Question

How much do wild mountain lions weigh? Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): 66 100 133 122 60 64 Assume that the population of x values has an approximately normal distribution.

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean weight x and sample standard deviation s. (Round your answers to one decimal place.)

x = _____ lb

s = _____ lb

(b) Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

lower limit = _______ lb

upper limit = _______ lb

Answer 1

Solution

Let xi = weight of ith adult wild mountain lion (18 months or older) captured and released for the first time in the San Andres Mountains, where i = 1 to n. [n = 6]

Back-up Theory

Sample Mean of x, xbar = (1/n) ∑[1,n]xi ……………………………………………. (1)

Sample Standard deviation of x, s = sqrt[{1/(n – 1)}∑[1,n]{xi – xbar}²]…………………. (2)

Or, equivalently, s = {1/(n – 1)}[{∑(1,n)(xi²)} – n.xbar²]………… …………………. (2a)

Given X ~ N(μ, σ²), 100(1 - α) % Confidence Interval for μ, when σ is not known is:

Xbar ± (t_{n- 1, α /2})s/√n ……………………………………………………………………(3) where Xbar = sample mean, t_{n – 1, α /2} = upper (α /2)% point of

t-distribution with (n - 1) degrees of freedom, s = sample standard deviation and n = sample size.

Now, to work out the answer,

Preparatory Work

i	1	2	3	4	5	6	Total
xi	100	66	133	122	60	64	545
xi2	10000	4356	17689	14884	3600	4096	54625

Part (a)

Sample mean weight, xbar = 545/6 = 90.8 lb ANSWER 1

Sample standard deviation, s = √[{54625 – (6 x 90.8²)}/5] = 32.0 lb ANSWER 2

Part (b)

Given, 75% confidence => α = 0.25, and so t_{n- 1, α /2} = upper 12.5% of t₅

= 1.301[using Excel Function: TINV, Probability, Degrees_freedom]

So, 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region = 90.8 ± (1.301x 32)/√5

= 90.8 ± 18.6.

Thus.

lower limit = 72.2 lb

upper limit = 109.4 lb ANSWER