Question

A random sample of 15 adult male wolves from the Canadian Northwest Territories gave an average...

A random sample of 15 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 96.0 pounds with estimated sample standard deviation s1 = 7.5 pounds. Another sample of 27 adult male wolves from Alaska gave an average weight x2 = 89.0 pounds with estimated sample standard deviation s2 = 7.5 pounds. (a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference of means μ1μ2, or difference of proportions p1p2. Then solve the problem.

1 μ1μ2

2p

3    μ

4 p1p2

(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 90% confidence interval for μ1μ2. (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 90% 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.

Homework Answers

Answer #1

(A) we are find the mean difference for Alaskan wolve and Canadian wolves.

so, point estimate =

(B) Using TI 84 calculator

press stat then tests then 2-sampTInt

enter the data  

Pooled: yes

C-level = 0.90

press ENTER

lower limit = 2.9

upper limit = 11.1

(C) Since the confidence interval is calculated for mean of Canadian wolves minus the mean of Alaska wolves and we get positive limits for the confidence interval

this means that the mean for canadian wolves is greater than mean for Alaska wolves

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

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A random sample of 17 adult male wolves from the Canadian Northwest Territories gave an average...
A random sample of 17 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 97.4 pounds with estimated sample standard deviation s1 = 5.5 pounds. Another sample of 22 adult male wolves from Alaska gave an average weight x2 = 89.6 pounds with estimated sample standard deviation s2 = 6.9 pounds. Please show all steps in getting the answer. Thanks (a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference...
A random sample of 22 adult male wolves from the Canadian Northwest Territories gave an average...
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...
A random sample of 18 male wolves from the canadian northwest territories gave an average weight...
A random sample of 18 male wolves from the canadian northwest territories gave an average weight of 98 with a standard deviation of 6.5 pounds. another sample of 24 adult male wolves from alaska gave an average weight of 99 pounds and a standard deviation of 7.3 pounds . test the claim that the average weight of the canadian wolves is lower than the average weight of the alaskan wolves. use a=0.05 A)Claim_____B)Null Hypothesis______ C)Alternative Hypo D) Test to be...
A random sample of 19 wolf litters in Ontario, Canada, gave an average of x1 =...
A random sample of 19 wolf litters in Ontario, Canada, gave an average of x1 = 4.8 wolf pups per litter, with estimated sample standard deviation s1 = 1.0. Another random sample of 10 wolf litters in Finland gave an average of x2 = 3.0 wolf pups per litter, with sample standard deviation s2 = 1.4. (a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference of means μ1 – μ2, or difference of proportions...
On the Navajo Reservation, a random sample of 228 permanent dwellings in the Fort Defiance region...
On the Navajo Reservation, a random sample of 228 permanent dwellings in the Fort Defiance region showed that 56 were traditional Navajo hogans. In the Indian Wells region, a random sample of 145 permanent dwellings showed that 18 were traditional hogans. Let p1 be the population proportion of all traditional hogans in the Fort Defiance region, and let p2 be the population proportion of all traditional hogans in the Indian Wells region. (a) Find a 90% confidence interval for p1...
How profitable are different sectors of the stock market? One way to answer such a question...
How profitable are different sectors of the stock market? One way to answer such a question is to examine profit as a percentage of stockholder equity. A random sample of 35 retail stocks such as Toys 'R' Us, Best Buy, and Gap was studied for x1, profit as a percentage of stockholder equity. The result was x1 = 15.0. A random sample of 38 utility (gas and electric) stocks such as Boston Edison, Wisconsin Energy, and Texas Utilities was studied...
A random sample of 364 married couples found that 280 had two or more personality preferences...
A random sample of 364 married couples found that 280 had two or more personality preferences in common. In another random sample of 570 married couples, it was found that only 40 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 99% confidence interval for...
Most married couples have two or three personality preferences in common. A random sample of 362...
Most married couples have two or three personality preferences in common. A random sample of 362 married couples found that 140 had three preferences in common. Another random sample of 564couples showed that 220 had two personality preferences in common. Let p1 be the population proportion of all married couples who have three personality preferences in common. Let p2 be the population proportion of all married couples who have two personality preferences in common. (a) Find a 95% confidence interval...
A random sample of 380 married couples found that 280 had two or more personality preferences...
A random sample of 380 married couples found that 280 had two or more personality preferences in common. In another random sample of 578 married couples, it was found that only 24 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 90% confidence interval for...
A random sample of 390 married couples found that 298 had two or more personality preferences...
A random sample of 390 married couples found that 298 had two or more personality preferences in common. In another random sample of 582 married couples, it was found that only 28 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 95% confidence interval for...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT