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.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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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