In: Statistics and Probability
15) Wolf society, packs, and ecology have been studied extensively at different locations throughout the world. Sex ratios for 7 study sites in Northern Europe are shown below.
Site No. |
% Males (Winter) |
%Males (Summer) |
1 |
71 |
54 |
2 |
48 |
51 |
3 |
89 |
72 |
4 |
57 |
46 |
5 |
62 |
55 |
6 |
50 |
49 |
7 |
58 |
44 |
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.
H0 : µd=0; H1 : µd>0
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: ud< 0
Alternative hypothesis: ud > 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = 14.48398
SE = s / sqrt(n)
S.E = 5.47443
DF = n - 1 = 7 -1
D.F = 6
t = [ (x1 - x2) - D ] / SE
t = 2.45
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than 2.45; that is, less than - 2.45 or greater than 2.45.
Thus, the P-value = 0.025
Interpret results. Since the P-value (0.025) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that the average percentage of males in a wolf pack is higher in winter.