/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 An investigation carried out to ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 13

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.

Short Answer

Expert verified
Based on the ANOVA test, we do not reject the null hypothesis. Therefore, it suggests that the mean dry weight is the same across all mercury concentrations.

Step by step solution

01

Understanding the Problem

The problem is a hypothesis testing of means using ANOVA. The null hypothesis states that all group means are equal, while the alternative hypothesis is that at least one group mean is different.
02

Defining Hypotheses

The null hypothesis \(H_0\) is that the mean dry weight is the same across all mercury concentration levels. It means all group means are equal. The Alternative Hypothesis \(H_a\) is that at least one concentration level has a mean dry weight different from others.
03

Interpret the ANOVA F statistic

ANOVA \(F\) value is given as \(1.895\) and the significance level is \(0.05\). The critical \(F\) value for \(df_1 = 9, df_2 = 30\) (10 groups - 1 = 9, 40 - 10 = 30) with significance level \(0.05\) is approximately \(2.26\).
04

Decision Making

Since the calculated \(F\) value \(1.895\) is less than the critical \(F\) value \(2.26\), we do not have enough evidence to reject the null hypothesis that the mean dry weight is the same for all concentration levels. So we accept the null hypothesis at the \(0.05\) significance level.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics, used to make decisions based on data. It involves proposing a hypothesis about a dataset and then determining whether there is enough statistical evidence to reject that hypothesis. In this example, we have the null hypothesis, \( H_0 \), which states that the mean dry weight of barley seedlings is the same across different concentrations of mercury. The alternative hypothesis, \( H_a \), proposes that at least one concentration level has a different mean dry weight than the others. Hypothesis testing is conducted by comparing calculated statistics against known thresholds, which help determine if observed differences are statistically significant or simply due to random chance.
Analyzing Mercury Toxicity
Mercury toxicity refers to the harmful effects of mercury exposure on living organisms. In the context of this study, different concentrations of mercury were applied to barley seedlings to observe their impact on growth, measured as the average dry weight of seedlings. Mercury can disrupt cellular processes and metabolism, leading to reduced growth or even death. This experiment helps assess at what concentration mercury begins to adversely affect plant health. This is crucial for understanding safe levels of mercury in environments to protect both ecosystems and human health, given that plants are a critical part of the food chain.
Essentials of Experimental Design
Experimental design is the blueprint for conducting a scientific study, ensuring that results are reliable and applicable. In this study, the design included ten different concentrations of mercury treatments, each replicated four times. By repeating the experiment multiple times, researchers can ensure consistency and account for variability in the results. Such design helps mitigate errors and increases confidence that any observed effects are due to changes in mercury levels, not by random factors. Therefore, a solid experimental design is essential for drawing accurate conclusions from hypothesis testing.
The Role of Statistical Significance
Statistical significance is the likelihood that a result from data analysis is not due to random chance. In hypothesis testing, a significance level (alpha) is predetermined; here, it is set at 0.05. This means there is a 5% risk of concluding that a difference exists when there is none. In this ANOVA test, the critical \( F \) value is 2.26. Since the calculated \( F \) statistic of 1.895 is less than the critical value, the result isn't statistically significant at this level. Thus, the experiment supports the null hypothesis, suggesting no significant difference in mean dry weights across mercury concentrations. Understanding statistical significance is crucial for making informed conclusions and ensuring results are trustworthy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

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: $$ \begin{array}{lccc} \text { Type of Car } & \text { American } & \text { German } & \text { Japanese } \\ \hline \text { Sample size } & 58 & 38 & 59 \\ \text { Sample mean rating } & 3.05 & 2.87 & 2.67 \\ \hline \end{array} $$ 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.

The article "Growth Response in Radish to Sequential and Simultaneous Exposures of \(\mathrm{NO}_{2}\) and \(\mathrm{SO}_{2} "\) ' (Environmental Pollution [1984]: 303-325) compared a control group (no exposure), a sequential exposure group (plants exposed to one pollutant followed by exposure to the second four weeks later), and a simultaneous-exposure group (plants exposed to both pollutants at the same time). The article states, "Sequential exposure to the two pollutants had no effect on growth compared to the control. Simultaneous exposure to the gases significantly reduced plant growth." Let \(\bar{x}_{1}, \bar{x}_{2}\), and \(\bar{x}_{3}\) represent the sample means for the control, sequential, and simultaneous groups, respectively. Suppose that \(\bar{x}_{1}>\bar{x}_{2}>\bar{x}_{3}\). Use the given information to construct a table where the sample means are listed in increasing order, with those that are judged not to be significantly different underscored.

The article "Utilizing Feedback and Goal Setting to Increase Performance Skills of Managers" (Academy of Management Journal \([1979]: 516-526)\) reported the results of an experiment to compare three different interviewing techniques for employee evaluations. One method allowed the employee being evaluated to discuss previous evaluations, the second involved setting goals for the employee, and the third did not allow either feedback or goal setting. After the interviews were concluded, the evaluated employee was asked to indicate how satisfied he or she was with the interview. (A numerical scale was used to quantify level of satisfaction.) The authors used ANOVA to compare the three interview techniques. An \(F\) statistic value of \(4.12\) was reported. a. Suppose that a total of 33 subjects were used, with each technique applied to 11 of them. Use this information to conduct a level \(.05\) test of the null hypothesis of no difference in mean satisfaction level for the three interview techniques. b. The actual number of subjects on which each technique was used was \(45 .\) After studying the \(F\) table, explain why the conclusion in Part (a) still holds.

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?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.