Question

A) Listed below are numbers of registered pleasure boats in Florida measured in tens of thousands (the x-values) and the numbers of manatee fatalities from encounters with boats in Florida (the y-values) for each of several recent years. Find the linear correlation coefficient (r). [Round to 3 decimal places]

x	y	xy	x2	y2
99	92	9108	9801	8464
99	73	7227	9801	5329
97	90	8730	9409	8100
95	97	9215	9025	9409
90	83	7470	8100	6889
90	88	7920	8100	7744
87	81	7047	7569	6561
90	73	6570	8100	5329
90	68	6120	8100	4624

B) Use the given information to conclude whether there is a linear correlation between the samples?

Group of answer choices:

There is not sufficient evidence to support a linear correlation?

There is sufficient evidence to support a linear correlation?

C) Find the regression line equation (y^). What is the slope of the regression line equation?

[Round to 3 decimal places]

D) For the same regression line equation, what is the y-intercept?

[Round to tenths place]

E) Find the best predicted y-value for the x-value of 79, which was the actual number of manatee fatalities. [Round to the nearest whole number]

Answer 1

Sol:

a)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
99	92	36.00	85.05	55.33
99	73	36.00	95.60	-58.67
97	90	16.00	52.16	28.89
95	97	4.00	202.27	28.44
90	83	9.00	0.05	-0.67
90	88	9.00	27.27	-15.67
87	81	36.00	3.16	10.67
90	73	9.00	95.60	29.33
90	68	9.00	218.38	44.33

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	837.0000	745.0000	164.0000	779.5556	122.0000
mean	93.0000	82.7778	SSxx	SSyy	SSxy

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.3412
      

Ho:   ρ = 0  
Ha:   ρ ╪ 0  
n=   9  
alpha,α =    0.05  
correlation , r=   0.3412  
t-test statistic = r*√(n-2)/√(1-r²) =        0.960
DF=n-2 =   7  
p-value =    0.3689  
Decison:   P value > α, So, Do not reject Ho  

there is no evidence of significant correlation

sample size ,   n =   9          
here, x̅ = Σx / n=   93.000   ,     ȳ = Σy/n =   82.778  
                  
SSxx =    Σ(x-x̅)² =    164.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   122.0          
                  
estimated slope , ß1 = SSxy/SSxx =   122.0   /   164.000   =   0.74390
                  
intercept,   ß0 = y̅-ß1* x̄ =   13.59485          
                  
so, regression line is   Ŷ =   13.595   +   0.743902   *x

Predicted Y at X=   97   is                  
Ŷ =   13.5949   +   0.7439   *   97   =   85.7534 ≈86

86

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