Question

You discover an isolated population of island squirrels and collect 200 of them, finding leucism 12. Perform a hypothesis test for a difference in the proportions of leucism among this island population and the previously considered population. Report your conclusion at both the a=.01 and a=.05 level.

Answer 1

a.
TRADITIONAL METHOD
given that,
sample one, x1 =12, n1 =200, p1= x1/n1=0.06
sample two, x2 =188, n2 =200, p2= x2/n2=0.94
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.06*0.94/200) +(0.94 * 0.06/200))
=0.0237
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.58
margin of error = 2.58 * 0.0237
=0.0613
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.06-0.94) ±0.0613]
= [ -0.9413 , -0.8187]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =12, n1 =200, p1= x1/n1=0.06
sample two, x2 =188, n2 =200, p2= x2/n2=0.94
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.06-0.94) ± 2.58 * 0.0237]
= [ -0.9413 , -0.8187 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ -0.9413 , -0.8187] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the difference between
true population mean P1-P2

b.
TRADITIONAL METHOD
given that,
sample one, x1 =12, n1 =200, p1= x1/n1=0.06
sample two, x2 =188, n2 =200, p2= x2/n2=0.94
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.06*0.94/200) +(0.94 * 0.06/200))
=0.0237
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.0237
=0.0465
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.06-0.94) ±0.0465]
= [ -0.9265 , -0.8335]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =12, n1 =200, p1= x1/n1=0.06
sample two, x2 =188, n2 =200, p2= x2/n2=0.94
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.06-0.94) ± 1.96 * 0.0237]
= [ -0.9265 , -0.8335 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.9265 , -0.8335] contains the difference between
true population proportion P1-P2
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 difference between
true population mean P1-P2

conclusion:
at level of significance =0.01 then confidence interval = [ -0.9413 , -0.8187 ]
at level of significance = 0.05 then confidence interval = [ -0.9265 , -0.8335 ]
there is slight change in the difference in the proportions of leucism among this island population and the previously considered population