Question

Wolf packs tend to be large extended family groups that have a well-defined hunting territory. Wolves not in the pack are driven out of the territory or killed. In ecologically similar regions, is the size of an extended wolf pack related to size of hunting region? Using radio collars on wolves, the size of the hunting region can be estimated for a given pack of wolves. Let x represent the number of wolves in an extended pack and y represent the size of the hunting region in km²/1000. The following data are representative of one of the national parks.

x wolves	29	31	22	67	96
y km2/1000	7.37	12.14	8.17	15.36	16.85

(a) Verify that

Σx = 245,

Σy = 59.89,

Σx² = 15,991,

Σy² = 788.2975,

Σxy = 3416.53,

and

r ≈ 0.9063.

Σx
Σy
Σx2
Σy2
Σxy
r

(b) Use a 1% level of significance to test the claim

ρ > 0.

(Use 2 decimal places.)

t
critical t

Conclusion

Reject the null hypothesis, there is sufficient evidence that ρ > 0.Reject the null hypothesis, there is insufficient evidence that ρ > 0.     Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.Fail to reject the null hypothesis, there is sufficient evidence that ρ > 0.

(c) Verify that

S_e ≈ 2.0550,

a ≈ 6.0537,

and

b ≈ 0.12090.

Se
a
b

(d) Find the predicted size of the hunting region for an extended pack of 56 wolves. (Use 2 decimal places.)
km²/1000

(e) Find an 85% confidence interval for your prediction of part (d). (Use 1 decimal place.)

lower limit	km2/1000
upper limit	km2/1000

(f) Use a 1% level of significance to test the claim that

β > 0.

(Use 2 decimal places.)

t
critical t

Conclusion

Reject the null hypothesis, there is sufficient evidence that β > 0.Reject the null hypothesis, there is insufficient evidence that β > 0.     Fail to reject the null hypothesis, there is insufficient evidence that β > 0.Fail to reject the null hypothesis, there is sufficient evidence that β > 0.

(g) Find a 95% confidence interval for β and interpret its meaning in terms of drift rate. (Use 2 decimal places.)

lower limit
upper limit

Interpretation

For every wolf joining the pack, the hunting territory decreases by an amount that falls within the confidence interval.For every wolf joining the pack, the hunting territory increases by an amount that falls within the confidence interval.     For every wolf joining the pack, the hunting territory decreases by an amount that falls outside the confidence interval.For every wolf joining the pack, the hunting territory increases by an amount that falls outside the confidence interval.

Answer 1

X	Y	XY	X²	Y²
29	7.37	213.73	841	54.3169
31	12.14	376.34	961	147.3796
22	8.17	179.74	484	66.7489
67	15.36	1029.12	4489	235.9296
96	16.85	1617.6	9216	283.9225

a)

Sample size, n =	5
Ʃ x =	245
Ʃ y =	59.89
Ʃ xy =	3416.53
Ʃ x² =	15991
Ʃ y² =	788.2975
Correlation coefficient, r = SSxy/√(SSxx*SSyy) =	0.9063

b) Null and alternative hypothesis:

Ho:   ρ = 0 ; Ha:   ρ > 0  

alpha, α = 0.01  

Test statistic:

t = r*√(n-2)/√(1-r²) = 3.71

df = 5-2 =3

Critical t = T.INV.RT(0.05, 3) = 1.25  

Conclusion:

Reject the null hypothesis, there is sufficient evidence that ρ > 0.

------------------------------------

c) b = SSxy/SSxx = 481.92/3986 =  0.12090
a = y̅ -b* x̅ = 11.978 - 0.12090*49 =  6.0537

Standard error, se = √((Syy -b*Ssxy)/(n-2)) = √((70.93508-0.12090*481.92)/3) =  2.0550

d) Predicted value at x = 56

ŷ = 6.0537 + 0.1209 * 56 = 12.82   

e) 85% confidence interval:

At α = 0.15 and df = n-2 = 3   , critical value, t_c = T.INV.2T(0.15, 3 ) =   1.9243

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx))) =	11.0
Upper limit = ŷ + tc*se*√( (1/n) + ((x-x̅)²/(SSxx))) =	14.6

f) Null and alternative hypothesis

Ho:   β₁ =   0
H1:   β₁ >   0

Test statistic:

Critical t = T.INV.RT(0.05, 3) = 1.25  

Conclusion:

Reject the null hypothesis, there is sufficient evidence that β > 0

g) 95%   Confidence interval for slope :

At α = 0.05 and df = n-2 = 3, critical value, t_c = T.INV.2T(0.05, 3) =   3.1824

Lower limit = b- tc*se/√SSxx =   0.02
Upper limit = b + tc*se/√SSxx =   0.22