/*! 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 38 $$ \begin{array}{lrrrrr} \te... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

$$ \begin{array}{lrrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array} $$ Is there evidence for a difference in the population means of the four groups? Justify your answer using specific value(s) from the output.

Short Answer

Expert verified
Yes, there is evidence for a difference in the population means of the four groups, as the P-value (0.003) is less than the significance level 0.05.

Step by step solution

01

Identify the P-value

The P-value, found in the p column of the ANOVA table under 'Groups', is 0.003. The p-value is the probability that you would obtain the observed data, or something more extreme, if the null hypothesis were true.
02

Interpret the P-value

A commonly used significance level is 0.05. If the p-value is less than or equal to this significance level, it indicates strong evidence against the null hypothesis, suggesting the null hypothesis is likely false, and there is a difference between at least two group means. In this case, the P-value is 0.003, which is less than 0.05.
03

Make the conclusion

Since the p-value is less than 0.05, we reject the null hypothesis. This means there is statistically significant evidence to suggest that there is a difference in the population means of the four groups.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding the p-value
The concept of the p-value is central to hypothesis testing in statistics and serves as a quantitative tool to determine the strength of the evidence against the null hypothesis. In the context of ANOVA, which stands for Analysis of Variance, the p-value helps us to decide whether the differences among group means are likely to be due to random chance or to real differences in the data.

When you perform an ANOVA, like in the given exercise, you are comparing more than two groups to see if there's a statistically significant difference between them. The p-value given in the exercise is 0.003. This number represents the probability that the observed differences among the group means occurred solely by random chance, assuming the null hypothesis is true—that is, that there's actually no real difference between the groups.

To interpret the p-value effectively, it's often compared against a predetermined significance level, commonly set at 0.05. If the p-value is less than this threshold, as it is in the exercise (0.003 < 0.05), we take it as evidence that at least one group mean is significantly different from the others, and thus, we have a reason to reject the null hypothesis.
The Role of Null Hypothesis in ANOVA
The null hypothesis is the default statement or position that there is no effect or no difference. It is the starting point for any statistical significance testing, including ANOVA. In the ANOVA context, specifically, the null hypothesis asserts that all groups have the same population means, implying that any observed differences in sample means are simply due to sampling variability and not due to actual differences in the populations.

In the given exercise, the null hypothesis would claim that the means of the four groups are equal. When we perform the ANOVA test, we're essentially assessing whether the data provides enough evidence to reject this hypothesis in favor of the alternative hypothesis, which states that at least one group mean is different.

The p-value guides us in this decision. Since the computed p-value in the exercise is extremely low (0.003), it suggests that the observed differences between group means are highly unlikely to have occurred under the null hypothesis, thus pointing towards rejecting the null hypothesis and concluding that not all group means are equal.
Statistical Significance in ANOVA
Statistical significance is a determination about the null hypothesis based on the evidence provided by the p-value. A statistically significant result means that we have enough evidence to conclude that the observed effect in our data is unlikely to have occurred due to random chance alone.

In our exercise, the ANOVA results produced a p-value of 0.003. Since this value is below the commonly accepted significance level of 0.05, we consider our results to be statistically significant. This determination allows us to confidently reject the null hypothesis and assert that there is a significant difference in the population means of the four groups being studied.

It's crucial to understand, however, that statistical significance does not necessarily imply practical significance. A statistically significant outcome merely indicates that a difference exists, but it does not measure the size or importance of that difference. That's what the 'effect size' measures, which is another important consideration in research, but separate from the concept of statistical significance.

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

Body Mass Gain The mice in the study had body mass measured throughout the study. Computer output showing an analysis of variance table to test for a difference in mean body mass gain (in grams) after four weeks between mice in the three different light conditions is shown. We see in Exercise 8.24 that the conditions for ANOVA are met, and we also find the summary statistics for each experimental group there. (a) State the null and alternative hypotheses. (b) What is the F-statistic? What is the p-value? What is the conclusion of the test? (c) Does there appear to be an association between the two variables (body mass gain and light condition)? If so, discuss the nature of that relationship. Under what light condition do mice appear to gain the most weight? (d) Can we conclude that there is a cause-and-effect relationship between the variables? Why or why not?

Exercises 8.33 to 8.37 refer to the data with analysis shown in the following computer output: $$ \begin{array}{lrrr} \text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387 \end{array} $$ $$ \begin{array}{lrrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array} $$ Find a \(95 \%\) confidence interval for the mean of population \(\mathrm{A}\).

Exercises 8.38 to 8.44 refer to the data with analysis shown in the following computer output: $$ \begin{array}{lrrr} \text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 6 & 86.833 & 5.231 \\ \text { B } & 6 & 76.167 & 6.555 \\ \text { C } & 6 & 80.000 & 9.230 \\ \text { D } & 6 & 69.333 & 6.154 \end{array} $$ $$ \begin{array}{lrrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array} $$ Test for a difference in population means between groups \(\mathrm{B}\) and \(\mathrm{D}\). Show all details of the test.

Some computer output for an analysis of variance test to compare means is given. (a) How many groups are there? (b) State the null and alternative hypotheses. (c) What is the p-value? (d) Give the conclusion of the test, using a \(5 \%\) significance level. $$ \begin{array}{lrrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\\ \text { Groups } & 2 & 540.0 & 270.0 & 8.60 \\ \text { Error } & 27 & 847.8 & 31.4 & \\ \text { Total } & 29 & 1387.8 & & \end{array} $$

Stress Levels and a Mother's Voice A recent study \(^{2}\) examined the impact of a mother's voice on stress levels in young girls. The study included 68 girls ages 7 to 12 who reported good relationships with their mothers. Each girl gave a speech and then solved mental arithmetic problems in front of strangers. Cortisol levels in saliva were measured for all girls and were high, indicating that the girls felt a high level of stress from these activities. (Cortisol is a stress hormone and higher levels indicate greater stress.) After the stress-inducing activities, the girls were randomly divided into four equal-sized groups: one group talked to their mothers in person, one group talked to their mothers on the phone, one group sent and received text messages with their mothers, and one group had no contact with their mothers. Cortisol levels were measured before and after the interaction with mothers and the change in the cortisol level was recorded for each girl. (a) What are the two main variables in this study? Identify each as categorical or quantitative. (b) Is this an experiment or an observational study? (c) The researchers are testing to see if there is a difference in the change in cortisol level depending on the type of interaction with mom. What are the null and alternative hypotheses? Define any parameters used. (d) What are the total degrees of freedom? The \(d f\) for groups? The \(d f\) for error? (e) The results of the study show that hearing mother's voice was important in reducing stress levels. Girls who talk to their mother in person or on the phone show decreases in cortisol significantly greater, at the \(5 \%\) level, than girls who text with their mothers or have no contact with their mothers. There was not a difference between in person and on the phone and there was not a difference between texting and no contact. Was the p-value of the original ANOVA test above or below \(0.05 ?\)

See all solutions

Recommended explanations on Math Textbooks

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.