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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.

Short Answer

Expert verified
There is a statistically significant difference in mean pulse rates between all the three pairs: (a) Academy Award vs Nobel Prize, (b) Academy Award vs Olympic gold medal, (c) Nobel Prize vs Olympic gold medal, as suggested by a high F-value and a low P-value.

Step by step solution

01

Understand ANOVA table

The ANOVA (Analysis of Variance) table is a statistical table that shows the source of variance in a data set. Here, the sources provided are corresponding to the three awards. And also the Error and Total are given. The F value (7.10) and P value (0.001) in the table indicates a significant difference in the pulse rates across the three groups.
02

Compare Academy Award vs Nobel Prize

To compare the mean pulse rates of two groups, look at the mean values given in the data. For the Academy Award, the mean pulse rate is 70.52, and for the Nobel Prize, it is 72.21. Although the nobel prize winners have a higher mean pulse rate, a definitive comparison cannot be made without considering the standard deviations. Here, since they are not particularly small in comparison to the means and considering the high F-value and low P-value, there is statistically significant difference between these two groups.
03

Compare Academy Award vs Olympic Gold Medal

For the Olympic Gold Medal recipients, the mean pulse rate is 67.25, which is lower than the Academy Award winners. In this comparison too, considering the F-value, P-value and standard deviations, there is statistically significant difference between these two groups.
04

Compare Nobel Prize vs Olympic Gold Medal

In the case of the comparison between Nobel Prize winners and Olympic Gold Medal winners, the mean pulse rate is higher for the Nobel Prize winners. Again, based on the significance level from the ANOVA table, there is statistically significant difference between these two groups as well.

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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) is a powerful statistical technique used to determine whether there are any statistically significant differences between the means of three or more independent (unrelated) groups. ANOVA works by analyzing the variance within each group compared to the variance between groups.

In the context of our pulse rate study, the ANOVA table displays different components like the between-groups variance (Award), within-groups variance (Error), and the total variance. The key values to look at in an ANOVA table include the 'F' and 'P'. The 'F' value is a ratio of the variance between the groups to the variance within the groups; a larger 'F' often indicates a higher likelihood of differences between group means. The 'P' value, on the other hand, tells us about the level of statistical significance. In our study, a 'P' value of 0.001 suggests that the differences observed in pulse rates are not due to chance, and are thus statistically significant.

It's important to note that while ANOVA tells us that there is a significant difference, it does not specify which groups are different from each other. For that purpose, post-hoc tests or pairwise comparisons are typically conducted.
Mean Comparison
When the ANOVA suggests a difference between group means, further investigation is required to compare the means of each pair of groups. This is known as mean comparison or pairwise comparison. It is essential in studies like our pulse rate analysis to understand which specific groups' means differ from one another.

To conduct a mean comparison, one would typically look at the sample means and standard deviations for the groups in question. In the pulse rate study, the mean for Academy Award lovers is 70.52 beats per minute (bpm), Nobel Prize lovers have a mean of 72.21 bpm, and Olympic gold medal enthusiasts have a mean of 67.25 bpm. By comparing these means, educators and researchers can pinpoint not just whether there are differences, but the nature and direction of those differences. It is critical, however, to keep in mind the variability within groups (indicated by standard deviations), as large variability can affect the confidence in the differences between group means.
Statistical Significance
Statistical significance is a term used to determine whether the result of an experiment or study is likely not due to chance. In the setting of our pulse rate study, the term refers to the likelihood that the differences observed between the mean pulse rates of the various award preference groups are real and not just random variations.

The role of statistical significance in ANOVA is crucial as it guides the decision-making process. In our case, the P value is 0.001, which is well below the typical cutoff of 0.05. This indicates a less than 0.1% probability that the observed differences in pulse rates happened by chance, giving us strong evidence against the null hypothesis (which would suggest there are no differences). The lower the P value, the more convincing the evidence that there is a meaningful difference between the groups.

Understanding statistical significance helps students and researchers avoid drawing conclusions from random noise. It's a safeguard against attributing differences to an effect or treatment when they may have arisen simply from random fluctuations within the data.

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Most popular questions from this chapter

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

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?

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 ±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?

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Is there evidence for a difference in the population means of the four groups? Justify your answer using specific value(s) from the output.

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 \(\mathrm{A}\).

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