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Chapter 10: Problem 7

Shrubs and fire Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. One study of resprouting took place in a dry area of Mexico. The investigators randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted.

Short Answer

Expert verified
The treatment group had a 100% resprouting rate, while the control group had 67%.

Step by step solution

01

Understanding the Context

The study concerns the ability of shrubs to resprout after fire damage. Shrubs were grouped into a control group and a treatment group to test resprouting capability.
02

Identify Key Information

We are given that all 12 shrubs in the treatment group resprouted after fire simulation, whereas only 8 of the 12 shrubs in the control group resprouted after just clipping.
03

Calculate the Proportion of Resprouting

For the treatment group, the resprouting proportion is 12/12, which is 1 or 100%. For the control group, the resprouting proportion is 8/12, which is approximately 0.67 or 67%.
04

Compare Proportions

Comparing the resprouting success rates of the two groups, the treatment group (100%) had a higher success rate than the control group (67%). This indicates the effect of the simulated fire treatment.

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Key Concepts

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

Resprouting Capability
Resprouting capability refers to a plant's ability to regenerate after extreme damage, such as fire or cutting. This ability is vital for survival in harsh environments where natural disasters like wildfires are common. In the study conducted in a dry area of Mexico, researchers specifically explored this capability among shrubs in dry climates. When a shrub is subjected to damage, its ability to sprout from its remaining roots or stems is tested. Those that can resprout will have a better chance to survive in environments susceptible to fire destruction. In the exercise, researchers simulated a fire by clipping the tops of shrubs and using a propane torch on the treatment group to assess how effectively the shrubs could regrow. Such experimental designs help us understand their survival strategies and resilience, ultimately helping in conservation efforts.
Control and Treatment Groups
In experimental design, the use of control and treatment groups is key to understanding the effects of a certain treatment. The control group is not subjected to the treatment that is being tested, while the treatment group is. In this way, researchers can compare results between the two groups to understand the true impact of the treatment applied. In the shrub resprouting study, the shrubs in the control group had their tops clipped but did not undergo the propane torch fire simulation. Meanwhile, the treatment group not only had their tops clipped but also endured the fire simulation. This difference allows researchers to isolate the fire's effect on the shrubs鈥� ability to resprout. By carefully observing the results from both groups, scientists can gather data that highlight the influence of the simulated fire, helping to reinforce or reject the hypotheses they are testing.
Proportion Comparison
Proportion comparison is an essential technique in data analysis, especially in experiments with binary outcomes like success (resprouting) or failure (no resprouting). It involves calculating and comparing the proportions of successful outcomes in different groups to infer the effectiveness of a treatment. In the exercise, we calculate the proportion of shrubs that resprouted in both the control and the treatment groups. For the treatment group, the proportion was calculated as 12 out of 12, equating to 100%. Meanwhile, the control group had a proportion of 8 out of 12, which amounts to approximately 67%. By comparing these proportions, researchers can deduce that the treatment dramatically increased the chance of resprouting. This comparison gives a clear indication of the effectiveness of the simulated fire treatment, emphasizing its role in promoting the resprouting capability of the shrubs. Such proportion comparisons are critical for interpreting the results of experiments and drawing meaningful conclusions.

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Most popular questions from this chapter

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A better drug? In a pilot study, a company's new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood \((\mathrm{mg} / \mathrm{dl})\) greater than with the current drug, the company will begin the expensive process of mass- producing the new drug. For the 14 subjects who were assigned at random to the current drug. the mean cholesterol reduction was \(54.1 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(11.93 \mathrm{mg} / \mathrm{dl}\). For the \(15 \mathrm{sub}\) jects who were randomly assigned to the new drug, the mean cholesterol reduction was \(68.7 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(13.3 \mathrm{mg} / \mathrm{dl}\). Graphs of the data reveal no outliers or strong skewness. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(\left.H_{0}: \mu_{1}-\mu_{2}=0 .\right)\) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

The researchers report that the results were statistically significant at the \(1 \%\) level. Which of the following is the most appropriate conclusion? (a) Because the \(P\) -value is less than \(1 \%\), fail to reject \(H_{0}\). There is not convincing evidence that the proportion of male college students in the study who worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer. (b) Because the \(P\) -value is less than \(1 \%\), fail to reject \(H_{0}\). There is not convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.

State which inference procedure from Chapter \(8,9,\) or 10 you would use. Be specific. For example, you might say, "Two-sample z test for the difference between two proportions." You do not need to carry out any procedures. Which inference method? (a) Drowning in bathtubs is a major cause of death in children less than 5 years old. A random sample of parents was asked many questions related to bathtub safety. Overall, \(85 \%\) of the sample said they used baby bathtubs for infants. Estimate the percent of all parents of young children who use baby bathtubs. (b) How seriously do people view speeding in comparison with other annoying behaviors? A large random sample of adults was asked to rate a number of behaviors on a scale of 1 (no problem at all) to 5 (very severe problem ). Do speeding drivers get a higher average rating than noisy neighbors? (c) You have data from interviews with a random sample of students who failed to graduate from a particular college in 7 years and also from a random sample of students who entered at the same time and did graduate. You will use these data to compare the percents of students from rural backgrounds among dropouts and graduates. (d) Do experienced computer game players earn higher scores when they play with someone present to cheer them on or when they play alone? Fifty teenagers with experience playing a particular computer game have volunteered for a study. We randomly assign 25 of them to play the game alone and the other 25 to play the game with a supporter present. Each player's score is recorded.
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