Question

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

In environmental studies, sex ratios are of great importance. Wolf society, packs, and ecology have been studied extensively at different locations in the U.S. and foreign countries. Sex ratios for eight study sites in northern Europe are shown below.

Location of Wolf Pack	% Males (Winter)	% Males (Summer)
Finland	62	67
Finland	60	65
Finland	84	53
Lapland	55	48
Lapland	64	55
Russia	50	50
Russia	41	50
Russia	55	45

It is hypothesized that in winter, "loner" males (not present in summer packs) join the pack to increase survival rate. Use a 5% level of significance to test the claim that the average percentage of males in a wolf pack is higher in winter. (Let d = winter − summer.)

(a) What is the level of significance?


State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ_d = 0; H₁: μ_d > 0; right-tailed H₀: μ_d = 0; H₁: μ_d ≠ 0; two-tailed     H₀: μ_d > 0; H₁: μ_d = 0; right-tailed H₀: μ_d = 0; H₁: μ_d < 0; left-tailed

(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that d has an approximately normal distribution. The standard normal. We assume that d has an approximately uniform distribution.     The Student's t. We assume that d has an approximately uniform distribution. The standard normal. We assume that d has an approximately normal distribution.

What is the value of the sample test statistic? (Round your answer to three decimal places.)


(c) Find (or estimate) the P-value.

P-value > 0.250 0.125 < P-value < 0.250     0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005

Answer 1

It is one tailed test

H₀: μ_d = 0; H₁: μ_d > 0

level of significance

The Student's t. We assume that d has an approximately normal distribution.

Test Statistic :-





t = 1.052

Test Criteria :-
Reject null hypothesis if


Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 1.0521 ) = 0.1639
Reject null hypothesis if P value < level of significance
P - value = 0.1639 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

0.125 < P-value < 0.250

There is insufficient evidence to support the claim that the average percentage of males in a wolf pack is higher in winter.