Question

A random sample of 22 adult male wolves from the Canadian Northwest Territories gave an average weight x₁ = 96.0 pounds with estimated sample standard deviation s₁ = 5.7 pounds. Another sample of 28 adult male wolves from Alaska gave an average weight x₂ = 88.0 pounds with estimated sample standard deviation s₂ = 6.2 pounds.

(a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference of means μ₁ – μ₂, or difference of proportions p₁ – p₂. Then solve the problem.

μ₁ – μ₂

p

p₁ – p₂

μ

(b) Let μ₁ represent the population mean weight of adult male wolves from the Northwest Territories, and let μ₂ represent the population mean weight of adult male wolves from Alaska. Find a 95% confidence interval for μ₁ – μ₂. (Use 1 decimal place.)

lower limit
upper limit

(c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 95% level of confidence, does it appear that the average weight of adult male wolves from the Northwest Territories is greater than that of the Alaska wolves?

Because the interval contains only positive numbers, we can say that Canadian wolves weigh more than Alaskan wolves.

Because the interval contains both positive and negative numbers, we can not say that Canadian wolves weigh more than Alaskan wolves.    

We can not make any conclusions using this confidence interval.

Because the interval contains only negative numbers, we can say that Alaskan wolves weigh more than Canadian wolves.

Answer 1

Part a)

μ₁ – μ₂

Part b)

Confidence interval :-

t(α/2, DF) = t(0.05 /2, 46 ) = 2.013

DF = 46

Lower Limit =
Lower Limit = 4.6019 ≈ 4.6
Upper Limit =
Upper Limit = 11.3981 ≈ 11.4
95% Confidence interval is ( 4.6 , 11.4 )

Part c)

Because the interval contains only positive numbers, we can say that Canadian wolves weigh more than Alaskan wolves.