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In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of...

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   74   57
Finland   60   67
Finland   64   45
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: μd = 0; H1: μd > 0; right-tailed
H0: μd = 0; H1: μd ≠ 0; two-tailed
H0: μd > 0; H1: μd = 0; right-tailed
H0: μd = 0; H1: μd < 0; left-tailed

(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that d has an approximately uniform distribution.
The Student's t. We assume that d has an approximately normal distribution.
The standard normal. We assume that d has an approximately normal distribution.
The Student's t. We assume that d has an approximately uniform 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

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) State your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence to claim that the average percentage of male wolves in winter is higher.
Reject the null hypothesis, there is sufficient evidence to claim that the average percentage of male wolves in winter is higher.
Fail to reject the null hypothesis, there is insufficient evidence to claim that the average percentage of male wolves in winter is higher.
Fail to reject the null hypothesis, there is sufficient evidence to claim that the average percentage of male wolves in winter is higher.

Homework Answers

Answer #1

a)

0.05 is alpha

H0: μd = 0; H1: μd > 0; right-tailed
b)

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

Test statistic,
t = (dbar - 0)/(s(d)/sqrt(n))
t = (5.75 - 0)/(10.3199/sqrt(8))
t = 1.576

c)


P-value Approach
P-value = 0.0795

0.050 < P-value < 0.125

d)

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

e)

Fail to reject the null hypothesis, there is insufficient evidence to claim that the average percentage of male wolves in winter is higher

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