Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A recent study \(^{2}\) examines 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 \(0.05 ?\)

Short Answer

Expert verified
The two variables are the type of contact with mothers (categorical) and the change in cortisol levels (quantitative). This is an experiment. Null hypothesis: No difference in cortisol level change between groups. Alternative hypothesis: Difference in cortisol level change between groups. Total degrees of freedom: 67, groups: 3, error: 64. Since the p-value is significant at the 5% level, it is below 0.05.

Step by step solution

01

Determination of Variables and Their Types

The two main variables in this study are: \n 1) Type of interaction with the mother (categories: in person, on the phone, text messages, no contact) - This is a categorical variable. \n 2) The change in the cortisol level - This is a quantitative variable as it can be measured and quantified.
02

Identification of the Study Type

This is an experiment rather than an observational study. The reason it is an experiment is that the researchers actively and randomly assigned the girls to different groups which defined the type of interaction they had with their mothers. Thus, the researchers had control over the variables under investigation.
03

Formation of Hypotheses

The null hypothesis (\(H_0\)) is: There is no difference in the change in cortisol levels among the four groups (in-person talk, phone talk, texting, and no contact). \n The alternative hypothesis (\(H_1\)) is: There is a difference in the change in cortisol levels among the four groups. The parameters here would be the mean change in cortisol levels for each of the four groups.
04

Calculation of Degrees of Freedom

The total degrees of freedom is number of observations, n, minus 1, which is 68-1=67. The degrees of freedom for groups is k-1 where k is number of groups, which is 4-1=3. The degrees of freedom for error is n-k, which is 68-4=64.
05

Interpretation of P-Value

If the result shows a significant decrease in cortisol levels in those who talked to their mothers in person or on the phone than those who texted or had no contact at the 5% level, this suggests that the p-value of the original ANOVA test was below 0.05. A p-value less than or equal to 0.05 typically demonstrates strong evidence against the null hypothesis, so it can be safely assumed that the p-value here was indeed less than or equal to 0.05.

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 Vaia!

Key Concepts

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

Categorical vs Quantitative Variables
When it comes to research and data analysis, it's crucial to distinguish between categorical and quantitative variables. Categorical variables, also known as qualitative variables, represent categories or groups that cannot be quantified. In our study on stress levels, the type of interaction with the mother is a categorical variable as it consists of distinct categories: in-person, on the phone, text messages, and no contact.

On the other hand, quantitative variables are numeric and can be measured. These can be further divided into discrete (whole numbers) or continuous (can take on any number within a range). In the same cortisol level study, the change in cortisol level is a quantitative variable because it can be measured and expressed numerically, allowing for more sophisticated statistical analysis like calculation of mean changes.
Experimental Study Design
The experimental study design is a powerful tool in research as it allows us to infer cause and effect relationships by manipulating independent variables. In our cortisol stress study, this design was used to randomize young girls into groups with different types of mother interactions. This random assignment aims to ensure that other factors do not influence the results, which is why it's a cornerstone of experimental studies. By controlling these variables, researchers can more confidently attribute changes in cortisol levels to the type of maternal interaction, rather than to some other external factor.
Hypothesis Testing
Hypothesis testing is a formal procedure that researchers use to determine if there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In hypothesis testing, we start with a null hypothesis (\(H_0\)) that implies no effect or no difference, and an alternative hypothesis (\(H_1\) or \(H_a\) that suggests there is an effect or a difference. Based on the cortisol study, the null hypothesis states that there's no difference in the change in cortisol levels across the different interaction types. The alternative hypothesis posits that there is a difference. Researchers then use data to test these hypotheses, and the outcome informs us whether or not the data provides significant evidence to support the alternative hypothesis.
Degrees of Freedom
In statistics, the concept of degrees of freedom relates to the number of independent values that can vary in an analysis without breaking any constraints. It is crucial in calculating the reliability of certain statistics and in determining the critical values of various test statistics. In ANOVA, for example, the degrees of freedom can be split into two parts: one part for how the groups differ among themselves (between-group variation) and another for how the individuals within those groups differ (within-group or error variation). In the cortisol levels study, with 68 subjects and four groups, we have 67 total degrees of freedom: 3 are attributed to the groups and 64 to the error.
ANOVA
The ANOVA (Analysis of Variance) is a statistical method that allows researchers to compare more than two groups at the same time to determine if there is a statistically significant difference among them. It extends the t-test when there are more than two groups being compared. ANOVA decomposes the observed variance in a variable into component parts associated with different sources of variation. In our cortisol stress study, the ANOVA test was carried out to see if the means of cortisol changes for the different types of maternal interaction groups were statistically significantly different from each other. Importantly, ANOVA tells us whether there's at least one group that differs but doesn't specify which groups differ from each other. To find that out, we would require post-hoc testing.
P-Value Interpretation
The p-value is a fundamental aspect of hypothesis testing. It tells us the probability of obtaining the observed results, or more extreme, if the null hypothesis is true. A low p-value indicates that the observed data is unlikely under the null hypothesis, leading to its rejection. Conventionally, a threshold of 0.05 (5%) is used, below which we say the results are statistically significant. In the cortisol levels study, the fact that there were significant differences in stress reduction based on interaction type suggests that the p-value was below 0.05. This means that there's less than a 5% chance that the observed differences in cortisol levels happened due to random fluctuations, providing strong evidence against the null hypothesis.

