Question

A random sample of 19 wolf litters in Ontario, Canada, gave an average of x₁ = 5.6 wolf pups per litter, with estimated sample standard deviation s₁ = 1.1. Another random sample of 6 wolf litters in Finland gave an average of x₂ = 3.8 wolf pups per litter, with sample standard deviation s₂ = 1.0.

(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

μ

p₁ – p₂

p

(b) Find an 99% confidence interval for μ₁ – μ₂, the difference in population mean litter size between Ontario and Finland. (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 99% level of confidence, does it appear that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland?

Because the interval contains only positive numbers, we can say that the average litter size of wolf pups is greater in Ontario.

Because the interval contains both positive and negative numbers, we can not say that the average litter size of wolf pups is greater in Ontario.

   We can not make any conclusions using this confidence interval.

Because the interval contains only negative numbers, we can say that the average litter size of wolf pups is greater in Finland.

Answer 1

a)

μ₁ – μ₂

_b)

_{Degree of freedom, DF=   n1+n2-2 =    23              
t-critical value =    t α/2 =    2.8073   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.0790              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.5053              
margin of error, E = t*SE =    2.8073   *   0.51   =   1.42  
                      
difference of means =    x̅1-x̅2 =    5.6000   -   3.800   =   1.8000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    1.8000   -   1.4186   =   0.4
Interval Upper Limit=   (x̅1-x̅2) + E =    1.8000   +   1.4186   =   3.2}

c)

Because the interval contains only positive numbers, we can say that the average litter size of wolf pups is greater in Ontario.