1) For neighborhoods with more than 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(0) = 142.5737 - 1.9215(Prop_change_income) (2) For neighborhoods with 33% to 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(1) = 123.36849 - 1.9215(Prop_change_income) (3) For neighborhoods with less than 33% of schools p
(1) For neighborhoods with more than 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(0) = 142.5737 - 1.9215(Prop_change_income)
(2) For neighborhoods with 33% to 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(1) = 123.36849 - 1.9215(Prop_change_income)
(3) For neighborhoods with less than 33% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(1) - 19.20521(0) = 103.69959 - 1.9215(Prop_change_income)
Questions:
1. What is difference between models in 1,2,and 3?
2. Which estimated model 1,2, or 3 results in the line with the highest estimated crime rates (the highest positioned line)?
3. Test whether the crime rate in neighborhoods with 33% to 66% of schools participating in free school lunch program is significantly different than the crime rate in neighborhoods with more than 66% of schools participating in free school lunch program when adjusting for percent change in household income over past several years using a PARTIAL F TEST. Don't forget to specify the null and alternative hypothesis, test statistic, degrees of freedom, p-value, decision and conclusion in the words of the problem. Use level of significance 5%.
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