In: Statistics and Probability
(2) Does the number of tadpoles vary between the relative altitudes at which the sites were located? If so, which of the altitude level means are significantly different? . The mean number of breeding ponds was found to be 48. The researcher wants to know if the mean number of breeding ponds found in NSW is different from the mean number found in the Victorian surveys?
Site | Rain | Temp | Altitude | Tadpoles | Breeding |
1 | 769.77 | 22.37 | Low | 97 | 61 |
2 | 664.85 | 19.46 | Low | 86 | 70 |
3 | 807.13 | 23.17 | Low | 104 | 61 |
4 | 620.05 | 17.85 | Low | 117 | 113 |
5 | 673.21 | 20.29 | Low | 98 | 92 |
6 | 734.18 | 20.58 | Low | 88 | 100 |
7 | 526.62 | 16.29 | Low | 95 | 102 |
8 | 647.86 | 18.51 | Low | 38 | 66 |
9 | 672.3 | 20.13 | Low | 64 | 64 |
10 | 666.67 | 18.91 | Low | 57 | 18 |
breeding data-
61 |
70 |
61 |
113 |
92 |
100 |
102 |
66 |
64 |
18 |
Ho : µ = 48
Ha : µ ╪ 48
(Two tail test)
Level of Significance , α =
0.050
sample std dev , s = √(Σ(X- x̅ )²/(n-1) )
= 27.7971
Sample Size , n = 10
Sample Mean, x̅ = ΣX/n =
74.7000
degree of freedom= DF=n-1= 9
Standard Error , SE = s/√n = 27.7971 / √
10 = 8.7902
t-test statistic= (x̅ - µ )/SE = ( 74.700
- 48 ) / 8.7902 =
3.037
p-Value = 0.0141 [Excel formula
=t.dist(t-stat,df) ]
Decision: p-value<α = 0.05 , Reject null hypothesis
Conclusion: There is enough evidence to conclude that
mean number of breeding ponds found in NSW is different
from the mean number found in the Victorian survey