A study of fox rabies in southern Germany gave the following information about d
ID: 3367191 • Letter: A
Question
A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region (Reference: B. Sayers et al., "A Pattern Analysis Study of a Wildlife Rabies Epizootic," Medical Informatics, Vol. 2, pp. 11-34). The data gives the number of cases of fox rabies found in 16 locations of region I and 15 locations of region II. Region I Data ? ? ? 3 3 2 1 4 4 1 Region II Data 4 9 6 6 ? ?? 6 a u 2 4 2 4 3 3 1 2 ?? ?? 10 Let ui, uz represent the mean number of cases of fox rabies in region I and in region II respectively. At significance level 0.01, does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions? Assume the distribution of rabies cases in both regions is mound-shaped and approximately normal. A) Using 2 decimal places for 1, 02, 81, and 82 from the two samples, find d and the combined standard error, SEC d = SE. = *. Round your answer to 4 decimal places. B) State the Null and Alternate Hypotheses? Ho: M1 = H2 (d = 0) H: M1 (d = 0) HO: M1 = 42 (d = 0) H1:41 > M2 (d > 0) Ho: Mi = H2 (d = 0) H: MiExplanation / Answer
Solution:
A.
d-bar = X-bar1 - X-bar2
d-bar = 4.3125 - 4.4667
d-bar = -0.1542
SE = ?SD1^2/n1 + SD2^2/n2
SE = ?2.8918/16 + 2.5875/15
SE = 0.9844
B.
Null Hypothesis (Ho): µ1 = µ2
Alternative Hypothesis (Ha): µ1 ? µ2
C. Test Statistics
t = d-bar/SE
t = -0.1542/0.9844
t = -0.157
D. df = [(s1^2/n1 + s2^2/n2)^2]^2/[1/(n1 - 1) (s1^2/n1)^2 + 1/(n2 - 1) (s2^2/n2)^2]
df = [(2.8918^2/16 + 2.5875^2/15]^2/[1/(16-1) (2.8918^2/16)^2 + 1/(15-1) (2.5875^2/15)^2]
df = 28.94 ~ 29
E. Using t-tables, with 29 degrees of freedom is
F. P-value = 0.8766
G. P-value > 0.01, No
H. tc = 1.617
CI: d-bar ± 1.59
Region I (X1) Region II (X2) 0 2 8 2 8 3 8 1 6 4 7 9 8 6 1 5 3 4 3 4 2 3 1 2 4 6 1 6 4 10 5 Mean = 4.3125 4.4667 SD = 2.8918 2.5875 n = 16 15Related Questions
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