/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Exercise \(15.8\) presented the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercise \(15.8\) presented the accompanying summary information on satisfaction levels for employees on three different work schedules: $$ \begin{array}{lll} n_{1}=24 & n_{2}=24 & n_{3}=20 \\ \bar{x}_{1}=6.60 & \bar{x}_{2}=5.37 & \bar{x}_{3}=5.20 \end{array} $$ MSE was computed to be 2.028. Calculate the \(95 \%\) T-K intervals, and identify any significant differences.

Short Answer

Expert verified
The steps to calculate the Tukey-Kramer intervals involve comparing each pair of group means, and constructing the intervals using the mean squared error, sample sizes and appropriate studentized range value. Significance is determined by whether or not the interval contains zero.

Step by step solution

01

Gather data and necessary equations

Get the group sample sizes (\(n_1\), \(n_2\), \(n_3\)), the group means (\(\bar{x}_1\), \(\bar{x}_2\), \(\bar{x}_3\)), and the mean squared error (MSE). The equation to calculate the Tukey-Kramer statistic is \(| \bar{x}_i - \bar{x}_j |\) and for the intervals it is \(\bar{x}_i - \bar{x}_j \pm \sqrt{MSE}*\sqrt{(1/n_i)+(1/n_j)} * Q(0.05, v, r)\) where v is the degrees of freedom (df error), r is the number of groups and the function Q are the studentized range values which you usually find in tables.
02

Calculate difference between group means

Calculate the difference between each pair of group means: \( \bar{x}_1 - \bar{x}_2, \bar{x}_1 - \bar{x}_3 \) and \( \bar{x}_2 - \bar{x}_3 \). This will serve as the center of the T-K intervals.
03

Calculate the T-K intervals

For each pair of group means, plug in the values into the T-K interval formula to get the intervals. If the interval does not contain zero, it means that the means of the two groups are significantly different.
04

Interpret the results

Check the intervals obtained in Step 3. If the interval does not contain zero, it means that there is a significant difference between the groups. If all intervals contain zero, then there are no significant differences between the groups.

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Key Concepts

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

Tukey-Kramer Method
The Tukey-Kramer Method is a statistical procedure used when comparing the means of different groups, especially after performing an ANOVA test. This method helps identify which specific group means are significantly different from each other. It is particularly useful when you have unequal sample sizes, as this is a common occurrence in real-world data.

The essence of the Tukey-Kramer procedure is to calculate an interval around the difference between each pair of group means. By assessing whether these intervals include zero, we can determine if the differences are statistically significant. The intervals are adjusted to take into account both the mean squared error (MSE) from the ANOVA and the number of groups being compared.
  • If an interval includes zero, the means can be considered not significantly different.
  • If an interval does not include zero, the means are significantly different.
Group Mean Comparison
Group Mean Comparison involves examining the average outcomes of different groups to determine if they are statistically distinct from one another. In the context of the provided exercise, these groups represent different work schedules with their respective satisfaction levels.

The process begins by taking the difference between the means of two groups. This step allows us to quantify how much the group averages vary from each other, serving as the foundation for further statistical analysis, such as constructing confidence intervals. By carefully examining these differences, we gain insights into where significant disparities in satisfaction might exist.
ANOVA Assumptions
ANOVA, or Analysis of Variance, is based on several key assumptions that need to be met for the results to be valid. These assumptions are important to ensure that the statistical conclusions drawn from the data accurately reflect reality.

  • Independence: Observations within each group must be independent of each other.
  • Normality: The data in each group should follow a normal distribution.
  • Homogeneity of Variances: The variance among the groups should be approximately equal. This is often assessed using Levene's Test.
The assumption of homogeneity of variances is particularly critical because unequal variances can lead to misleading results. In cases where these assumptions are violated, adjustments or alternative statistical methods may be required.
Confidence Intervals
Confidence Intervals are a fundamental concept in statistics, providing an estimated range for a population parameter. In the context of the Tukey-Kramer Method, confidence intervals help determine if the mean differences between groups are statistically significant.

In this exercise, a 95% confidence interval was calculated for each pair of group means. This interval aims to capture the true difference in means with 95% confidence. When an interval does not include zero, the difference is considered statistically significant, implying a real difference in satisfaction levels between those groups.

Confidence intervals provide a more nuanced view of data, allowing for a better understanding of variability and reliability in the estimates of group differences.
Satisfaction Levels
Satisfaction Levels often refer to how satisfied individuals are with a particular experience or aspect of their work. In this exercise, satisfaction levels are the central measure of interest, assessed among employees on different work schedules.

Measuring and comparing satisfaction levels across groups can offer insights into employee well-being and productivity. Higher satisfaction levels might correlate with better job performance and lower turnover, making it a crucial metric for organizational success.

Through statistical analysis, such as the Tukey-Kramer Method, comparing the satisfaction across different schedules can help identify optimal conditions that promote happier and more effective work environments.

