Question

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 485 eggs in group I boxes, of which a field count showed about 276 hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 780 eggs in group II boxes, of which a field count showed about 270 hatched.

(a) Find a point estimate p̂₁ for p₁, the proportion of eggs that hatch in group I nest box placements. (Round your answer to three decimal places.)
p̂₁ =

Find a 99% confidence interval for p₁. (Round your answers to three decimal places.)

lower limit
upper limit

(b) Find a point estimate p̂₂ for p₂, the proportion of eggs that hatch in group II nest box placements. (Round your answer to three decimal places.)
p̂₂ =

Find a 99% confidence interval for p₂. (Round your answers to three decimal places.)

lower limit
upper limit

(c) Find a 99% confidence interval for p₁ − p₂. (Round your answers to three decimal places.)

lower limit
upper limit

Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes?

Because the interval contains only negative numbers, we can say that a higher proportion of eggs hatched in highly visible, closely grouped nesting boxes.Because the interval contains both positive and negative numbers, we can not say that a higher proportion of eggs hatched in well-separated and well-hidden nesting boxes.    We can not make any conclusions using this confidence interval.Because the interval contains only positive numbers, we can say that a higher proportion of eggs hatched in well-separated and well-hidden nesting boxes.

(d) What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

No conclusion can be made.A greater proportion of wood duck eggs hatch if the eggs are laid in well-separated, well-hidden nesting boxes.    The eggs hatch equally well in both conditions.A greater proportion of wood duck eggs hatch if the eggs are laid in highly visible, closely grouped nesting boxes.

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=276
sample size(n)=485
success rate ( p )= x/n = 0.569
I.
sample proportion = 0.569
standard error = Sqrt ( (0.569*0.431) /485) )
= 0.022
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.022
= 0.058
III.
CI = [ p ± margin of error ]
confidence interval = [0.569 ± 0.058]
= [ 0.511 , 0.627]
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DIRECT METHOD
given that,
possible chances (x)=276
sample size(n)=485
success rate ( p )= x/n = 0.569
CI = confidence interval
confidence interval = [ 0.569 ± 2.576 * Sqrt ( (0.569*0.431) /485) ) ]
= [0.569 - 2.576 * Sqrt ( (0.569*0.431) /485) , 0.569 + 2.576 * Sqrt ( (0.569*0.431) /485) ]
= [0.511 , 0.627]
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interpretations:
1. We are 99% sure that the interval [ 0.511 , 0.627] contains the true population proportion
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 true population proportion
b.
TRADITIONAL METHOD
given that,
possible chances (x)=270
sample size(n)=780
success rate ( p )= x/n = 0.346
I.
sample proportion = 0.346
standard error = Sqrt ( (0.346*0.654) /780) )
= 0.017
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.017
= 0.044
III.
CI = [ p ± margin of error ]
confidence interval = [0.346 ± 0.044]
= [ 0.302 , 0.39]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=270
sample size(n)=780
success rate ( p )= x/n = 0.346
CI = confidence interval
confidence interval = [ 0.346 ± 2.576 * Sqrt ( (0.346*0.654) /780) ) ]
= [0.346 - 2.576 * Sqrt ( (0.346*0.654) /780) , 0.346 + 2.576 * Sqrt ( (0.346*0.654) /780) ]
= [0.302 , 0.39]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.302 , 0.39] contains the true population proportion
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 true population proportion
c.
TRADITIONAL METHOD
given that,
sample one, x1 =276, n1 =485, p1= x1/n1=0.569
sample two, x2 =270, n2 =780, p2= x2/n2=0.346
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( (0.569*0.431/485) +(0.346 * 0.654/780))
=0.028
II.
margin of error = Z a/2 * (standard 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 * 0.028
=0.073
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.569-0.346) ±0.073]
= [ 0.15 , 0.296]
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DIRECT METHOD
given that,
sample one, x1 =276, n1 =485, p1= x1/n1=0.569
sample two, x2 =270, n2 =780, p2= x2/n2=0.346
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 = [ (0.569-0.346) ± 2.58 * 0.028]
= [ 0.15 , 0.296 ]
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interpretations:
1) we are 99% sure that the interval [ 0.15 , 0.296] 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
conclusion:
interval indicate that the proportion of eggs hatched from group I nest boxes is
lower than the proportion of eggs hatched from group II nest boxes.
Because the interval contains only positive numbers, we can say that a higher proportion of eggs hatched
in well-separated and well-hidden nesting boxes.
d.

conclusions about placement of nest boxes can be drawn
A greater proportion of wood duck eggs hatch if the eggs are laid in well-separated, well-hidden nesting boxes.