Question

Let x be a random variable representing percentage change in neighborhood population in the past few years, and let y be a random variable representing crime rate (crimes per 1000 population). A random sample of six Denver neighborhoods gave the following information.

x	26	2	11	17	7	6
y	179	34	132	127	69	53

In this setting we have Σx = 69, Σy = 594, Σx² = 1175, Σy² = 74,320, and Σxy = 9134.

(a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your least-squares estimates to four decimal places.)

x	=
y	=
b	=
ŷ	=	+  x

Find the sample correlation coefficient r and the coefficient of determination. (Round your answers to three decimal places.)

r =
r2 =

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)
%

(d) Test the claim that the population correlation coefficient ρ is not zero at the 1% level of significance. (Round your test statistic to three decimal places.)

t =

Find or estimate the P-value of the test statistic.

P-value > 0.2500.125 < P-value < 0.250    0.100 < P-value < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005

For a neighborhood with x = 19% change in population in the past few years, predict the change in the crime rate (per 1000 residents). (Round your answer to one decimal place.)
crimes per 1000 residents

(f) Find S_e. (Round your answer to three decimal places.)
S_e =

(g) Find an 80% confidence interval for the change in crime rate when the percentage change in population is x = 19%. (Round your answers to one decimal place.)

lower limit	crimes per 1000 residents
upper limit	crimes per 1000 residents

(h) Test the claim that the slope β of the population least-squares line is not zero at the 1% level of significance. (Round your test statistic to three decimal places.)

t =

Find or estimate the P-value of the test statistic.

P-value > 0.2500.125 < P-value < 0.250    0.100 < P-value < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005

Find an 80% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit
upper limit

Answer 1

ΣX = 69
ΣY = 594
ΣX * Y = 9134
ΣX2 = 1175
ΣY2 = 74320

Part a)

X̅ = Σ( Xi / n ) = 69/6 = 11.5
Y̅ = Σ( Yi / n ) = 594/6 = 99

Equation of regression line is Ŷ = a + bX


b = 6.037
a =( Σ Y - ( b * Σ X) ) / n
a =( 594 - ( 6.0367 * 69 ) ) / 6
a = 29.578
Equation of regression line becomes Ŷ = 29.578 + 6.037 X

Part b)

r = 0.947

= 0.896

Explained variation = 0.896* 100 = 89.6%

Part d)

To Test :-
H0 :- ρ = 0
H1 :- ρ ≠ 0

Test Statistic :-
t = (r * √(n - 2) / (√(1 - r2))
t = ( 0.9466 * √(6 - 2) ) / (√(1 - 0.8961) )
t = 5.873

To Test :-
H0 :- ρ = 0
H1 :- ρ ≠ 0

Test Statistic :-
t = (r * √(n - 2) / (√(1 - r2))
t = ( 0.9466 * √(6 - 2) ) / (√(1 - 0.8961) )
t = 5.87339

P - value = P ( t > 5.8734 ) = 0.0042
0.0005 < P-value < 0.005

When X = 19
Ŷ = 29.578 + 6.037 X
Ŷ = 29.578 + ( 6.037 * 19 )
Ŷ = 144.3

Part f)

Standard Error of Estimate S = √ ( Σ (Y - Ŷ ) / n - 2) = √(1611.4862 / 4) = 20.0717

Part g)

Confidence Interval of
Ŷ = 29.578 + 6.0367 X
Ŷ = 144.281

t(α/2) = t(0.2/2) =1.533
80% confidence interval is ( 127.0 < < 161.5 ).

Part h)

To Test :-
H0 :- ß = 0
H1 :- ß ≠ 0

Test Statistic :-
t = ( b - ß) / ( S / √(S(xx)))
t = ( 6.0367 - 0 ) / ( 20.0717 / √(381.5))
t = 5.874

P - value = P ( t > 5.8744 ) = 0.00419
0.0005 < P-value < 0.005

Confidence Interval for ß

t(α/2) = t(0.2/2) =1.533
80% confidence interval is ( 6.037 < β̂ < 1.028  )

We are 80% confidence that the slope lies within the interval.