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 had the following weights (pounds):
69 103 126 122 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.
B) Find a 75% confidence interval for the population average weight of all adult mountain lions in the specified region.

Answer 1

Answer:

Step-by-step explanation:

From the information given:

Mean $\overline x = \dfrac{\sum x_i}{n}$

Mean $\overline x = \dfrac{69+103+126+122+60+64}{6}$

Mean $\overline x = \dfrac{544}{6}$

Mean $\overline x = 90.67$ pounds

Standard deviation $s = \sqrt{\dfrac {\sum (x_i - \overline x) ^2}{n-1}$

Standard deviation $s = \sqrt{\dfrac {(69 - 90.67)^2+(103 - 90.67)^2+ (126- 90.67) ^2+ ..+ (64 - 90.67)^2}{6-1}}$

Standard deviation s = 30.011 pounds

B) Find a 75% confidence interval for the population average weight of all adult mountain lions in the specified region.

At 75% confidence interval ; the level of significance ∝ = 1 - 0.75 = 0.25

$t_{(\alpha/2)}$ = 0.25/2

$t_{(\alpha/2)}$ = 0.125

t(0.125,5)=1.30

Degree of freedom = n - 1

Degree of freedom = 6 - 1

Degree of freedom = 5

Confidence interval  = $(\overline x - t_{(\alpha/2)(n-1)}(\dfrac{s}{\sqrt{n}})< \mu < (\overline x + t_{(\alpha/2)(n-1)}(\dfrac{s}{\sqrt{n}})$

Confidence interval  = $(90.67 - 1.30(\dfrac{30.011}{\sqrt{6}})< \mu < (90.67+ 1.30(\dfrac{30.011}{\sqrt{6}})$

Confidence interval  = $(90.67 - 1.30(12.252})< \mu < (90.67+ 1.30(12.252})$

Confidence interval  = $(90.67 - 15.9276 < \mu < (90.67+ 15.9276)$

Confidence interval  =  $(74.7424 < \mu <106.5976)$

i.e the lower limit = 74.74 pounds

    the upper limit = 106.60 pounds