A random sample of 16 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 97 lb with estimated sample standard deviation s1 = 6.3 lb. Another sample of 22 adult male wolves from Alaska gave an average weight x2 = 88 lb with estimated sample standard deviation s2 = 7.1 lb. (a) Let ?1 represent the population mean weight of adult male wolves from the Northwest Territories, and let ?2 represent the population mean weight of adult male wolves from Alaska. Find a 75% confidence interval for ?1 – ?2. (Round your answers to one decimal place.) lower limit upper limit (b) 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 75% 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. (c) Test the claim that the average weight of adult male wolves from the Northwest Territories is different from that of Alaska wolves. Use ? = 0.01. (i) What is the level of significance? State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 ≠ ?2H0: ?1 = ?2; H1: ?1 < ?2 H0: ?1 = ?2; H1: ?1 > ?2H0: ?1 ≠ ?2; H1: ?1 = ?2 (ii) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference ?1 − ?2. Round your answer to three decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Fail to reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions. Fail to reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.

A random sample of 16 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 97 lb with estimated sample standard deviation s1 = 6.3 lb. Another sample of 22 adult male wolves from Alaska gave an average weight x2 = 88 lb with estimated sample standard deviation s2 = 7.1 lb. (a) Let ?1 represent the population mean weight of adult male wolves from the Northwest Territories, and let ?2 represent the population mean weight of adult male wolves from Alaska. Find a 75% confidence interval for ?1 – ?2. (Round your answers to one decimal place.) lower limit upper limit (b) 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 75% 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. (c) Test the claim that the average weight of adult male wolves from the Northwest Territories is different from that of Alaska wolves. Use ? = 0.01. (i) What is the level of significance? State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 ≠ ?2H0: ?1 = ?2; H1: ?1 < ?2 H0: ?1 = ?2; H1: ?1 > ?2H0: ?1 ≠ ?2; H1: ?1 = ?2 (ii) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference ?1 − ?2. Round your answer to three decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Fail to reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions. Fail to reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

A random sample of 16 adult male wolves from the Canadian Northwest Territories gave an average weight x₁ = 97 lb with estimated sample standard deviation s₁ = 6.3 lb. Another sample of 22 adult male wolves from Alaska gave an average weight x₂ = 88 lb with estimated sample standard deviation s₂ = 7.1 lb.

(a) Let ?₁ represent the population mean weight of adult male wolves from the Northwest Territories, and let ?₂ represent the population mean weight of adult male wolves from Alaska. Find a 75% confidence interval for ?₁ – ?₂. (Round your answers to one decimal place.)

lower limit
upper limit

(b) 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 75% 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.

(c) Test the claim that the average weight of adult male wolves from the Northwest Territories is different from that of Alaska wolves. Use ? = 0.01.

(i) What is the level of significance?

State the null and alternate hypotheses.

H₀: ?₁ = ?₂; H₁: ?₁ ≠ ?₂H₀: ?₁ = ?₂; H₁: ?₁ < ?₂ H₀: ?₁ = ?₂; H₁: ?₁ > ?₂H₀: ?₁ ≠ ?₂; H₁: ?₁ = ?₂

(ii) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations.

What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference ?₁ − ?₂. Round your answer to three decimal places.)

(iii) Find (or estimate) the P-value.

P-value > 0.2500.125 < P-value < 0.250 0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

Sketch the sampling distribution and show the area corresponding to the P-value.

(iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ??

At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.

(v) Interpret your conclusion in the context of the application.

Reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Fail to reject the null hypothesis, there is insufficient evidence to suggest that the mean weights of the wolves are different in the two regions. Fail to reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.Reject the null hypothesis, there is sufficient evidence to suggest that the mean weights of the wolves are different in the two regions.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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