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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 represents the probability of observing the given result under the null hypothesis. It's calculated based on the F-value and the degrees of freedom for the treatment and error in the test. Due to the complexity of the calculation, it's recommended to use a statistical software or F-distribution table for finding the p-value.

Step by step solution

01

Calculate the degrees of freedom

Determine the degrees of freedom for treatments which is \(df_{treatment}= k-1\). Next, calculate the error degrees of freedom which is \(df_{error}=N-k\), where \(N\) is the total number of observations in all groups, and \(k\) is the number of groups.
02

Calculate the p-value

Apply the F-distribution to calculate the p-value. For the F-distribution, the F value calculated from the F test and the degrees of freedom (for treatment and error) are required. Use a statistical software or a F-distribution table to find the p-value. If the calculated p-value is less than 0.05, we reject the null hypothesis that the group means are equal.
03

Follow step 1 and 2 for each situation

Repeat above steps for each situation given in the exercise.

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Most popular questions from this chapter

Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data reported in Consumer Reports.) $$ \begin{array}{llllll} \text { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \\ \text { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \\ \text { Blue Bonnet } & 13.5 & 13.4 & 14.1 & 14.3 & \\ \text { Chiffon } & 13.2 & 12.7 & 12.6 & 13.9 & \\ \text { Mazola } & 16.8 & 17.2 & 16.4 & 17.3 & 18.0 \\ \text { Fleischmann's } & 18.1 & 17.2 & 18.7 & 18.4 & \end{array} $$ a. Test for differences among the true average PAPUFA percentages for the different brands. Use \(\alpha=.05\). b. Use the \(\mathrm{T}-\mathrm{K}\) procedure to compute \(95 \%\) simultaneous confidence intervals for all differences between means and give the corresponding underscoring pattern.

The authors of the paper "Beyond the Shooter Game: Examining Presence and Hostile Outcomes among Male Game Players" (Communication Research \([2006]: 448-466\) ) studied how video-game content might influence attitudes and behavior. Male students at a large Midwestern university were assigned at random to play one of three action-oriented video games. Two of the games involved some violence -one was a shooting game and one was a fighting game. The third game was a nonviolent race car driving game. After playing a game for 20 minutes, participants answered a set of questions. The responses were used to determine values of three measures of aggression: (1) a measure of aggressive behavior; (2) a measure of aggressive thoughts; and (3) a measure of aggressive feelings. The authors hypothesized that the means for the three measures of aggression would be greatest for the fighting game and lowest for the driving game. a. For the measure of aggressive behavior, the paper reports that the mean score for the fighting game was significantly higher than the mean scores for the shooting and driving game, but that the mean scores for the shooting and driving games were not significantly different. The three sample means were: $$ \begin{array}{llll} & \text { Driving } & \text { Shooting } & \text { Fighting } \\ \text { Sample mean } & 3.42 & 4.00 & 5.30 \end{array} $$ Use the underscoring procedure of this section to construct a display that shows any significant differences in mean aggressive behavior score among the three games. b. For the measure of aggressive thoughts, the three sample means were: $$ \begin{array}{llll} & \text { Driving } & \text { Shooting } & \text { Fighting } \\ \text { Sample mean } & 2.81 & 3.44 & 4.01 \end{array} $$ The paper states that the mean score for the fighting game only significantly differed from the mean score for the driving game and that the mean score for the shooting game did not significantly differ from either the fighting or driving games. Use the underscoring procedure of this section to construct a display that shows any significant differences in mean aggressive thoughts score among the three games.

The paper referenced in the previous exercise also gave data for 12 - to 13 -year-old girls. Data consistent with summary values in the paper are shown below. Do the data provide convincing evidence that the mean rating associated with the game description for 12 - to 13 -year-old girls is not the same for all four age restrictive rating labels? Test the appropriate hypotheses using \(\alpha=.05\). $$ \begin{array}{cccc} \text { 7+ label } & \text { 12+ label } & \text { 16+ label } & \text { 18+ label } \\ \hline 4 & 4 & 6 & 8 \\ 7 & 5 & 4 & 6 \\ 6 & 4 & 8 & 6 \\ 5 & 6 & 6 & 5 \\ 3 & 3 & 10 & 7 \\ 6 & 5 & 8 & 4 \\ 4 & 3 & 6 & 10 \\ 5 & 8 & 6 & 6 \\ 10 & 5 & 8 & 8 \\ 5 & 9 & 5 & 7 \\ \hline \end{array} $$

In the introduction to this chapter, we considered a study comparing three groups of college students (soccer athletes, nonsoccer athletes, and a control group consisting of students who did not participate in intercollegiate sports). The following information on scores from the Hopkins Verbal Learning Test (which measures immediate memory recall) was $$ \begin{array}{lccc} \text { Group } & \begin{array}{c} \text { Soccer } \\ \text { Athletes } \end{array} & \begin{array}{c} \text { Nonsoccer } \\ \text { Athletes } \end{array} & \text { Control } \\ \hline \text { Sample size } & 86 & 95 & 53 \\ \text { Sample mean score } & 29.90 & 30.94 & 29.32 \\ \text { Sample standard } & 3.73 & 5.14 & 3.78 \\ \text { deviation } & & & \\ \hline \end{array} $$ In addition, \(\overline{\bar{x}}=30.19 .\) Suppose that it is reasonable to regard these three samples as random samples from the three student populations of interest. Is there sufficient evidence to conclude that the mean Hopkins score is not the same for the three student populations? Use \(\alpha=.05\).

The authors of the paper "Age and Violent Content Labels Make Video Games Forbidden Fruits for Youth" (Pediatrics [2009]: \(870-876\) ) carried out an experiment to determine if restrictive labels on video games actually increased the attractiveness of the game for young game players. Participants read a description of a new video game and were asked how much they wanted to play the game. The description also included an age rating. Some participants read the description with an age restrictive label of \(7+\), indicating that the game was not appropriate for children under the age of \(7 .\) Others read the same description, but with an age restrictive label of \(12+, 16+\), or \(18+.\) The data below for 12 - to 13 -year-old boys are fictitious, but are consistent with summary statistics given in the paper. (The sample sizes in the actual experiment were larger.) For purposes of this exercise, you can assume that the boys were assigned at random to one of the four age label treatments \((7+, 12+, 16+\), and \(18+\) ). Data shown are the boys' ratings of how much they wanted to play the game on a scale of 1 to 10 . Do the data provide convincing evidence that the mean rating associated with the game description by 12 - to 13 -year-old boys is not the same for all four restrictive rating labels? Test the appropriate hypotheses using a significance level of \(.05 .\) $$ \begin{array}{cccc} \hline 7+\text { label } & 12+\text { label } & 16+\text { label } & 18+\text { label } \\ \hline 6 & 8 & 7 & 10 \\ 6 & 7 & 9 & 9 \\ 6 & 8 & 8 & 6 \\ 5 & 5 & 6 & 8 \\ 4 & 7 & 7 & 7 \\ 8 & 9 & 4 & 6 \\ 6 & 5 & 8 & 8 \\ 1 & 8 & 9 & 9 \\ 2 & 4 & 6 & 10 \\ 4 & 7 & 7 & 8 \\ \hline \end{array} $$
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