Question

In a scientific study that tries 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 conform approximately to a normal population and that the variances can be considered equal.

b. Assess whether the evidence supports the presumption of equal variances

c. Would it have been better to use a 95% range? Explain in detail

Answer 1

a)

Sample #1   ---->   1              
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

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   =   0.6280
Interval Upper Limit=   (x̅1-x̅2) + E =    1.0700   +   0.4420   =   1.5120

b)

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
Critical value(s):  
Lower tailed critical value, FL = F.INV(0.1/2, 11, 11) =    0.3549
Upper tailed critical value, FU = F.INV(1-0.1/2, 11, 11) =    2.8179
P-value :  
P-value = 2*F.DIST.RT(2.9618, 11, 11) =    0.0853
Conclusion:  
As p-value < α, we reject the null hypothesis.  

c)

Yes it would be good to have 95% range so that we can consider both sample equal variance.

Please let me know in case of any doubt.

Thanks in advance!

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