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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?

Short Answer

Expert verified
The pooled standard deviation is approximately 2.312439, and the degrees of freedom used are 14.

Step by step solution

01

Calculate the contribution of each group to the pooled standard deviation.

For group A: \((n_1 - 1) * s1^2 = (5-1) * 2.864^2 = 24.454656\). For group B: \((n_2 - 1) * s2^2 = (5-1) * 2.168^2 = 17.759744\). For group C: \((n_3 - 1) * s3^2 = (5-1) * 2.387^2 = 22.781769\).
02

Sum the group contributions

Total sum: \(24.454656 + 17.759744 + 22.781769 = 64.996169\).
03

Calculate the denominator of the pooled standard deviation formula

This is found by summing the sample sizes and subtracting the number of groups: \((n1 + n2 + n3 - k) = (5 + 5 + 5 - 3) = 12\).
04

Compute the pooled standard deviation

We do the square root of the contribution sum divided by the sum of the degrees of freedom: \(Sp = \sqrt{64.996169/12} = 2.312439\)
05

Find the Degrees of freedom

The degrees of freedom for inferences will equal to the total degrees of freedom shown in the table. Therefore, the degrees of freedom used for these means and differences in means is 14.

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Key Concepts

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

Degrees of Freedom
The concept of 'degrees of freedom' (DF) is essential in statistical analysis, acting as an integral part of various test and estimation procedures. In the context of pooled standard deviation, as in the textbook exercise, degrees of freedom represent the number of independent pieces of information available to estimate a statistic. When we calculate the variance or standard deviation for a single sample, we use DF = n - 1, where n is the sample size. This subtraction accounts for the fact that one value (the sample mean) has already been estimated, which restricts the variability of the dataset.

For ANOVA (Analysis of Variance), which compares the means of three or more groups, the calculation of degrees of freedom becomes partitioned. One part pertains to the variation between the groups (DF groups), and the other is concerned with the variation within the groups (DF error or DF within).
  • DF groups is calculated as the number of groups minus one (k - 1).
  • DF error is the total number of observations across all groups minus the number of groups (N - k).
The sum of these provides the total degrees of freedom, which is used to determine the critical value or p-value from the relevant distribution, be it a t-distribution or F-distribution, for the hypothesis being tested.
ANOVA
When analyzing datasets involving multiple groups, as seen in the textbook exercise, 'ANOVA' or Analysis of Variance is a statistical method used to test if there are significant differences between the means of three or more independent groups. ANOVA does this by comparing the ratio of between-group variance to within-group variance. The F-statistic, which results from this ratio, allows us to see if any of the group averages differ more than we'd expect by chance alone.

In the exercise, we are given an ANOVA table which includes several important metrics:

DF (Degrees of Freedom)

Which splits into 'Groups' and 'Error', helping us understand the number of independent values that can vary in the calculation of the statistics.

SS (Sum of Squares)

Which measures the total variation and splits again into 'Groups' (variation due to the interaction between groups) and 'Error' (variation within groups).

MS (Mean Square)

Which is calculated by dividing the 'SS' by their respective 'DF'. This average square gets larger if the observations are more spread out.

F

Which is the calculated test statistic from dividing 'MS Groups' by 'MS Error', and it tells us how much the group means vary relative to the variation within the groups themselves.
The 'P' value is then derived from the F-statistic, informing us whether the observed between-group differences are statistically significant. If the P-value is below a predetermined threshold (typically 0.05), we reject the null hypothesis that all group means are equal, suggesting there are significant differences among them.
Statistical Inference
Statistical inference involves drawing conclusions about a population's properties based on a sample. It encompasses a variety of procedures, including hypothesis testing, determining confidence intervals, and making predictions. In the case of our textbook exercise, statistical inference is used to decide whether the observed differences in sample means reflect true differences in population means or are merely a result of random sampling variation.

In the step-by-step solution, 'degrees of freedom' are particularly important for statistical inference as they influence the shape of the distribution used to determine the significance of our test statistic. For instance, a t-distribution, which is often used when estimating the mean of a normally-distributed population, becomes more similar to the standard normal distribution as the degrees of freedom increase.
  • In our pooled standard deviation calculation, we use the degrees of freedom associated with the error term from the ANOVA table to pool the variance estimates.
  • In hypothesis testing (such as in our ANOVA), degrees of freedom are also used to find the critical value or p-value associated with the test statistic.
Understanding statistical inference is crucial because it helps us make data-driven decisions and determine the reliability of our estimates and tests. Essentially, it's a measure of how confident we can be in the results obtained from our sample when applying them to the broader population.

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Most popular questions from this chapter

Effects of Synchronization and Exertion on Closeness Exercise 8.16 on page 554 looks at possible differences in ratings of closeness to a group after doing a physical activity that involves either high or low levels of synchronization (HS or LS) and high or low levels of exertion (HE or LE). Students were randomly assigned to one of four groups with different combinations of these variables, and the change in their ratings of closeness to their group (on a 1 to 7 scale) were recorded. The data are stored in SynchronizedMovement and the means for each treatment group are given below, along with an ANOVA table that indicates a significant difference in the means at a \(5 \%\) level. \(\begin{array}{llrr}\text { Group } & \text { N } & \text { Mean } & \text { StDev } \\ \text { HS+HE } & 72 & 0.319 & 1.852 \\ \text { HS+LE } & 64 & 0.328 & 1.861 \\ \text { LS+HE } & 66 & 0.379 & 1.838 \\ \text { LS+LE } & 58 & -0.431 & 1.623\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Analysis or varlance } & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F-Value } & \text { P-Value } \\ \text { Group } & 3 & 27.04 & 9.012 & 2.77 & 0.042 \\ \text { Error } & 256 & 831.52 & 3.248 & & \\ \text { Total } & 259 & 858.55 & & & \end{array}\) The first three means look very similar, but the LS+LE group looks quite a bit different from the others. Is that a significant difference? Test this by comparing the mean difference in change in closeness ratings between the synchronized, high exertion activity group (HS+HE) and the nonsvnchronized. low exertion activity group (LS+LE).

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Is there sufficient evidence of a difference in the population means of the three groups? Justify your answer using specific value(s) from the output.

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 \(0.05 ?\)

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 \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\ \text { Light } & 2 & 116.18 & 58.09 & 8.96 & 0.001 \\ \text { Error } & 25 & 162.10 & 6.48 & & \\ \text { Total } & 27 & 278.28 & & & \end{array}\)

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.\(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 2503 & 1999 \\ \text { LD } & 8 & 2433 & 2266 \\ \text { LL } & 9 & 2862 & 2418\end{array}\)

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