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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. \(k=5, n_{1}=n_{2}=n_{3}=n_{4}=n_{5}=4, F=5.37\) 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-value can't be calculated exactly without a table or statistical programming function. However, given the calculated degrees of freedom and the F-statistic, one could look up the P-values in a F-distribution table. After calculating the P-values for each scenario, you must compare the P-value for each with a significance level (assuming 0.05, unless otherwise specified) to potentially decide on the validity of the null hypothesis.

Step by step solution

01

Calculation of Degree of Freedom for Each Scenario

We calculate the degree of freedom in each scenario. For each scenario, the degree of freedom between groups (df1) is \(k-1\) and the degree of freedom within groups (df2) is \(N-k\), where \(k\) is the number of groups and \(N\) is sample size.
02

Finding P-value from F-distribution Table

We find the P-value corresponding to the F-statistic for each case using the F-distribution table (or a programming function if you're using a software) with the calculated df1 and df2. If the calculated F-statistic is not in the table, we find the closest one. Please note that this can add some minor inaccuracies and is the reason we might use statistical software.
03

Interpreting the P-value

Compare the P-value with the significance level (commonly 0.05). If the P-value is smaller than the significance level, we reject the null hypothesis that the groups are the same.\nThis step isn't strictly calculated information from an F-test, but it is information you often infer from the P-value. It could be a quick way to judge the outcomes from each test.\nThis step should take notice that without an actual significance level stated in the problem, any interpretations made are merely suggestions for how one might typically interpret the results.

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Most popular questions from this chapter

15.19 The Gunning Fog index is a measure of reading difficulty based on the average number of words per sentence and the percentage of words with three or more syllables. High values of the Gunning Fog index are associated with difficult reading levels. Independent random samples of six advertisements were taken from three different magazines, and Gunning Fog indices were computed to obtain the data given in the accompanying table ("Readability Levels of Magazine Advertisements"'Journal of Advertising Research [1981]: 45-50). Construct an ANOVA table, and then use a significance level of .01 to test the null hypothesis of no difference between the mean Gunning Fog index levels for advertisements appearing in the three magazines. $$ \begin{array}{lrrrrrr} \text { Scientific American } & 15.75 & 11.55 & 11.16 & 9.92 & 9.23 & 8.20 \\ \text { Fortune } & 12.63 & 11.46 & 10.77 & 9.93 & 9.87 & 9.42 \\ \text { New Yorker } & 9.27 & 8.28 & 8.15 & 6.37 & 6.37 & 5.66 \end{array} $$

An investigation carried out to study the toxic effects of mercury was described in the article "Comparative Responses of the Action of Different Mercury Compounds on Barley" (International Journal of Environmental Studies [1983]: 323-327). Ten different concentrations of mercury \((0,1,5,10,50,100,200,300,400\), and \(500 \mathrm{mg} / \mathrm{L})\) were compared with respect to their effects on average dry weight (per 100 seven- day-old seedlings). The basic experiment was replicated four times for a total of 40 dryweight observations (four for each treatment level). The article reported an ANOVA \(F\) statistic value of \(1.895 .\) Using a significance level of \(.05\), test the null hypothesis that the true mean dry weight is the same for all 10 concentration levels.

The nutritional quality of shrubs commonly used for feed by rabbits was the focus of a study summarized in the article "Estimation of Browse by Size Classes for Snowshoe Hare" (Journal of Wildlife Management [1980]: 34-40). The energy content (cal/g) of three sizes ( \(4 \mathrm{~mm}\) or less, \(5-7 \mathrm{~mm}\), and \(8-10 \mathrm{~mm}\) ) of serviceberries was studied. Let \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) denote the true energy content for the three size classes. Suppose that \(95 \%\) simultaneous confidence intervals for \(\mu_{1}-\mu_{2}, \mu_{1}-\mu_{3}\), and \(\mu_{2}-\mu_{3}\) are \((-10,290),(150,450)\), and \((10,310)\), respectively. How would you interpret these intervals?

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. $$ \begin{array}{lrl} \text { Status } & \begin{array}{l} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{l} \text { Sample } \\ \text { Mean } \end{array} \\ \hline \text { Continuous smoker } & 96 & 2.15 \\ \text { Recent ex-smoker } & 34 & 2.21 \\ \text { Long-term ex-smoker } & 86 & 1.47 \\ \text { Never smoked } & 206 & 1.69 \\ & & \\ \hline \end{array} $$

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.
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