Allen\'s hummingbird ( Selasphorus sasin ) has been studied by zoologist Bill Al
ID: 3336054 • Letter: A
Question
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 20 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 = 0.38 gram.
(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)
(b) What conditions are necessary for your calculations? (Select all that apply.)
n is large is unknown is knownuniform distribution of weightsnormal distribution of weights
(c) Interpret 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 a 20% 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 an 80% 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 margin of error E = 0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds
Explanation / Answer
TRADITIONAL METHOD
given that,
standard deviation, =0.38
sample mean, x =3.15
population size (n)=20
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.38/ sqrt ( 20) )
= 0.085
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 =1.282
since our test is two-tailed
value of z table is 1.282
margin of error = 1.282 * 0.085
= 0.1089
III.
CI = x ± margin of error
confidence interval = [ 3.15 ± 0.1089 ]
= [ 3.0411,3.2589 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.38
sample mean, x =3.15
population size (n)=20
level of significance, = 0.2
from standard normal table, two tailed z /2 =1.282
since our test is two-tailed
value of z table is 1.282
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 ( 0.38/ Sqrt ( 20) ) ]
= [ 3.15 - 1.282 * (0.085) , 3.15 + 1.282 * (0.085) ]
= [ 3.0411,3.2589 ]
-----------------------------------------------------------------------------------------------
[ANSWERS]
A.
Lower limit 3.0411
Upper limit 3.2589
Margi of error = 0.1089
B.
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 = 1.282 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.38
ME =0.16
n = ( 1.282*0.38/0.16) ^2
= (0.4872/0.16 ) ^2
= 9.2705 ~ 10
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.