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Chapter 15: Problem 14

Leaf surface area is an important variable in plant gas-exchange rates. Dry matter per unit surface area \(\left(\mathrm{mg} / \mathrm{cm}^{3}\right)\) was measured for trees raised under three different growing conditions. Let \(\mu_{1}, \mu_{2},\) and \(\mu_{3}\) represent the mean dry matter per unit surface area for the growing conditions \(1,2,\) and \(3,\) respectively. The given \(95 \%\) simultaneous confidence intervals are: Difference \(\quad \mu_{1}-\mu_{2} \quad \mu_{1}-\mu_{3} \quad \mu_{2}-\mu_{3}\) Interval \(\quad(-3.11,-1.11)(-4.06,-2.06)(-1.95, .05)\) 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.

Short Answer

Expert verified
The correct answer is c: \(\mu_{2}\) equals \(\mu_{3}\), and \(\mu_{1}\) is significantly different from \(\mu_{2}\) and \(\mu_{3}\)

Step by step solution

01

Interpret the confidence intervals

Look at the simultaneous confidence intervals given for the difference in dry matter per unit surface area under three growing conditions. If a confidence interval contains 0, it means that the differences between the two means is undiscernible and they can be stated as equal. If a confidence interval doesn't contain 0, it implies the mean growths are significantly different from each other. Here \(\mui_{1}-\mu_{2}\) and \(\mu_{1}-\mu_{3}\) are negative which indicates \(\mui_{1}\) is lesser than \(\mu_{2}\) and \(\mu_{3}\). The interval \(\mu_{2}-\mu_{3}\) contains 0, meaning \(\mu_{2}\) and \(\mu_{3}\) could be equal.
02

Match the results with the options

Option c. states that \(\mu_{2} = \mu_{3}\), which is plausible as per our analysis since the confidence interval contains 0, and \(\mu_{1}\) is distinct, which is also true as both \(\mu_{1}-\mu_{2}\) and \(\mu_{1}-\mu_{3}\) ranges were negative implying \(\mu_{1}\) is less than either \(\mu_{2}\) or \(\mu_{3}\). Therefore, option c. correctly describes the relationship between \(\mu_{1}, \mu_{2},\) and \(\mu_{3}\).

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Key Concepts

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

Confidence Intervals
Confidence intervals are a fundamental concept in statistics, providing a range within which we can expect the true value of a parameter to lie. In our problem, we are given 95% confidence intervals for the differences between means of dry matter per unit surface area under different growing conditions. This means we are 95% confident that the true differences in means are within these intervals.
  • If the interval includes zero, it suggests no significant difference between the means.
  • If zero is not included, it implies a significant difference.
In this exercise, the interval for \(\mu_{2}-\mu_{3}\) includes zero, indicating a possibility that these means are equal. This insight helps us understand which means might be the same and which are distinct based on the intervals given.
Mean Comparison
Mean comparison is all about analyzing the differences between group means to identify any significant variations. In the context of this exercise, we have three groups represented by \(\mu_{1}, \mu_{2},\) and \(\mu_{3}\). The analysis of the intervals sheds light on these relationships:
  • The intervals for \(\mu_{1}-\mu_{2}\) and \(\mu_{1}-\mu_{3}\) show values less than zero, suggesting \(\mu_{1}\) is less than both \(\mu_{2}\) and \(\mu_{3}\).
  • The \(\mu_{2}-\mu_{3}\) interval includes zero, indicating these means could potentially be equal.
Understanding these comparisons is crucial when analyzing data from experiments, as it informs us whether the conditions under study produce different outcomes.
Statistical Analysis
Statistical analysis is the backbone of making informed decisions from data. It involves collecting, exploring, and interpreting data to uncover patterns and trends. In our problem, the provided confidence intervals are a tool of statistical analysis that help determine relationships between group means.
  • Accurate statistical analysis allows researchers to draw meaningful conclusions about the effects of different treatments or conditions.
  • By examining whether intervals contain zero, we gain insights into hypotheses about equality or differences among means.
Incorporating correct statistical analysis helps in reliably understanding the differences among the growing conditions, leading to more informed decisions in research and application.

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Most popular questions from this chapter

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 broadleyil." 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. $$ \begin{array}{lccc} \text { Treatment } & n & \bar{x} & s \\ \hline \text { Uneaten figs } & 20 & 2.40 & .30 \\ \text { Lizard dung } & 20 & 2.35 & .33 \\ \text { Bird dung } & 20 & 1.70 & .34 \\ \text { Hyrax dung } & 20 & 1.45 & .28 \\ \hline \end{array} $$ 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 support your answer.

Give as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations. a. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=5.37\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=1.90\) c. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=4.89\) d. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=14.48\) e. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=2.69\) f. \(\quad \mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

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: Driving Shooting \(\quad\) Fighting | 2.81 Sample mean 3.44 4.01 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 "Trends in Blood Lead Levels and Blood Lead Testing among U.S. Children Aged 1 to 5 Years" (Pediatrics [2009]: e376-e385) gave data on blood lead levels (in \(\mu \mathrm{g} / \mathrm{dL}\) ) for samples of children living in homes that had been classified either at low, medium, or high risk of lead exposure based on when the home was constructed. After using a multiple comparison procedure, the authors reported the following: 1\. The difference in mean blood lead level between low-risk housing and medium-risk housing was significant. 2\. The difference in mean blood lead level between low-risk housing and high- risk housing was significant. 3\. The difference in mean blood lead level between medium-risk housing and high-risk housing was significant. Which of the following sets of T-K intervals (Set 1,2 , or 3) is consistent with the authors' conclusions? Explain your choice. \(\mu_{L}=\) mean blood lead level for children living in low-risk housing \(\mu_{M}=\) mean blood lead level for children living in medium-risk housing \(\mu_{H}=\) mean blood lead level for children living in high-risk housing $$ \begin{array}{cccc} \text { Difference } & \text { Set } 1 & \text { Set } 2 & \text { Set } 3 \\ \hline \mu_{L}-\mu_{M} & (-0.6,0.1) & (-0.6,-0.1) & (-0.6,-0.1) \\ \mu_{L}-\mu_{H} & (-1.5,-0.6) & (-1.5,-0.6) & (-1.5,-0.6) \\ \mu_{M}-\mu_{H} & (-0.9,-0.3) & (-0.9,0.3) & (-0.9,-0.3) \\ \hline \end{array} $$

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. \(k=4, n_{1}=n_{2}=15, n_{3}=12, n_{4}=10, F=1.75\)
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