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Give as much information as you can about the \(P\) -value of the single-factor ANOVA \(F\) test in each of the following situations. a. \(k=5, n_{1}=n_{2}=n_{3}=n_{4}=n_{5}=4, F=5.37\) b. \(k=5, n_{1}=n_{2}=n_{3}=5, n_{4}=n_{5}=4, F=2.83\) c. \(k=3, n_{1}=4, n_{2}=5, n_{3}=6, F=5.02\) d. \(k=3, n_{1}=n_{2}=4, n_{3}=6, F=15.90\) e. \(k=4, n_{1}=n_{2}=15, n_{3}=12, n_{4}=10, F=1.75\)

Short Answer

Expert verified
To find P-value for each scenario, calculate the degrees of freedom using group and residual counts, then identify the corresponding P-value using an F-distribution table or statistical software against the provided F-value. Repeat for each scenario.

Step by step solution

01

Calculate Degrees of Freedom (DOF)

Calculating the degrees of freedom is crucial in determining p-value. Degrees of Freedom are calculated using the formula: DOF groups = k - 1 and DOF residuals = N - k, where N is the total number of observations.
02

Find P-value for a

Using the given values, k = 5 and each group (n1 = n2 = n3 = n4 = n5) has 4 observations. Thus, DOF groups = k - 1 = 5 - 1 = 4. DOF residuals = N - k = (4 * 5) - 5 = 15. Now, looking these values (4, 15) up in the F-distribution table or using statistical software with F = 5.37, we can find the corresponding P-value.
03

Repeat for b, c, d, e

Repeat the process for each situation given in the exercise. For each, start by finding the degrees of freedom, then look up the F-value in the F-distribution table to find the corresponding P-value. Always take into account the number of groups and number of observations in each group.

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Most popular questions from this chapter

The accompanying data resulted from a flammability study in which specimens of five different fabrics were tested to determine burn times. \(\begin{array}{rrrrrr} & 1 & 17.8 & 16.2 & 15.9 & 15.5 \\ & 2 & 13.2 & 10.4 & 11.3 & \\ \text { Fabric } & 3 & 11.8 & 11.0 & 9.2 & 10.0 \\ 4 & 16.5 & 15.3 & 14.1 & 15.0 & 13.9 \\ 5 & 13.9 & 10.8 & 12.8 & 11.7 & \\ & \text { MSTr } & =23.67 & & & \\ & \text { MSE } & =1.39 & & & \end{array}\) \(F=17.08\) \(P\) -value \(=.000\) The accompanying output gives the T-K intervals as calculated by Minitab. Identify significant differences and give the underscoring pattern. Individual error rate \(=0.00750\) Critical value \(=4.37\) Intervals for (column level mean) - (row level mean) \(\begin{array}{lrrrr} & 1 & 2 & 3 & 4 \\ & 1.938 & & & \\ 2 & 7.495 & & & \\ & 3.278 & -1.645 & & \\ 3 & 8.422 & 3.912 & & \\ & -1.050 & -5.983 & -6.900 & \\ 4 & 3.830 & -0.670 & -2.020 & \\ & 1.478 & -3.445 & -4.372 & 0.220 \\ 5 & 6.622 & 2.112 & 0.772 & 5.100\end{array}\)

Do lizards play a role in spreading plant seeds? Some research carried out in South Africa would suggest so ("Dispersal of Namaqua Fig [Ficus cordata cordata] Seeds by the Augrabies Flat Lizard [Platysaurus broadleyil." Journal of Herpetology [1999]: 328-330). The researchers collected 400 seeds of this particular type of fig, 100 of which were from each treatment: lizard dung, bird dung, rock hyrax dung, and uneaten figs. They planted these seeds in batches of 5 , and for each group of 5 they recorded how many of the seeds germinated. This resulted in 20 observations for each treatment. The treatment means and standard deviations are given in the accompanying table. $$ \begin{array}{lccc} \text { Treatment } & n & \bar{x} & s \\ \hline \text { Uneaten figs } & 20 & 2.40 & .30 \\ \text { Lizard dung } & 20 & 2.35 & .33 \\ \text { Bird dung } & 20 & 1.70 & .34 \\ \text { Hyrax dung } & 20 & 1.45 & .28 \\ \hline \end{array} $$ a. Construct the appropriate ANOVA table, and test the hypothesis that there is no difference between mean number of seeds germinating for the four treatments. b. Is there evidence that seeds eaten and then excreted by lizards germinate at a higher rate than those eaten and then excreted by birds? Give statistical evidence to support your answer.

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