there is significant evidence to conclude that a relationship exists b/t lead and zinc content in fish.

Performing a Linear Regression t-Test from the Calculator Is there a relationship in the amount of lead verses the amount of Zinc present in fish? Lead Zinc     0.73 45.3   1.14 50.8   0.60 40.2   1.59 64.0 4.34 150.0   1.98 106.0   3.12 90.8   1.80 58.8 STAT  Test  LinRegTTest Because the p-val < 0.05 there is significant evidence to conclude that a relationship exists b/t lead and zinc content in fish.

Is there a relationship between the amount of lead verses the amount of Zinc present in fish? State Hypothesis and Conclusion . Metal Concentrations in 1999 Spokane River Fish Samples (concentrations in parts per million (ppm) Source: WA State Dept. of Ecology report) Lead Zinc     0.73 45.3   1.14 50.8   0.60 40.2   1.59 64.0 4.34 150.   1.98 106.   3.12 90.8   1.80 58.8 Seb = 4.5837 P-value = 0.0010 Because the p-val < 0.05 there is significant evidence to conclude that a relationship exists b/t lead and zinc content in fish.