Allen\'s hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alth
ID: 3309766 • Letter: A
Question
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 16 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with = 0.26 gram.
(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)
margin of error?
(b) What conditions are necessary for your calculations? (Select all that apply.)
uniform distribution of weights
is unknown
normal distribution of weights
n is large
is known
(c) Give a brief interpretation of your results in the context of this problem.
The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.08 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
_______hummingbirds
margin of error?
(b) What conditions are necessary for your calculations? (Select all that apply.)
uniform distribution of weights
is unknown
normal distribution of weights
n is large
is known
(c) Give a brief interpretation of your results in the context of this problem.
The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.08 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
_______hummingbirds
Explanation / Answer
a.
TRADITIONAL METHOD
given that,
standard deviation, =0.26
sample mean, x =3.15
population size (n)=16
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.26/ sqrt ( 16) )
= 0.07
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.2
from standard normal table, two tailed z /2 =1.28
since our test is two-tailed
value of z table is 1.28
margin of error = 1.28 * 0.07
= 0.08
III.
CI = x ± margin of error
confidence interval = [ 3.15 ± 0.08 ]
= [ 3.07,3.23 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.26
sample mean, x =3.15
population size (n)=16
level of significance, = 0.2
from standard normal table, two tailed z /2 =1.28
since our test is two-tailed
value of z table is 1.28
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 3.15 ± Z a/2 ( 0.26/ Sqrt ( 16) ) ]
= [ 3.15 - 1.28 * (0.07) , 3.15 + 1.28 * (0.07) ]
= [ 3.07,3.23 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 80% sure that the interval [3.07 , 3.23 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 3.15
standard error =0.07
z table value = 1.28
margin of error = 0.08
confidence interval = [ 3.07 , 3.23 ]
b.
normal distribution of weights
n is large
is known
c.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
d.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.2% LOS is = 1.28 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.26
ME =0.08
n = ( 1.28*0.26/0.08) ^2
= (0.33/0.08 ) ^2
= 17.31 ~ 18 hummingbirds
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