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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.)

Short Answer

Expert verified
a. Apply an F test on the given data and decide on rejecting or accepting the null hypothesis based on the test result compared to the F-distribution critical value. b. Application of the T-K procedure may result in a contradiction to the F test result because these tests are based on different distributions and the null hypothesis might be 'barely' rejected.

Step by step solution

01

Calculate the group means and overall mean

For each plant type (1 to 5), calculate the mean growth by summing up all growth values and dividing by the number of measurements (in this case, 4). Also, compile all growth measurements, irrespective of plant type, to compute the overall mean (grand mean).
02

Compute the Between-group variance

This is done by summing the squared differences between each group's mean and the grand mean, and multiplying that by the number of observations per group (n). Then, divide by the degrees of freedom between groups (which is k-1, where k is the number of groups).
03

Calculate the Within-group variance

This is calculated using each measurement's square deviation from its group mean, summing them up, and then dividing by the degrees of freedom within the group (which is N-k, where N is the total number of observations and k is the number of groups).
04

Perform the F-test

An F-test compares the variances between and within groups. This is achieved by calculating the F statistic, which is the ratio of 'Between-group variance' and 'Within-group variance'. If the calculated F-value exceeds the critical F-value from the F-distribution table (for df1 = k-1 and df2 = N-k at the alpha level of 0.05), we reject the null hypothesis of equal means.
05

Conduct the T-K procedure

The Tukey-Kramer method involves comparing all possible pairs of group means to find out which ones have significant differences. Consult standard tables for this procedure. If a contradiction from the F-test occurs, it might be because the F-test just barely rejected the null hypothesis. These contradictions occur because the test and the multiple comparison method are based on different distributions.

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Key Concepts

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

ANOVA
Analysis of Variance, commonly referred to as ANOVA, is a statistical method used for comparing the means of three or more groups to see if at least one mean is different from the others. It essentially allows us to analyze if the differences seen in sample means are due to actual differences between groups or simply due to chance.

In performing an ANOVA, we first formulate two hypotheses: the null hypothesis which states that all group means are equal, and the alternative hypothesis which states that at least one mean is different. We then calculate the F-statistic to determine if any differences between sample means are statistically significant.

The F-test examines two sources of variance: between-group variance and within-group variance. The ratio of these variances is the F-statistic. If the F-statistic is sufficiently large, we reject the null hypothesis, implying that not all group means are equal and further investigation is needed.
Between-group Variance
Between-group variance, also known as 'between-group mean square', is a measure of the variation of the group means around the grand mean. It's calculated by taking each group's mean and subtracting the grand mean, then squaring the result. This squared deviation is then multiplied by the number of observations in each group to account for group size.

To find the average of these squared deviations, we divide by the degrees of freedom for the between-group variance, which is the number of groups minus 1 (k - 1). A higher between-group variance suggests that the group means differ significantly from the overall mean, which could be indicative of a meaningful difference between groups.
Within-group Variance
Within-group variance, or 'within-group mean square', measures how much the individual data points within each group deviate from their respective group mean. It provides a sense of how spread out the data are within each group. To calculate it, we subtract the group mean from each individual data point, square the result, and sum these up for each group.

To get the average within-group variance, we divide the total by the within-group degrees of freedom, which is the total number of observations across all groups minus the number of groups (N - k). If the within-group variance is small relative to the between-group variance, the distinctive characteristics of the groups become more pronounced, and the F-test may lead us to reject the null hypothesis.
Tukey-Kramer Method
The Tukey-Kramer method, also known as the Tukey's Honest Significant Difference (HSD) test when group sizes are equal, is a post-hoc analysis procedure used after an ANOVA. It is applied to perform multiple pairwise comparisons between groups to determine which mean differences are significant.

After rejecting the null hypothesis in an ANOVA, uncertainties may still exist about which group means differ from one another. The Tukey-Kramer method calculates a range or interval for mean differences. If the mean difference between any two groups is greater than this calculated range, the difference is considered statistically significant.

It should be noted that this method adjusts for the fact that multiple comparisons are being made, which can increase the likelihood of erroneous findings. As such, the Tukey-Kramer method helps to control the overall rate of Type I errors across these comparisons.

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Most popular questions from this chapter

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 $$\begin{array}{l|ccc} \text { Group } & \text { Soccer Athletes } & \text { Nonsoccer Athletes } & \text { Control } \\ \hline \text { Sample size } & 86 & 95 & 53 \\ \text { Sample mean score } & 29.90 & 30.94 & 29.32 \\ \begin{array}{l} \text { Sample standard } \\ \text { deviation } \end{array} & 3.73 & 5.14 & 3.78 \\ \hline \end{array}$$ 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 \(\alpha=.05\).

An investigation carried out to study purchasers of luxury automobiles reported data on a number of different attributes that might affect purchase decisions, including comfort, safety, styling, durability, and reliability ("Measuring Values Can Sharpen Segmentation in the Luxury Car Market," Journal of Advertising Research \([1995]: 9-\) 22). Here is summary information on the level of importance of speed, rated on a seven-point scale: $$ \begin{array}{lccc} \text { Type of Car } & \text { American } & \text { German } & \text { Japanese } \\ \hline \text { Sample size } & 58 & 38 & 59 \\ \text { Sample mean rating } & 3.05 & 2.87 & 2.67 \end{array} $$ In addition, \(\mathrm{SSE}=459.04\). Carry out a hypothesis test to determine if there is sufficient evidence to conclude that the mean importance rating of speed is not the same for owners of these three types of cars.

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 broadleyi)," 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. $$\begin{array}{lccc} \text { Treatment } & \boldsymbol{n} & \overline{\boldsymbol{x}} & \boldsymbol{s} \\ \hline \text { Uneaten figs } & 20 & 2.40 & .30 \\ \text { Lizard dung } & 20 & 2.35 & .33 \\ \text { Bird dung } & 20 & 1.70 & .34 \\ \text { Hyrax dung } & 20 & 1.45 & .28 \end{array}$$ 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 sup- port your answer.

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?

High productivity and carbohydrate storage ability of the Jerusalem artichoke make it a promising agricultural crop. The article "Leaf Gas Exchange and Tuber Yield in Jerusalem Artichoke Cultivars" (Field Crops Research [1991]: \(241-252\) ) reported on various plant characteristics. Consider the accompanying data on chlorophyll concentration \(\left(\mathrm{gm} / \mathrm{m}^{2}\right)\) for four varieties of Jerusalem artichoke: $$\begin{array}{lccll} \text { Variety } & \text { BI } & \text { RO } & \text { WA } & \text { TO } \\\ \text { Sample mean } & .30 & .24 & .41 & .33 \end{array}$$ Suppose that the sample sizes were \(5,5,4\), and 6, respectively, and also that MSE \(=.0130 .\) Do the data suggest that true average chlorophyll concentration depends on the variety? State and test the appropriate hypotheses using a significance level of \(.05\).

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