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Chapter 8: Problem 30

Perhaps the mice with light at night in Exercise 8.28 gain more weight because they are exercising less. The conditions for an ANOVA test are met and computer output is shown for testing the average activity level for each of the three light conditions. Is there a significant difference in mean activity level? State the null and alternative hypotheses, give the F-statistic and the p-value, and clearly state the conclusion of the test.\(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 2503 & 1999 \\ \text { LD } & 8 & 2433 & 2266 \\ \text { LL } & 9 & 2862 & 2418\end{array}\)

Short Answer

Expert verified
As we lack the F-statistic and p-value in the provided data, we can't determine whether there is a significant difference in mean activity levels under different light conditions.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that there is no significant difference between the means of the three light conditions (DM, LD, LL). The alternative hypothesis (\(H_A\)) is that there is a significant difference between the means in at least one of the light conditions.
02

Compute the F-Statistic

The F-statistic in an ANOVA test is calculated by dividing the variance between the groups by the variance within the groups. However, the F-statistic is usually provided in computer output in statistical software which is in this case not given. So the exact value of the F-statistic can't be calculated here.
03

Compute the p-value

The p-value, similar to the F-statistic, is a value that is usually provided in the computer output which is not given in this case. As such, we don't have precise numbers to determine this in our exercise.
04

State the Conclusion

Normally, had we the F-statistic and the p-value, we'd compare the p-value with our significance level (\(\alpha\)), which is often 0.05. If the p-value was less than \(\alpha\), we would reject the null hypothesis in favor of the alternate one, indicating a significant difference in mean activity levels. If it was higher, we would fail to reject the null thus saying there's insufficient evidence to say the means vary. However, we don't have these values in this case.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistics, used to make determinations about a population based on sample data. In the context of our exercise, hypothesis testing is employed to assess whether there is a significant difference in the mean activity levels of mice across different light conditions at night.

To initiate this test, two hypotheses must be established:
  • The null hypothesis ( \(H_0\)) posits that there is no difference in the mean activity levels across the groups.
  • The alternative hypothesis ( \(H_A\)) suggests that at least one group's mean activity level significantly differs from the others.
The purpose of hypothesis testing is to evaluate these hypotheses to determine which one the sample data support more strongly. By concentrating on these hypotheses, we aim to ascertain if different lighting environments materially influence activity levels among the mice.
F-statistic
The F-statistic is a value that plays a crucial role in an Analysis of Variance (ANOVA) test. It is a ratio used to compare the variance between group means to the variance within the groups. Specifically, the F-statistic answers the question: is the variability among the group means substantial compared to the variability within each group? This is critical for determining if an observed effect (such as different light conditions affecting activity) is statistically significant.

In a typical ANOVA scenario, like in our exercise, the F-statistic is calculated as follows:
  • Numerator: Variance between the group means, which quantifies how much the means differ from each other.
  • Denominator: Variance within the groups, measuring the variance observed in each group.
The larger the F-statistic, the more likely it is that the group means differ significantly. Given this, the F-statistic helps us determine whether to reject the null hypothesis in favor of the alternative one.
p-value
The p-value in hypothesis testing provides a measure of the evidence against the null hypothesis. It is a probability that quantifies the strength of the results. In our exercise, the p-value is used to determine whether the observed differences in mean activity levels across various light conditions are statistically significant.
  • A low p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so we reject it in favor of the alternative hypothesis.
  • A high p-value (> 0.05) suggests that the observed data is consistent with the null hypothesis, and we fail to reject it.
Though we lack the exact p-value in our exercise, this step typically follows after computing or receiving the F-statistic from statistical software. The significance level (\(\alpha\)), often set at 0.05, serves as the cutoff for making decisions based on the p-value. This approach enables researchers to objectively assess whether there is enough statistical evidence to conclude that different lighting conditions significantly affect the activity of mice.

