Question

# A regional planner employed by a public university is studying the demographics of nine counties in...

A regional planner employed by a public university is studying the demographics of nine counties in the eastern region of an Atlantic seaboard state. She has gathered the following data:

 County Median Income Median Age Coastal A \$ 49,374 58.5 0 B 46,850 46.5 1 C 47,586 48.5 1 D 47,781 45.5 1 E 33,738 37.3 0 F 35,553 43.4 0 G 39,910 45.3 0 H 37,266 34.2 0 I 34,571 36.5 0

1. Is there a linear relationship between the median income and median age? (Round your answer to 3 decimal places.)
1. Which variable is the "dependent" variable?
• Median Age

• Median Income

1. c-1. Use regression analysis to determine the relationship between median income and median age. (Round your answers to 2 decimal places.)

1. c-2. Interpret the value of the slope in a simple regression equation. (Round your answers to 2 decimal places.)

1. Include the aspect that the county is "coastal" or not in a multiple linear regression analysis using a "dummy" variable. (Negative amounts should be indicated by a minus sign. Round your answers to 2 decimal places.)
1. Test each of the individual coefficients to see if they are significant. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)
1. Make a histogram of the residuals. Which plot is correct?
 Plot 1 Plot 2 Plot 3
• Plot 1

• Plot 2

• Plot 3

1. Make a scatter diagram of the residual values versus the fitted values. Which plot is correct?
 Plot 1 Plot 2 Plot 3
• Plot 1

• Plot 2

• Plot 3

Please like if u helps you ?

a)using excel formula correl(y,x)

correlation of Income and Median Age is =0.817

B)Income =10170.98+710.36*median age

c)

For each year increase in age, the income increases 710.36 on average

d)

Income =13611.39+582.85*median age+6497.41*Coastal

E)

 t Stat P-value Intercept 2.70 0.0357 ~0.04 median age 5.00 0.0024~ 0.00 coastal 3.77 0.0093~ 0.01