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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. \(k=5, n_{1}=n_{2} \equiv n_{3}=n_{4}=n_{5}=4, F=5.37\) 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
The P-values can't be exactly calculated here without use of the F-distribution table or statistical software but are going to be a measure showing if the evidence against the null hypothesis is strong or not for each provided situation.

Step by step solution

01

Understanding the problem and definitions

The P-value is a probability that provides a measure of the evidence against the null hypothesis provided by the data. The null hypothesis (H0) for ANOVA test is that all group means are equal. The smaller the P-value, the stronger the evidence against H0. We are given the number of groups (k), the number of observations in each group (n1, n2, ..nk), and the F value, a factor calculated from the data which follows an F-distribution.
02

Determine degrees of freedom

The degrees of freedom in the denominator (df1) is equal to the total number of groups (k) minus 1. The degrees of freedom in the numerator (df2) is equal to the total number of observations (n1 + n2 +...+ nk) minus k. The F value provided is a critical value and is used along with the degrees of freedom to find the P-value using the F-distribution table or statistical software.
03

Compute P-values

Now, having understood the F distribution and calculated degrees of freedom, we can find the P-values for each situation by looking them up in the F-distribution table or by calculating using statistical software.
04

Interpret the results

Now that we have calculated the P-values, the last step is to interpret the result. If the P-value is less than 0.05 (commonly chosen significance level), we reject the null hypothesis, indicating at least one group mean is different from the others. If the P-value is greater than 0.05, we do not reject the null hypothesis, indicating no significant difference in the group means. By providing P-values, we give an indication of the strength of evidence against the null hypothesis in each of the situations we are analyzing.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

F-distribution
When performing an ANOVA test, or Analysis of Variance, it's crucial to understand the concept of the F-distribution. This is a probability distribution that arises when we consider the ratio of two independent chi-squared variables, each divided by their respective degrees of freedom.
Think of the F-distribution as a way to measure how much variation there is within groups compared to between groups in an experiment. It's an essential part of hypothesis testing and is used to calculate the critical F-value during an ANOVA test.
In simpler terms, if you obtain an F-value from your data that falls into the extreme tail of the F-distribution, it suggests that the observed variation is unlikely to be due to random chance alone.
Degrees of Freedom
Degrees of freedom (df) are vital in statistical tests as they describe the number of independent pieces of information used to calculate a statistic. In the context of ANOVA, degrees of freedom break down into two types: within-groups and between-groups.
For between-groups (numerator df), subtract 1 from the number of groups. For within-groups (denominator df), subtract the number of groups from the total number of observations in all groups.
Understanding how to correctly calculate degrees of freedom is essential, as it affects the shape of the F-distribution and, consequently, the validity of the test results.
Null Hypothesis
The null hypothesis, often symbolized as H0, is a fundamental concept in statistical testing. In the realm of ANOVA, it states that there is no difference in the population means of the groups being tested.
Upholding the null hypothesis suggests that any differences observed in the sample means are simply the result of random variance, whereas rejecting it implies that at least one group mean is statistically significantly different from the others.
Understanding the null hypothesis is crucial because it provides a baseline or standard against which the observed data is compared. It's the presumed truth that your experiment seeks to challenge with evidence from sample data.
Evidence against Null Hypothesis
The p-value is the protagonist in the story of evidence against the null hypothesis. It quantifies the strength of the evidence against H0, or the likelihood of obtaining a test statistic at least as extreme as the one observed if the null hypothesis were true.
A low p-value indicates strong evidence against the null hypothesis, suggesting that the sample data is not consistent with H0. In contrast, a high p-value suggests weak evidence against H0, implying the sample data does not provide sufficient proof to conclude there is a significant difference in group means.
As a result, the p-value is the pivot around which the decision to reject or fail to reject the null hypothesis revolves. It encapsulates the uncertainty, allowing researchers to make informed decisions about the validity of their hypotheses.

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Most popular questions from this chapter

Leaf surface area is an important variable in plant gas-exchange rates. The article "Fluidized Bed Coating of Conifer Needles with Glass Beads for Determination of Leaf Surface Area" (Forest Science [1980]: 29-32) included an analysis of dry matter per unit surface area \(\left(\mathrm{mg} / \mathrm{cm}^{3}\right)\) for trees raised under three different growing conditions. Let \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) represent the true mean dry matter per unit surface area for the growing conditions 1 , 2 , and 3 , respectively. The given \(95 \%\) simultaneous confidence intervals are based on summary quantities that appear in the article: \(\begin{array}{llll}\text { Difference } & \mu_{1}-\mu_{2} & \mu_{1}-\mu_{3} & \mu_{2}-\mu_{3}\end{array}\) \(\begin{array}{llll}\text { Interval } & (-3.11,-1.11) & (-4.06,-2.06) & (-1.95, .05)\end{array}\) Which of the following four statements do you think describes the relationship between \(\mu_{1}, \mu_{2}\), and \(\mu_{3} ?\) Explain your choice. a. \(\mu_{1}=\mu_{2}\), and \(\mu_{3}\) differs from \(\mu_{1}\) and \(\mu_{2}\). b. \(\mu_{1}=\mu_{3}\), and \(\mu_{2}\) differs from \(\mu_{1}\) and \(\mu_{3}\). c. \(\mu_{2}=\mu_{3}\), and \(\mu_{1}\) differs from \(\mu_{2}\) and \(\mu_{3}\). d. All three \(\mu\) 's are different from one another.

