A random sample of 21 wolf litters in Ontario, Canada, gave an average of x1 = 4.7 wolf pups per litter, with estimated sample standard deviation s1 = 1.0. Another random sample of 7 wolf litters in Finland gave an average of x2 = 2.8 wolf pups per litter, with sample standard deviation s2 = 1.4. (a) Find an 85% confidence interval for ?1 – ?2, the difference in population mean litter size between Ontario and Finland. (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 85% 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. (ii) What sampling distribution will you use? What assumptions are you making? 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 unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known 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.

A random sample of 21 wolf litters in Ontario, Canada, gave an average of x1 = 4.7 wolf pups per litter, with estimated sample standard deviation s1 = 1.0. Another random sample of 7 wolf litters in Finland gave an average of x2 = 2.8 wolf pups per litter, with sample standard deviation s2 = 1.4. (a) Find an 85% confidence interval for ?1 – ?2, the difference in population mean litter size between Ontario and Finland. (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 85% 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. (ii) What sampling distribution will you use? What assumptions are you making? 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 unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known 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.

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 21 wolf litters in Ontario, Canada, gave an average of x₁ = 4.7 wolf pups per litter, with estimated sample standard deviation s₁ = 1.0. Another random sample of 7 wolf litters in Finland gave an average of x₂ = 2.8 wolf pups per litter, with sample standard deviation s₂ = 1.4.

(a) Find an 85% confidence interval for ?₁ – ?₂, the difference in population mean litter size between Ontario and Finland. (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 85% 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.

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

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 unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known 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.

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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