The following data represent petal lengths (in cm) for independent random samples of two species of iris. Petal length (in cm) of Iris virginica: x1; n1 = 36 5.0 5.8 6.1 6.1 5.1 5.5 5.3 5.5 6.9 5.0 4.9 6.0 4.8 6.1 5.6 5.1 5.6 4.8 5.4 5.1 5.1 5.9 5.2 5.7 5.4 4.5 6.4 5.3 5.5 6.7 5.7 4.9 4.8 5.8 5.1 5.1 Petal length (in cm) of Iris setosa: x2; n2 = 38 1.6 1.8 1.4 1.5 1.5 1.6 1.4 1.1 1.2 1.4 1.7 1.0 1.7 1.9 1.6 1.4 1.5 1.4 1.2 1.3 1.5 1.3 1.6 1.9 1.4 1.6 1.5 1.4 1.6 1.2 1.9 1.5 1.6 1.4 1.3 1.7 1.5 1.7 (a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to four decimal places.) x1= s1= x2= s2= (b) Let ?1 be the population mean for x1 and let ?2 be the population mean for x2. Find a 99% confidence interval for ?1 − ?2. (Round your answers to two decimal places.) lower limitupper limit (c) Explain what the confidence interval 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, is the population mean petal length of Iris virginica longer than that of Iris setosa? Because the interval contains only positive numbers, we can say that the mean petal length of Iris virginica is longer.Because the interval contains only negative numbers, we can say that the mean petal length of Iris virginica is shorter. Because the interval contains both positive and negative numbers, we cannot say that the mean petal length of Iris virginica is longer. (d) Which distribution did you use? Why? The standard normal distribution was used because ?1 and ?2 are known.The standard normal distribution was used because ?1 and ?2 are unknown. The Student's t-distribution was used because ?1 and ?2 are unknown.The Student's t-distribution was used because ?1 and ?2 are known. Do you need information about the petal length distributions? Explain. Both samples are large, so information about the distributions is needed.Both samples are large, so information about the distributions is not needed. Both samples are small, so information about the distributions is needed.Both samples are small, so information about the distributions is not needed.

The following data represent petal lengths (in cm) for independent random samples of two species of iris. Petal length (in cm) of Iris virginica: x1; n1 = 36 5.0 5.8 6.1 6.1 5.1 5.5 5.3 5.5 6.9 5.0 4.9 6.0 4.8 6.1 5.6 5.1 5.6 4.8 5.4 5.1 5.1 5.9 5.2 5.7 5.4 4.5 6.4 5.3 5.5 6.7 5.7 4.9 4.8 5.8 5.1 5.1 Petal length (in cm) of Iris setosa: x2; n2 = 38 1.6 1.8 1.4 1.5 1.5 1.6 1.4 1.1 1.2 1.4 1.7 1.0 1.7 1.9 1.6 1.4 1.5 1.4 1.2 1.3 1.5 1.3 1.6 1.9 1.4 1.6 1.5 1.4 1.6 1.2 1.9 1.5 1.6 1.4 1.3 1.7 1.5 1.7 (a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to four decimal places.) x1= s1= x2= s2= (b) Let ?1 be the population mean for x1 and let ?2 be the population mean for x2. Find a 99% confidence interval for ?1 − ?2. (Round your answers to two decimal places.) lower limitupper limit (c) Explain what the confidence interval 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, is the population mean petal length of Iris virginica longer than that of Iris setosa? Because the interval contains only positive numbers, we can say that the mean petal length of Iris virginica is longer.Because the interval contains only negative numbers, we can say that the mean petal length of Iris virginica is shorter. Because the interval contains both positive and negative numbers, we cannot say that the mean petal length of Iris virginica is longer. (d) Which distribution did you use? Why? The standard normal distribution was used because ?1 and ?2 are known.The standard normal distribution was used because ?1 and ?2 are unknown. The Student's t-distribution was used because ?1 and ?2 are unknown.The Student's t-distribution was used because ?1 and ?2 are known. Do you need information about the petal length distributions? Explain. Both samples are large, so information about the distributions is needed.Both samples are large, so information about the distributions is not needed. Both samples are small, so information about the distributions is needed.Both samples are small, so information about the distributions is not needed.

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

The following data represent petal lengths (in cm) for independent random samples of two species of iris.

Petal length (in cm) of Iris virginica: x₁; n₁ = 36

5.0	5.8	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.8	5.1	5.1

Petal length (in cm) of Iris setosa: x₂; n₂ = 38

1.6	1.8	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.7

(a)

Use a calculator with mean and standard deviation keys to calculate x₁, s₁, x₂, and s₂. (Round your answers to four decimal places.)

x₁=

s₁=

x₂=

s₂=

(b)

Let ?₁ be the population mean for x₁ and let ?₂ be the population mean for x₂. Find a 99% confidence interval for ?₁ − ?₂. (Round your answers to two decimal places.)

lower limitupper limit

(c)

Explain what the confidence interval 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, is the population mean petal length of Iris virginica longer than that of Iris setosa?

Because the interval contains only positive numbers, we can say that the mean petal length of Iris virginica is longer.Because the interval contains only negative numbers, we can say that the mean petal length of Iris virginica is shorter. Because the interval contains both positive and negative numbers, we cannot say that the mean petal length of Iris virginica is longer.

(d)

Which distribution did you use? Why?

The standard normal distribution was used because ?₁ and ?₂ are known.The standard normal distribution was used because ?₁ and ?₂ are unknown. The Student's t-distribution was used because ?₁ and ?₂ are unknown.The Student's t-distribution was used because ?₁ and ?₂ are known.

Do you need information about the petal length distributions? Explain.

Both samples are large, so information about the distributions is needed.Both samples are large, so information about the distributions is not needed. Both samples are small, so information about the distributions is needed.Both samples are small, so information about the distributions is not needed.

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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5.0	5.8	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.8	5.1	5.1

1.6	1.8	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.7

5.0	5.8	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.8	5.1	5.1

1.6	1.8	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.7

5.0	5.8	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.8	5.1	5.1

1.6	1.8	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.7