Some investigators think that the concentration \((\mathrm{mg} / \mathrm{mL})\) of a particular antigen in supernatant fluids could be related to onset of meningitis in infants. The accompanying data are typical of that given in plots in the article "Type-Specific Capsular Antigen Is Associated with Virulence in Late-Onset Group B Streptococcal Type III Disease" (Infection and Immunity [1984]: 124-129). Construct an ANOVA table, and use it to test the null hypothesis of no difference in mean antigen concentrations for the groups. $$\begin{array}{lllllll} \text { Asymptomatic infants } & 1.56 & 1.06 & 0.87 & 1.39 & 0.71 & 0.87 & \\ \text { Infants with late onset sepsis } & 1.51 & 1.78 & 1.45 & 1.13 & 1.87 & 1.89 & 1.071 .72 \\ \text { Infants with late onset meningitis } & 1.21 & 1.34 & 1.95 & 2.27 & 0.88 & 1.67 & 2.57 \end{array}$$

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

The experiment described in Example \(15.4\) also gave data on change in body fat mass for men ("Growth Hormone and Sex Steroid Administration in Healthy Aged Women and Men," Journal of the American Medical Association [2002]: 2282-2292). Each of 74 male subjects who were over age 65 was assigned at random to one of the following four treatments: (1) placebo "growth hormone" and placebo "steroid" (denoted by \(\mathrm{P}+\mathrm{P}),(2)\) placebo "growth hormone" and the steroid testosterone (denoted by \(\mathrm{P}+\mathrm{S}\) ), (3) growth hormone and placebo "steroid" (denoted by G + P), and (4) growth hormone and the steroid testosterone (denoted by \(\mathrm{G}+\mathrm{S}\) ). The accompanying table lists data on change in body fat mass over the 26-week period following the treatment that are consistent with summary quantities given in the article $$\begin{array}{rrrr} \text { Treatment } \quad \mathbf{P}+\mathbf{P} & \mathbf{P}+\mathbf{S} & \mathbf{G}+\mathbf{P} & \mathbf{G}+\mathbf{S} \\ \hline 0.3 & -3.7 & -3.8 & -5.0 \\ 0.4 & -1.0 & -3.2 & -5.0 \\ -1.7 & 0.2 & -4.9 & -3.0 \\ -0.5 & -2.3 & -5.2 & -2.6 \\ -2.1 & 1.5 & -2.2 & -6.2 \\ 1.3 & -1.4 & -3.5 & -7.0 \\ 0.8 & 1.2 & -4.4 & -4.5 \\ 1.5 & -2.5 & -0.8 & -4.2 \\ -1.2 & -3.3 & -1.8 & -5.2 \\ -0.2 & 0.2 & -4.0 & -6.2 \\ 1.7 & 0.6 & -1.9 & -4.0 \\ 1.2 & -0.7 & -3.0 & -3.9 \end{array}$$ $$\begin{array}{rrrrr} \text { Treatment } & \mathbf{P}+\mathbf{P} & \mathbf{P}+\mathbf{S} & \mathbf{G}+\mathbf{P} & \mathbf{G}+\mathbf{S} \\ \hline & 0.6 & -0.1 & -1.8 & -3.3 \\ & 0.4 & -3.1 & -2.9 & -5.7 \\ & -1.3 & 0.3 & -2.9 & -4.5 \\ & -0.2 & -0.5 & -2.9 & -4.3 \\ & 0.7 & -0.8 & -3.7 & -4.0 \\ & & -0.7 & & -4.2 \\ & & -0.9 & & -4.7 \\ & & -2.0 & & \\ & & -0.6 & & \\ n & 17 & 21 & 17 & 19 \\ \bar{x} & 0.100 & -0.933 & -3.112 & -4.605 \\ s & 1.139 & 1.443 & 1.178 & 1.122 \\ s^{2} & 1.297 & 2.082 & 1.388 & 1.259 \end{array}$$ Also, \(N=74\), grand total \(=-158.3\), and \(\overline{\bar{x}}=\frac{-158.3}{74}=\) \(-2.139 .\) Carry out an \(F\) test to see whether true mean change in body fat mass differs for the four treatments.

The accompanying underscoring pattern appeared in the article "Effect of \(\mathrm{SO}_{2}\) on Transpiration, Chlorophyll Content, Growth, and Injury in Young Seedlings of Woody Angiosperms" ( \mathrm{\\{} C a n a d i a n ~ J o u r n a l ~ o f ~ F o r e s t ~ R e s e a r c h ~ [1980]: 78-81). Water loss of plants (Acer saccharinum) exposed to \(0,2,4,8\), and 16 hours of fumigation was recorded, and a multiple comparison procedure was used to detect differences among the mean water losses for the different fumigation durations. How would you interpret this pattern? $$\begin{array}{rrrrr} &\text { Duration of fumigation } &16 & 0 & 8 & 2 & 4 \\ &\text { Sample mean water loss } &27.57 & 28.23 & 30.21 & 31.16 & 36.21 \end{array}$$

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