How much do wild mountain lions weigh? The 77th Annual Report of the New Mexico

Question

How much do wild mountain lions weigh? The 77th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Six adult wild mountain lions (18 months and older) captured and released for the first time in the San Andres Mountains gave the average weight of 91.0 pounds and a standard deviation of 30.7 pounds.

a. Based on the given data what is the point estimate for the average weight of all wild mountain lions?

b. What is the standard deviation of the point estimate x?

c. Find the critical value that will be used in constructing 75% confidence interval for the population mean?

d. Calculate 75% confidence interval for the average weight of all wild mountain lions.

e. Interpret the confidence interval found in part (d).

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, =30.7
sample mean, x =91
population size (n)=6
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 30.7/ sqrt ( 6) )
= 12.533
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.25
from standard normal table, two tailed z /2 =1.15
since our test is two-tailed
value of z table is 1.15
margin of error = 1.15 * 12.533
= 14.413
III.
CI = x ± margin of error
confidence interval = [ 91 ± 14.413 ]
= [ 76.587,105.413 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =30.7
sample mean, x =91
population size (n)=6
level of significance, = 0.25
from standard normal table, two tailed z /2 =1.15
since our test is two-tailed
value of z table is 1.15
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 = [ 91 ± Z a/2 ( 30.7/ Sqrt ( 6) ) ]
= [ 91 - 1.15 * (12.533) , 91 + 1.15 * (12.533) ]
= [ 76.587,105.413 ]
-----------------------------------------------------------------------------------------------

[ANSWERS]
a.
best point of estimate = mean = 91
b.
stanadard deviation = 30.7
standard error =12.533
c. critical value that will be used in constructing 75% confidence interval for the population mean
z table value = 1.15
margin of error = 14.413
d.
confidence interval = [ 76.587 , 105.413 ]
e.
interpretations:
1. we are 75% sure that the interval [76.587 , 105.413 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 75% of these intervals will contains the true population mean

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