MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Question

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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 Sample Sample Population size mean standard deviation 1 Species 1 ni = 20 17.153 cm/year SI = 2.666 cm/year s2 = 3.605 cm/year S3 = 3.774 cm/year 2 Species 2 n2 = 20 X2 — 13.983 сm/year 3 Species 3 n3 = 20 x3 = 15.120 cm/year 4 Species 4 n4 = 20 X4 = 14.328 cm/year S4 = 3.011 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 f P-value f-critical of variation Between groups 120.988 3 40.329 3.716 0.015 2.725 Within groups 824.710 76 10.851 Total 945.697 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to reject the null hypothesis at a significance level of a = 0.05. There is is sufficient evidence to conclude that the population mean of species 1 different from all of the other population means
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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 Sample Sample Population size mean standard deviation 1 Species 1 ni = 20 17.153 cm/year SI = 2.666 cm/year s2 = 3.605 cm/year S3 = 3.774 cm/year 2 Species 2 n2 = 20 X2 — 13.983 сm/year 3 Species 3 n3 = 20 x3 = 15.120 cm/year 4 Species 4 n4 = 20 X4 = 14.328 cm/year S4 = 3.011 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 f P-value f-critical of variation Between groups 120.988 3 40.329 3.716 0.015 2.725 Within groups 824.710 76 10.851 Total 945.697 79 Complete the sentences to state the decision and conclusion of the horticulturist's test. The decision is to reject the null hypothesis at a significance level of a = 0.05. There is is sufficient evidence to conclude that the population mean of species 1 different from all of the other population means
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