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

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) What is the pooled standard deviation? What degrees of freedom are used in doing inferences for these means and differences in means?

Pulse Rate and Award Preference In Example 8.5 on page 548 we find evidence from the ANOVA of a difference in mean pulse rate among students depending on their award preference. The ANOVA table and summary statistics for pulse rates in each group are shown below. \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Award } & 2 & 2047 & 1024 & 7.10 & 0.001 \\ \text { Error } & 359 & 51729 & 144 & & \\ \text { Total } & 361 & 53776 & & & \\ \text { Level } & \text { N } & \text { Mean } & \text { StDev } & & \\ \text { Academy } & 31 & 70.52 & 12.36 & & \\ \text { Nobel } & 149 & 72.21 & 13.09 & & \\ \text { Olympic } & 182 & 67.25 & 10.97 & & \end{array}\) Use this information and/or the data in StudentSurvey to compare mean pulse rates, based on the ANOVA, between each of three possible pairs of groups: (a) Academy Award vs Nobel Prize. (b) Academy Award vs Olympic gold medal. (c) Nobel Prize vs Olympic gold medal.

Two sets of sample data, \(\mathrm{A}\) and \(\mathrm{B}\), are given. Without doing any calculations, indicate in which set of sample data, \(\mathrm{A}\) or \(\mathrm{B}\), there is likely to be stronger evidence of a difference in the two population means. Give a brief reason, comparing means and variability, for your answer. $$ \begin{array}{cc|cc} \hline {\text { Dataset A }} & {\text { Dataset B }} \\ \hline \text { Group 1 } & \text { Group 2 } & \text { Group 1 } & \text { Group 2 } \\ \hline 13 & 18 & 13 & 48 \\ 14 & 19 & 14 & 49 \\ 15 & 20 & 15 & 50 \\ 16 & 21 & 16 & 51 \\ 17 & 22 & 17 & 52 \\ \bar{x}_{1}=15 & \bar{x}_{2}=20 & \bar{x}_{1}=15 & \bar{x}_{2}=50 \end{array} $$

Drug Resistance and Dosing Exercise 8.39 on page 561 explores the topic of drug dosing and drug resistance by randomizing mice to four different drug treatment levels: untreated (no drug), light ( \(4 \mathrm{mg} / \mathrm{kg}\) for 1 day), moderate \((8 \mathrm{mg} / \mathrm{kg}\) for 1 day), or aggressive ( \(8 \mathrm{mg} / \mathrm{kg}\) for 5 or 7 days). Exercise 8.39 found that, contrary to conventional wisdom, higher doses can actually promote drug resistance, rather than prevent it. Here, we further tease apart two different aspects of drug dosing: duration (how many days the drug is given for) and amount per day. Recall that four different response variables were measured; two measuring drug resistance (density of resistant parasites and number of days infectious with resistant parasites) and two measuring health (body mass and red blood cell density). In Exercise 8.39 we don't find any significant differences in the health responses (Weight and \(R B C)\) so we concentrate on the drug resistance measures (ResistanceDensity and DaysInfectious) in this exercise. The data are available in DrugResistance and we are not including the untreated group. (a) Investigate duration by comparing the moderate treatment with the aggressive treatment (both of which gave the same amount of drug per day, but for differing number of days). Which of the two resistance response variables (ResistanceDensity and DaysInfectious) have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (b) Investigate amount per day by comparing the light treatment with the moderate treatment (both of which lasted only 1 day, but at differing amounts). Which of the two resistance response variables have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (c) Does duration or amount seem to be more influential (at least within the context of this study)? Why?

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 360.0 & 120.0 & 1.60 \\ \text { Error } & 16 & 1200.0 & 75.0 & \\ \text { Total } & 19 & 1560.0 & & \end{array}\)

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.

Sign-up for free