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Most popular questions from this chapter

The accompanying data on calcium content of wheat are consistent with summary quantities that appeared in the article "Mineral Contents of Cereal Grains as Affected by Storage and Insect Infestation" (Journal of Stored Products Research [1992]: 147-151). Four different storage times were considered. Partial output from the SAS computer package is also shown. $$ \begin{array}{lllllll} \begin{array}{l} \text { Storage } \\ \text { Period } \end{array} & \multicolumn{6}{c} {\text { Observations }} \\ \hline 0 \text { months } & 58.75 & 57.94 & 58.91 & 56.85 & 55.21 & 57.30 \\ 1 \text { month } & 58.87 & 56.43 & 56.51 & 57.67 & 59.75 & 58.48 \\ 2 \text { months } & 59.13 & 60.38 & 58.01 & 59.95 & 59.51 & 60.34 \\ 4 \text { months } & 62.32 & 58.76 & 60.03 & 59.36 & 59.61 & 61.95 \\ \hline \end{array} $$ $$ \begin{aligned} &\text { Dependent Variable: CALCIUM }\\\ &\begin{array}{lrrrrr} & & \text { Sum of } & \text { Mean } & & \\ \text { Source } & \text { DF } & \text { Squares } & \text { Square } & \text { F Value } & \text { Pr>F } \\ \text { Model } & 3 & 32.13815000 & 10.71271667 & 6.51 & 0.0030 \\ \text { Error } & 20 & 32.90103333 & 1.64505167 & & \\ \text { Corrected Total } & 23 & 65.03918333 & & & \\ & \text { R-Square } & \text { C.V. } & \text { Root MSE } & \text { CALCIUM Mean } \\ & 0.494135 & 2.180018 & 1.282596 & 58.8341667 \end{array} \end{aligned} $$ a. Verify that the sums of squares and df's are as given in the ANOVA table. b. Is there sufficient evidence to conclude that the mean calcium content is not the same for the four different storage times? Use the value of \(F\) from the ANOVA table to test the appropriate hypotheses at significance level .05.

15.19 The Gunning Fog index is a measure of reading difficulty based on the average number of words per sentence and the percentage of words with three or more syllables. High values of the Gunning Fog index are associated with difficult reading levels. Independent random samples of six advertisements were taken from three different magazines, and Gunning Fog indices were computed to obtain the data given in the accompanying table ("Readability Levels of Magazine Advertisements"'Journal of Advertising Research [1981]: 45-50). Construct an ANOVA table, and then use a significance level of .01 to test the null hypothesis of no difference between the mean Gunning Fog index levels for advertisements appearing in the three magazines. $$ \begin{array}{lrrrrrr} \text { Scientific American } & 15.75 & 11.55 & 11.16 & 9.92 & 9.23 & 8.20 \\ \text { Fortune } & 12.63 & 11.46 & 10.77 & 9.93 & 9.87 & 9.42 \\ \text { New Yorker } & 9.27 & 8.28 & 8.15 & 6.37 & 6.37 & 5.66 \end{array} $$

Do lizards play a role in spreading plant seeds? Some research carried out in South Africa would suggest so ("Dispersal of Namaqua Fig (Ficus cordata cordata) Seeds by the Augrabies Flat Lizard (Platysaurus broadleyi)," Journal of Herpetology [1999]: 328-330). The researchers collected 400 seeds of this particular type of fig, 100 of which were from each treatment: lizard dung, bird dung, rock hyrax dung, and uneaten figs. They planted these seeds in batches of 5, and for each group of 5 they recorded how many of the seeds germinated. This resulted in 20 observations for each treatment. The treatment means and standard deviations are given in the accompanying table. $$ \begin{array}{lccc} \text { Treatment } & \boldsymbol{n} & \overline{\boldsymbol{x}} & \boldsymbol{s} \\ \hline \text { Uneaten figs } & 20 & 2.40 & .30 \\ \text { Lizard dung } & 20 & 2.35 & .33 \\ \text { Bird dung } & 20 & 1.70 & .34 \\ \text { Hyrax dung } & 20 & 1.45 & .28 \\ \hline \end{array} $$ a. Construct the appropriate ANOVA table, and test the hypothesis that there is no difference between mean number of seeds germinating for the four treatments. b. Is there evidence that seeds eaten and then excreted by lizards germinate at a higher rate than those eaten and then excreted by birds? Give statistical evidence to support your answer.

The article "Growth Response in Radish to Sequential and Simultaneous Exposures of \(\mathrm{NO}_{2}\) and \(\mathrm{SO}_{2} "\) ' (Environmental Pollution [1984]: 303-325) compared a control group (no exposure), a sequential exposure group (plants exposed to one pollutant followed by exposure to the second four weeks later), and a simultaneous-exposure group (plants exposed to both pollutants at the same time). The article states, "Sequential exposure to the two pollutants had no effect on growth compared to the control. Simultaneous exposure to the gases significantly reduced plant growth." Let \(\bar{x}_{1}, \bar{x}_{2}\), and \(\bar{x}_{3}\) represent the sample means for the control, sequential, and simultaneous groups, respectively. Suppose that \(\bar{x}_{1}>\bar{x}_{2}>\bar{x}_{3}\). Use the given information to construct a table where the sample means are listed in increasing order, with those that are judged not to be significantly different underscored.

Research carried out to investigate the relationship between smoking status of workers and short-term absenteeism rate (hr/mo) yielded the accompanying summary information ("Work-Related Consequences of Smoking Cessation," Academy of Management Journal \([1989]:\) 606-621). In addition, \(F=2.56\). Construct an ANOVA table, and then state and test the appropriate hypotheses using a .01 significance level. $$ \begin{array}{lrl} \text { Status } & \begin{array}{l} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{l} \text { Sample } \\ \text { Mean } \end{array} \\ \hline \text { Continuous smoker } & 96 & 2.15 \\ \text { Recent ex-smoker } & 34 & 2.21 \\ \text { Long-term ex-smoker } & 86 & 1.47 \\ \text { Never smoked } & 206 & 1.69 \\ & & \\ \hline \end{array} $$

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