Question

After examining these data for all the jurisdictions, someone notes that certain areas have an unusually high “percent of 18-64 yr-olds with no high school diploma.” Based on this finding, this individual concludes that the high percentages are due to the rising population of immigrants in those areas. Further, the individual argues that any estimates of the associated “percent of low-income working families” in those areas should be recalculated after removing this sub-population from the data set, as they are causing the area to “look bad”. In addition to thinking critically, use the key rules about linear regression and extrapolation to write a statistically appropriate and socially responsible response to the individual’s conclusion and argument.

2011 Data
Jurisdiction	Percent of low income working families (<200% poverty level)	Percent of 18-64 year olds with no HS diploma
Alabama	37.3	15.3
Alaska	25.9	8.6
Arizona	38.9	14.8
Arkansas	41.8	14
California	34.3	17.6
Colorado	27.6	10.1
Connecticut	21.1	9.5
Delaware	27.8	11.9
District of Columbia	23.2	10.8
Florida	37.3	13.1
Georgia	36.6	14.9
Hawaii	25.8	7.2
Idaho	38.6	10.7
Illinois	30.4	11.5
Indiana	31.9	12.2
Iowa	28.8	8.1
Kansas	32	9.7
Kentucky	34.1	13.6
Louisiana	36.3	16.1
Maine	30.4	7.1
Maryland	19.5	9.7
Massachusetts	20.1	9.1
Michigan	31.6	10
Minnesota	24.2	7.3
Mississippi	43.6	17
Missouri	32.7	11.1
Montana	36	7
Nebraska	31.1	8.7
Nevada	37.4	16.6
New Hampshire	19.7	7.3
New Jersey	21.2	10.1
New Mexico	43	16.2
New York	30.2	13
North Carolina	36.2	13.6
North Dakota	27.2	5.9
Ohio	31.8	10.3
Oklahoma	37.4	13.2
Oregon	33.9	10.8
Pennsylvania	26	9.4
Rhode Island	26.9	12
South Carolina	38.3	14.2
South Dakota	31	8.7
Tennessee	36.6	12.7
Texas	38.3	17.8
Utah	32.3	9.9
Vermont	26.2	6.6
Virginia	23.3	10.2
Washington	26.4	10.2
West Virginia	36.1	12.9
Wisconsin	28.7	8.5
Wyoming	28.1	8

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1595.1	574.8	1943.091765	486.6	681.10
mean	31.28	11.27	SSxx	SSyy	SSxy

sample size ,   n =   51          
here, x̅ = Σx / n=   31.28   ,     ȳ = Σy/n =   11.27  
                  
SSxx =    Σ(x-x̅)² =    1943.0918          
SSxy=   Σ(x-x̅)(y-ȳ) =   681.1          
                  
estimated slope , ß1 = SSxy/SSxx =   681.1   /   1943.092   =   0.3505
                  
intercept,   ß0 = y̅-ß1* x̄ =   0.3074          
                  
so, regression line is   Ŷ =   0.3074   +   0.3505   *x
                  
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    247.861          
                  
std error ,Se =    √(SSE/(n-2)) =    2.249          
                  
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.7005         

Relation is positive and moderate.

THANKS

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