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Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary her data. Population Sample description Sample Sample standard deviation Population size mean x1 = 15.123 cm/year x2 = 14.957 cm/year x3 = 14.289 cm/year X4 = 12.615 cm/year s1 = 2.376 cm/year s2 = 3.856 cm/year $3 = 3.566 cm/year S4 = 2.651 cm/year 1 Species 1 nį = 20 2 Species 2 n2 = 20 Species 3 Species 4 3 n3 = 20 %3D 4 n4 = 20 The growth rate distributions of each sample are approximately normal, and the data do not contain outliers. The horticulturi uses a one-way analysis of variance (ANOVA) at a significance level of a = 0.05 to test if the mean growth rates of all four species are equal. Her results are shown in the table. Source S df MS f P-value f-critical of variation Between groups 78.798 3 26.266 2.610 0.058 2.725 Within groups 764.871 76 10.064 Total 843.667 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to the at a significance level of a = 0.05. There is evidence to conclude that is

Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary her data. Population Sample description Sample Sample standard deviation Population size mean x1 = 15.123 cm/year x2 = 14.957 cm/year x3 = 14.289 cm/year X4 = 12.615 cm/year s1 = 2.376 cm/year s2 = 3.856 cm/year $3 = 3.566 cm/year S4 = 2.651 cm/year 1 Species 1 nį = 20 2 Species 2 n2 = 20 Species 3 Species 4 3 n3 = 20 %3D 4 n4 = 20 The growth rate distributions of each sample are approximately normal, and the data do not contain outliers. The horticulturi uses a one-way analysis of variance (ANOVA) at a significance level of a = 0.05 to test if the mean growth rates of all four species are equal. Her results are shown in the table. Source S df MS f P-value f-critical of variation Between groups 78.798 3 26.266 2.610 0.058 2.725 Within groups 764.871 76 10.064 Total 843.667 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to the at a significance level of a = 0.05. There is evidence to conclude that is

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary of her data. Population Sample description Species 1 Sample Sample Population size mean standard deviation x1 = 15.123 cm/year x2 = 14.957 cm/year 1 s1 = 2.376 cm/year s2 = 3.856 cm/year s3 = 3.566 cm/year ni = 20 Species 2 nz = 20 3 Species 3 n3 = 20 X3 = 14.289 cm/year 4 Species 4 n4 = 20 x4 = 12.615 cm/year $4 = 2.651 cm/year The growth rate distributions of each sample are approximately normal, and the data do not contain outliers. The horticulturist uses a one-way analysis of variance (ANOVA) at a significance level of a = 0.05 to test if the mean growth rates of all four species are equal. Her results are shown in the table. Source SS df MS P-value f-critical of variation Between groups 78.798 3 26.266 2.610 0.058 2.725 Within groups 764.871 10.064 76 Total 843.667 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to the at a significance level of a = 0.05. There is evidence to conclude that is Answer Bank sufficient the same as at least one of the population means null hypothesis alternative hypothesis insufficient the population mean of species 1 each of the tion means the population mean of species 4 at least one other population mean fail to accept different from the population mean of species 2 ассept all of the other population means reject fail to reject the population mean of species 3
Transcribed Image Text:Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a greenhouse. The shrubs were grown from a random sample of seeds, and they were all grown in the same soil mixture and in the same size pot. To ensure that any slight differences in the environmental conditions throughout the greenhouse are not confounded with species, she randomizes the location of the pots throughout the greenhouse. The table contains a summary of her data. Population Sample description Species 1 Sample Sample Population size mean standard deviation x1 = 15.123 cm/year x2 = 14.957 cm/year 1 s1 = 2.376 cm/year s2 = 3.856 cm/year s3 = 3.566 cm/year ni = 20 Species 2 nz = 20 3 Species 3 n3 = 20 X3 = 14.289 cm/year 4 Species 4 n4 = 20 x4 = 12.615 cm/year $4 = 2.651 cm/year The growth rate distributions of each sample are approximately normal, and the data do not contain outliers. The horticulturist uses a one-way analysis of variance (ANOVA) at a significance level of a = 0.05 to test if the mean growth rates of all four species are equal. Her results are shown in the table. Source SS df MS P-value f-critical of variation Between groups 78.798 3 26.266 2.610 0.058 2.725 Within groups 764.871 10.064 76 Total 843.667 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to the at a significance level of a = 0.05. There is evidence to conclude that is Answer Bank sufficient the same as at least one of the population means null hypothesis alternative hypothesis insufficient the population mean of species 1 each of the tion means the population mean of species 4 at least one other population mean fail to accept different from the population mean of species 2 ассept all of the other population means reject fail to reject the population mean of species 3
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