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In addition to monitoring weight gain, food consumed, and activity level, the study measured stress levels in the mice by measuring corticosterone levels in the blood (higher levels indicate more stress). Conditions for ANOVA are met and computer output for corticosterone levels for each of the three light conditions is shown. \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 73.40 & 67.49 \\ \text { LD } & 8 & 70.02 & 54.15 \\ \text { LL } & 9 & 50.83 & 42.22\end{array}\) (a) What is the conclusion of the analysis of variance test? (b) One group of mice in the sample appears to have very different corticosterone levels than the other two. Which group is different? What aspect of the data explains why the ANOVA test does not find this difference significant? How is this aspect reflected in both the summary statistics and the ANOVA table?

Short Answer

Expert verified
Without the ANOVA data, a general conclusion can't be made but differences in group means suggest the 'LL' group as being significantly different. However, large within-group variances, reflected by high standard deviations, might cause ANOVA not to detect this significance.

Step by step solution

01

Understanding the ANOVA test

Analyze the result of the ANOVA test. The Analysis of Variance (ANOVA) is a statistical method used to test if there are significant differences between the means of multiple groups. The ANOVA test first calculates the sum of squares (variance) between each group's mean and the total mean of all measurements (SSB), and the sum of squares within each group (SSW). The F-statistic is then calculated as the ratio of mean square between to mean square within. The p-value is the probability of observing the F-statistic given the null hypothesis.
02

Interpreting the conclusion of ANOVA test

As no numerical data is given for ANOVA, it is not possible to draw a concrete conclusion. However, generally, if the p-value from the ANOVA test is less than the significance level (conventionally 0.05), the null hypothesis is rejected, indicating that there is a statistical difference in corticosterone levels across the three different light conditions.
03

Identifying the different group

Analyzing the provided sample means and standard deviations for each group suggests that the 'LL' group might be the significantly different group. They have a noticeably lower mean corticosterone level than the 'DM' and 'LD' groups.
04

Reasoning the insignificance in ANOVA

The reason that ANOVA test might not find the difference in corticosterone levels of the 'LL' group significant could be due to large within-group variance (high standard deviation values), which increases the F value and hence the p-value. This could be reflected in the ANOVA table as high variability within each group, leading to inability to significantly differentiate groups.

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Key Concepts

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

Analysis of Variance
The Analysis of Variance (ANOVA) test is a cornerstone of statistical analysis when comparing the means of multiple groups. It helps determine if there is any statistically significant difference between the means. It does this by separating the total variation observed in the data into two parts: variation between the groups and variation within the groups. By comparing these variances, ANOVA can tell us if the differences among group means are more than what could be expected by chance alone. For example, in studying the corticosterone levels in mice under various light conditions, ANOVA can establish if the variation in stress levels is associated with the light exposure or if it's likely due to random variability.
Statistical Significance
Statistical significance is a term used to indicate that the results of an analysis are unlikely to have occurred by chance. In the context of ANOVA, when we say that the differences between group means are statistically significant, we mean there's a high confidence that these differences are real and not just random noise. This concept hinges on a predefined threshold called the alpha level (often 0.05), which is the probability limit we set to reject the null hypothesis. If the calculated p-value from ANOVA is below this level, we conclude that there's sufficient evidence to say that not all group means are equal, suggesting a significant effect of the independent variables (like light conditions) on the dependent variable (corticosterone levels).
F-statistic
The F-statistic is the test statistic calculated in an ANOVA test, which is used to determine whether the group means are significantly different. It's a ratio of the variance between the groups to the variance within the groups. A higher F-statistic indicates a greater degree of variance between group means than within groups, suggesting that the variable being tested for (like the intensity of the light conditions) has a significant impact on the outcome variable (stress levels in the mice). Nevertheless, its interpretation is not in isolation and should always be considered in conjunction with the p-value to make informed conclusions about the statistical significance of the results.
p-value
The p-value is a crucial concept in statistical hypothesis testing, and it represents the probability of observing the results, or more extreme results, assuming that the null hypothesis is true. In the context of ANOVA, the null hypothesis typically states that all group means are equal, and the alternative is that at least one group mean is different. A p-value lower than the chosen significance level (usually 0.05) empowers us to reject the null hypothesis with confidence, endorsing the presence of significant differences between the groups. In the mice study, if the p-value is below 0.05 for the light conditions, we would assert that the type of light environment influences stress levels in mice as measured by corticosterone.
Within-group variance
Within-group variance is a measure of dispersion that reflects how spread out the individual observations are around their group mean. High within-group variance can obscure differences between group means because it indicates that the data points within each group are widely scattered. This could be the case in our mice study, where the standard deviation, a measure of within-group variance, may be large. If the within-group variances are substantially higher, it can increase the p-value, making it more difficult to find a statistically significant difference between the group means. Hence, the ANOVA's ability to detect a true difference is hampered, illustrating the importance of not only looking at the means but also considering the variability within each group.

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Most popular questions from this chapter

Researchers hypothesized that the increased weight gain seen in mice with light at night might be caused by when the mice are eating. (As we have seen in the previous exercises, it is not caused by changes in amount of food consumed or activity level.) Perhaps mice with light at night eat a greater percentage of their food during the day, when they normally should be sleeping. Conditions for ANOVA are met and computer output for the percentage of food consumed during the day for each of the three light conditions is shown.\(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 55.516 & 10.881 \\ \text { LD } & 8 & 36.028 & 8.403 \\ \text { LL } & 9 & 76.573 & 9.646\end{array}\) (a) For mice in this sample on a standard light/dark cycle, what is the average percent of food consumed during the day? What percent is consumed at night? What about mice that had dim light at night? (b) Is there evidence that light at night influences when food is consumed by mice? Justify your answer with a p-value. Can we conclude that there is a cause-and-effect relationship?

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Test for a difference in population means between groups \(A\) and \(B\). 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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 2 & 540.0 & 270.0 & 8.60 \\ \text { Error } & 27 & 847.8 & 31.4 & \\ \text { Total } & 29 & 1387.8 & & \end{array}\)

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(95 \%\) confidence interval for the mean of population \(\mathrm{A}\).

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 \(0.05 ?\)

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