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. to. 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. (10 pts)

b. Assess whether the evidence supports the presumption of equal variances (5 pts)

c. Would it have been better to use a 95% range? Explain in detail (5 pts)

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=3.11
standard deviation , s.d1=0.771
number(n1)=12
y(mean)=2.04
standard deviation, s.d2 =0.448
number(n2)=12
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (11*0.594 + 11*0.201) / (24- 2 )
s^2 = 0.398
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 0.398 * (1/12+1/12) )
=0.258
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 22 d.f is 1.717
margin of error = 1.717 * 0.258
= 0.442
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (3.11-2.04) ± 0.442 ]
= [0.628 , 1.512]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=3.11
standard deviation , s.d1=0.771
sample size, n1=12
y(mean)=2.04
standard deviation, s.d2 =0.448
sample size,n2 =12
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 3.11-2.04) ± t a/2 * sqrt( 0.398 * (1/12+1/12) ]
= [ (1.07) ± 0.442 ]
= [0.628 , 1.512]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [0.628 , 1.512]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
Given that,
mean(x)=3.11
standard deviation , s.d1=0.771
number(n1)=12
y(mean)=2.04
standard deviation, s.d2 =0.448
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.717
since our test is two-tailed
reject Ho, if to < -1.717 OR if to > 1.717
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (11*0.594 + 11*0.201) / (24- 2 )
s^2 = 0.398
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=3.11-2.04/sqrt((0.398( 1 /12+ 1/12 ))
to=1.07/0.258
to=4.154
| to | =4.154
critical value
the value of |t α| with (n1+n2-2) i.e 22 d.f is 1.717
we got |to| = 4.154 & | t α | = 1.717
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 4.1545 ) = 0.0004
hence value of p0.1 > 0.0004,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 4.154
critical value: -1.717 , 1.717
decision: reject Ho
p-value: 0.0004
we have enough evidence to support the claim that difference of means between two locations.
c.TRADITIONAL METHOD
given that,
mean(x)=3.11
standard deviation , s.d1=0.771
number(n1)=12
y(mean)=2.04
standard deviation, s.d2 =0.448
number(n2)=12
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (11*0.594 + 11*0.201) / (24- 2 )
s^2 = 0.398
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 0.398 * (1/12+1/12) )
=0.258
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 22 d.f is 2.074
margin of error = 2.074 * 0.258
= 0.534
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (3.11-2.04) ± 0.534 ]
= [0.536 , 1.604]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=3.11
standard deviation , s.d1=0.771
sample size, n1=12
y(mean)=2.04
standard deviation, s.d2 =0.448
sample size,n2 =12
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 3.11-2.04) ± t a/2 * sqrt( 0.398 * (1/12+1/12) ]
= [ (1.07) ± 0.534 ]
= [0.536 , 1.604]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.536 , 1.604]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
yes,
better use in 95% range is [0.536 , 1.604]