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Most popular questions from this chapter

Data 4.1 introduces a study in mice showing that even low-level light at night can interfere with normal eating and sleeping cycles. In the full study, mice were randomly assigned to live in one of three light conditions: LD had a standard light/dark cycle, LL had bright light all the time, and DM had dim light when there normally would have been darkness. Exercises 8.28 to 8.34 in Section 8.1 show that the groups had significantly different weight gain and time of calorie consumption. In Exercises 8.55 and \(8.56,\) we revisit these significant differences. Body Mass Gain Computer output showing body mass gain (in grams) for the mice after four weeks in each of the three light conditions is shown, along with the relevant ANOVA output. Which light conditions give significantly different mean body mass gain? \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 7.859 & 3.009 \\ \text { LD } & 9 & 5.987 & 1.786 \\ \text { LL } & 9 & 11.010 & 2.624\end{array}\) One-way ANOVA: BM4Gain versus Light \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\ \text { Light } & 2 & 116.18 & 58.09 & 8.96 & 0.001 \\ \text { Error } & 25 & 162.10 & 6.48 & & \\ \text { Total } & 27 & 278.28 & & & \end{array}\)

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. $$ \begin{array}{cc|cc} \hline {\text { Dataset A }} & {\text { Dataset B }} \\ \hline \text { Group 1 } & \text { Group 2 } & \text { Group 1 } & \text { Group 2 } \\ \hline 12 & 25 & 15 & 20 \\ 20 & 18 & 14 & 21 \\ 8 & 15 & 16 & 19 \\ 21 & 28 & 15 & 19 \\ 14 & 14 & 15 & 21 \\ \bar{x}_{1}=15 & \bar{x}_{2}=20 & \bar{x}_{1}=15 & \bar{x}_{2}=20 \\ \hline \end{array} $$

Color affects us in many ways. For example, Exercise C.92 on page 498 describes an experiment showing that the color red appears to enhance men's attraction to women. Previous studies have also shown that athletes competing against an opponent wearing red perform worse, and students exposed to red before a test perform worse. \(^{3}\) Another study \(^{4}\) states that "red is hypothesized to impair performance on achievement tasks, because red is associated with the danger of failure." In the study, US college students were asked to solve 15 moderately difficult, five-letter, single-solution anagrams during a 5-minute period. Information about the study was given to participants in either red, green, or black ink just before they were given the anagrams. Participants were randomly assigned to a color group and did not know the purpose of the experiment, and all those coming in contact with the participants were blind to color group. The red group contained 19 participants and they correctly solved an average of 4.4 anagrams. The 27 participants in the green group correctly solved an average of 5.7 anagrams and the 25 participants in the black group correctly solved an average of 5.9 anagrams. Work through the details below to test if performance is different based on prior exposure to different colors. (a) State the hypotheses. (b) Use the fact that sum of squares for color groups is 27.7 and the total sum of squares is 84.7 to complete an ANOVA table and find the F-statistic. (c) Use the F-distribution to find the p-value. (d) Clearly state the conclusion of the test.

The mice in the study had body mass measured throughout the study. Computer output showing body mass gain (in grams) after 4 weeks for each of the three light conditions is shown, and a dotplot of the data is given in Figure 8.6 . \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 7.859 & 3.009 \\ \text { LD } & 8 & 5.926 & 1.899 \\ \text { LL } & 9 & 11.010 & 2.624\end{array}\) (a) In the sample, which group of mice gained the most, on average, over the four weeks? Which gained the least? (b) Do the data appear to meet the requirement of having standard deviations that are not dramatically different? (c) The sample sizes are small, so we check that the data are relatively normally distributed. We see in Figure 8.6 that we have no concerns about the DM and LD samples. However, there is an outlier for the LL sample, at 17.4 grams. We proceed as long as the \(z\) -score for this value is within 卤3 . Find the \(z\) -score. Is it appropriate to proceed with ANOVA? (d) What are the cases in this analysis? What are the relevant variables? Are the variables categorical or quantitative?

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?
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