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 sufficient evidence to conclude that the population mean of species 1 is different from all of the other population means

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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
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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