In: Statistics and Probability
In a scientific study that attempts to relate the biodiversity of macro invertebrates to the acidity of the water near mines, samples of macro invertebrates were taken at the rate of a monthly sample for 12 months in two different locations. For the first locality, an average biodiversity index of 3.11 and a sample standard deviation of 0.771 were obtained. For the second locality the average biodiversity index was 2.04 with a standard deviation of 0.448.
a) BUILD and INTERPRET a 90% confidence interval for the difference in the means of the two localities assuming that the populations fit approximately a normal population and that the variances can be considered equal.
b) Evaluate whether the evidence supports the presumption of equal variances.
c) Would it have been better to use a 95% interval? Explain in detail.
a)
mean of sample 1, x̅1=
3.110
standard deviation of sample 1, s1 =
0.771
size of sample 1, n1= 12
Sample #2 ----> 2
mean of sample 2, x̅2= 2.040
standard deviation of sample 2, s2 =
0.448
size of sample 2, n2= 12
difference in sample means = x̅1-x̅2 =
3.1100 - 2.0 =
1.07
Level of Significance , α = 0.1
Degree of freedom, DF= n1+n2-2 =
22
t-critical value = t α/2 =
1.7171 (excel formula =t.inv(α/2,df)
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 0.6305
std error , SE = Sp*√(1/n1+1/n2) =
0.2574
margin of error, E = t*SE = 1.7171
* 0.26 = 0.44
difference of means = x̅1-x̅2 =
3.1100 - 2.040 =
1.0700
confidence interval is
Interval Lower Limit= (x̅1-x̅2) - E =
1.0700 - 0.4420 =
0.628
Interval Upper Limit= (x̅1-x̅2) + E =
1.0700 + 0.4420 =
1.512
we are 90% confident that the true difference in the means of the two localities lies within these limits
.....................
b)
Sample 1
s₁² = 0.594441
n₁ = 12
Sample 2
s₂² = 0.200704
n₂ = 12
α = 0.05
Null and alternative hypothesis:
Hₒ : σ₁² = σ₂²
H₁ : σ₁² ≠ σ₂²
Test statistic:
F = s₁² / s₂² = 0.594441 / 0.200704 = 2.96
Degree of freedom:
df₁ = n₁-1 = 11
df₂ = n₂-1 = 11
P-value :
P-value = 2*F.DIST.RT(2.9618, 11, 11) = 0.0853
Conclusion:
As p-value > α, we fail to reject the null
hypothesis.
the evidence supports the presumption of equal variances.
....................
c)
Level of Significance , α = 0.05
Degree of freedom, DF= n1+n2-2 =
22
t-critical value = t α/2 =
2.0739 (excel formula =t.inv(α/2,df)
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 0.6305
std error , SE = Sp*√(1/n1+1/n2) =
0.2574
margin of error, E = t*SE = 2.0739
* 0.26 = 0.53
difference of means = x̅1-x̅2 =
3.1100 - 2.040 =
1.0700
95% confidence interval is
Interval Lower Limit= (x̅1-x̅2) - E =
1.0700 - 0.5338 =
0.536
Interval Upper Limit= (x̅1-x̅2) + E =
1.0700 + 0.5338 =
1.604
95% CI will be wider than 90% CI
..................
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