In: Math
A study of fox rabies in a country gave the following information about different regions and the occurrence of rabies in each region. A random sample of
n1 = 16
locations in region I gave the following information about the number of cases of fox rabies near that location.
x1:
Region I Data
2 | 9 | 9 | 9 | 7 | 8 | 8 | 1 |
3 | 3 | 3 | 2 | 5 | 1 | 4 | 6 |
A second random sample of
n2 = 15
locations in region II gave the following information about the number of cases of fox rabies near that location.
x2:
Region II Data
2 | 2 | 5 | 2 | 6 | 8 | 5 | 4 |
4 | 4 | 2 | 2 | 5 | 6 | 9 |
(i) Use a calculator with sample mean and sample standard deviation keys to calculate x1 and s1 in region I, and x2 and s2 in region II. (Round your answers to two decimal places.)
x1 | = |
s1 | = |
x2 | = |
s2 | = |
(ii) Does this information indicate that there is a difference
(either way) in the mean number of cases of fox rabies between the
two regions? Use a 5% level of significance. (Assume the
distribution of rabies cases in both regions is mound-shaped and
approximately normal.)
(a) What is the level of significance?
SolutionI)
use R software
mean and sd functions to find mean and standard deviations
Region1 <- c(2, 9 ,9, 9 ,7, 8, 8, 1, 3, 3 ,3, 2 ,5, 1, 4, 6)
x1 <- round(mean(Region1),2)
x1
s1 <- round(sd(Region1),2)
s1
x2 <- round(mean(Region2),2)
x2
s2 <- round(sd(Region2),2)
s2
output:
x1
[1] 5
> s1 <- round(sd(Region1),2)
> s1
[1] 2.99
> x2 <- round(mean(Region2),2)
> x2
[1] 4.4
> s2 <- round(sd(Region2),2)
> s2
[1] 2.23
x1 | 5 |
s1 | 2.99 |
x2 | 4.4 |
s2 | 2.23 |
(ii) Does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions
perform t test for difference in means
use t.test function in R software.
default level of signifance=0.05
t.test(Region1,Region2)
Output:
Welch Two Sample t-test
data: Region1 and Region2
t = 0.63608, df = 27.653, p-value = 0.53
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-1.333296 2.533296
sample estimates:
mean of x mean of y
5.0 4.4
t=0.636
p=0.53
p>0.05
Fail to reject H0.
There is no sufficient statistical evidence at 5% leve of signifcance to conclude that
there is a difference (either way) in the mean number of cases of fox rabies between the two regions
(a) What is the level of significance?
alpha=0.05