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Chapter 15: Problem 2

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

Consider the accompanying data on plant growth after the application of different types of growth hormone. $$\begin{array}{llrlrl} & \mathbf{1} & 13 & 17 & 7 & 14 \\ & \mathbf{2} & 21 & 13 & 20 & 17 \\ \text { Hormone } & \mathbf{3} & 18 & 14 & 17 & 21 \\ & \mathbf{4} & 7 & 11 & 18 & 10 \\ & \mathbf{5} & 6 & 11 & 15 & 8 \end{array}$$ a. Carry out the \(F\) test at level \(\alpha=.05\). b. What happens when the T-K procedure is applied? (Note: This "contradiction" can occur when is "barely" rejected. It happens because the test and the multiple comparison method are based on different distributions. Consult your friendly neighborhood statistician for more information.)

The article "Heavy Drinking and Problems Among Wine Drinkers" (Journal of Studies on Alcohol [1999]: 467-471) analyzed drinking problems among Canadians. For each of several different groups of drinkers, the mean and standard deviation of "highest number of drinks consumed" were calculated: \(\bar{x}\) $$\begin{array}{lccc} & \overline{\boldsymbol{x}} & \boldsymbol{s} & {n} \\ \hline \text { Beer only } & 7.52 & 6.41 & 1256 \\ \text { Wine only } & 2.69 & 2.66 & 1107 \\ \text { Spirits only } & 5.51 & 6.44 & 759 \\ \text { Beer and wine } & 5.39 & 4.07 & 1334 \\ \text { Beer and spirits } & 9.16 & 7.38 & 1039 \\ \text { Wine and spirits } & 4.03 & 3.03 & 1057 \\ \text { Beer, wine, and spirits } & 6.75 & 5.49 & 2151 \end{array}$$ Assume that each of the seven samples studied can be viewed as a random sample for the respective group. Is there sufficient evidence to conclude that the mean value of highest number of drinks consumed is not the same for all seven groups?

Research carried out to investigate the relationship between smoking status of workers and short-term absenteeism rate (hr/mo) yielded the accompanying summary information ("Work-Related Consequences of Smoking Cessation," Academy of Management Journal [1989]: \(606-621) .\) In addition, \(F=2.56\). Construct an ANOVA table, and then state and test the appropriate hypotheses using a .01 significance level. $$\begin{array}{lrl} \text { Status } & \begin{array}{l} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{l} \text { Sample } \\ \text { Mean } \end{array} \\ \hline \text { Continuous smoker } & 96 & 2.15 \\ \text { Recent ex-smoker } & 34 & 2.21 \\ \text { Long-term ex-smoker } & 86 & 1.47 \\ \text { Never smoked } & 206 & 1.69 \end{array}$$

Suppose that a random sample of size \(n=5\) was selected from the vineyard properties for sale in Sonoma County, California, in each of three years. The following data are consistent with summary information on price per acre (in dollars, rounded to the nearest thousand) for disease-resistant grape vineyards in Sonoma County (Wines and Vines, November 1999). $$\begin{array}{llllll} 1996 & 30,000 & 34,000 & 36,000 & 38,000 & 40,000 \\ 1997 & 30,000 & 35,000 & 37,000 & 38,000 & 40,000 \\ 1998 & 40,000 & 41,000 & 43,000 & 44,000 & 50,000 \end{array}$$ a. Construct boxplots for each of the three years on a common axis, and label each by year. Comment on the similarities and differences. b. Carry out an ANOVA to determine whether there is evidence to support the claim that the mean price per acre for vineyard land in Sonoma County was not the same for the three years considered. Use a significance level of \(.05\) for your test.

It has been reported that varying work schedules can lead to a variety of health problems for workers. The article "Nutrient Intake in Day Workers and Shift Workers" (Work and Stress [1994]: 332-342) reported on blood glucose levels (mmol/L) for day-shift workers and workers on two different types of rotating shifts. The sample sizes were \(n_{1}=37\) for the day shift, \(n_{2}=34\) for the second shift, and \(n_{3}=25\) for the third shift. A single- factor ANOVA resulted in \(F=3.834\). At a significance level of .05, does true average blood glucose level appear to depend on the type of shift?
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