Question

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 Sample Sample Population description size standard deviation 17.153 cm/year 2.666 cm/year x2 = 13.983 cm/year s2-3.605 cm/year 15.120 cm/year3 3.774 cm/year x4-14.328 cm/year s.-3.011cm/year mean 1 Species 1 n1-20 Species 2 n2=20 3 Species 3 n 20 Species 4 n4-20 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 of variation Between groups Within groups Total ss MS p-valuef-critical 120.988 3 40.329 3.7160.0152.725 824.710 76 10851 945.697 79Complete the sentence to state the decision and conclusion of horticulturist test.The decision is to ________the ___________at a significant level of________ α=0.05 There is _______ evidence to conclude that __________ is__________.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean  is different from  at least of the population  

Step-by-step explanation:

From the question we are told that the claim is

     The mean growth rates of all four species are equal.

The  null hypothesis is  

             $H_o : \mu _1 = \mu_2 = \mu_3 = \mu_4$

Th alternative hypothesis is    

             $H_a: at \ least \ one \ of \ the \ means \ is \not\ equal$

From question the p-value is $p-value = 0.015$

  And since the $p-value < \alpha$ so the null hypothesis will be rejected

So  

   The decision is to reject the null hypothesis at a significant level of significance $\alpha = 0.05$ There is sufficient evidence to conclude that at least one of the population mean  is different from  at least of the population