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Chapter 15: Problem 10

It has been reported that varying work schedules can lead to a variety of health problems for workers. The article "Nutrient Intake in Day Workers and Shift Workers" (Work and Stress [1994]: 332-342) reported on blood glucose levels (mmol/L) for day-shift workers and workers on two different types of rotating shifts. The sample sizes were \(n_{1}=37\) for the day shift, \(n_{2}=34\) for the second shift, and \(n_{3}=25\) for the third shift. A single- factor ANOVA resulted in \(F=3.834\). At a significance level of .05, does true average blood glucose level appear to depend on the type of shift?

Short Answer

Expert verified
Yes, at the given significance level of 0.05, the true average blood glucose level appears to depend on the type of shift, since our calculated F-statistic (3.834) exceeds the F critical value (approximately 3.11). This suggests that the null hypothesis has been rejected in favour of the alternative hypothesis.

Step by step solution

01

Understanding the Hypotheses

First, it's important to set up our null and alternative hypotheses. We assume initially (null hypothesis, \(H_0\)) that all the shifts have the same average glucose level. The alternative hypothesis (\(H_A\)) is that at least one of the shifts does not have the same average glucose level.
02

Calculate Degrees of Freedom

To perform an F-test, we will first need to calculate the degrees of freedom with the formula: \(df_1 = k - 1\) and \(df_2 = N - k\), where \(k\) is the number of groups (in this case, 3 shifts), and \(N\) is the total number of observations (total sample size, which is \(37 + 34 + 25 = 96\)). So, \(df_1 = 3 - 1 = 2\) and \(df_2 = 96 - 3 = 93\).
03

Find the Critical Value

Next, using an F-distribution table or a calculator with an F-distribution feature, find the critical value of F. Given the degrees of freedom (\(df1 = 2, df2 = 93\)) and significance level (\(\alpha = 0.05\)), assuming the null hypothesis, the critical value for F will be found from the F-distribution table. For the \(0.05\) level and degrees of freedom \(2\) and \(93\), the critical value is approximately \(3.11\).
04

Decision and Interpretation

Finally, we can now compare the calculated F statistic \(3.834\) to the critical F value \(3.11\) we found in the previous step. Since \(3.834 > 3.11\), we reject the null hypothesis. This suggests that the true average glucose level appears to depend on the type of shift.

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Key Concepts

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

F-test
The F-test is a statistical procedure used to determine if there are significant differences between the means of several groups. In the context of Analysis of Variance (ANOVA), which is the method applied in the exercise, the F-test is used to compare the variance among group means to the variance within the individual groups.

Using the F-test involves calculating an F-statistic, which is a ratio of two variances: the variance between group means (Mean Square Between or MSB) and the variance within the groups (Mean Square Error or MSE). Mathematically, the F-statistic is represented as \[\begin{equation}F = \frac{MSB}{MSE}\end{equation}\].In the exercise, after conducting an ANOVA on the blood glucose levels of workers across different shifts, an F-statistic of 3.834 was calculated. This F-statistic is then compared with a critical value from an F-distribution table, which is based on a pre-determined significance level (commonly 0.05) and the degrees of freedom for the variance estimates.

If the calculated F-statistic is greater than the critical value, this suggests that the observed variation in group means is unlikely to be due to random chance. The result would indicate that there is a statistically significant difference among the means, which would lead to rejecting the null hypothesis, as the exercise demonstrates.
Null Hypothesis
The null hypothesis, denoted as \[\begin{equation}H_{0}\end{equation}\], is a fundamental concept in hypothesis testing and serves as a default or baseline assumption. It posits that there is no effect or no difference between groups or treatments in the context of the research question. The purpose of employing the null hypothesis is to establish a standard against which the actual observed data can be compared.

In the given exercise, the null hypothesis proposes that the average blood glucose levels for workers are the same across all types of shifts. To test this hypothesis, the ANOVA and F-test methodology is applied. If the evidence (in this case, the calculated F-statistic) suggests that the observed data are inconsistent with the null hypothesis, then the null hypothesis would be rejected, indicating that there are significant differences between the groups. Conversely, if the data are consistent with the null hypothesis, it would not be rejected, implying that any observed differences might be due to random variation rather than a systematic effect.
Degrees of Freedom
Degrees of freedom often abbreviated as 'df', play a crucial role in the computation of statistical tests, including the F-test in ANOVA. Degrees of freedom are a measure of the amount of independent information available to estimate a parameter or calculate a statistic. They can be interpreted as the number of values in a calculation that are free to vary.

In a single-factor ANOVA, as presented in the exercise, two kinds of degrees of freedom are essential: \[\begin{equation}df_{1} = k - 1\end{equation}\] for the numerator and \[\begin{equation}df_{2} = N - k\end{equation}\] for the denominator, where \[\begin{equation}k\end{equation}\] is the number of groups, and \[\begin{equation}N\end{equation}\] is the total sample size. These degrees of freedom are used to specify the particular F-distribution to determine the critical F-value for the significance level chosen.

Having the correct degrees of freedom is vital because they ensure that the test's results are accurate and reflect the variability within the data set. If the degrees of freedom were calculated incorrectly, it might lead to interpretation errors regarding the statistical significance of the findings.

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Most popular questions from this chapter

Give as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations. a. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=5.37\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=1.90\) c. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=4.89\) d. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=14.48\) e. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=2.69\) f. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

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The article "Utilizing Feedback and Goal Setting to Increase Performance Skills of Managers" (Academy of Management Journal \([1979]: 516-526\) ) reported the results of an experiment to compare three different interviewing techniques for employee evaluations. One method allowed the employee being evaluated to discuss previous evaluations, the second involved setting goals for the employee, and the third did not allow either feedback or goal setting. After the interviews were concluded, the evaluated employee was asked to indicate how satisfied he or she was with the interview. (A numerical scale was used to quantify level of satisfaction.) The authors used ANOVA to compare the three interview techniques. An \(F\) statistic value of \(4.12\) was reported. a. Suppose that a total of 33 subjects were used, with each technique applied to 11 of them. Use this information to conduct a level \(.05\) test of the null hypothesis of no difference in mean satisfaction level for the three interview techniques. b. The actual number of subjects on which each technique was used was \(45 .\) After studying the \(F\) table, explain why the conclusion in Part (a) still holds.

In the introduction to this chapter, we considered a study comparing three groups of college students (soccer athletes, nonsoccer athletes, and a control group consisting of students who did not participate in intercollegiate sports). The following information on scores from the Hopkins Verbal Learning Test (which measures immediate memory recall) was $$\begin{array}{l|ccc} \text { Group } & \text { Soccer Athletes } & \text { Nonsoccer Athletes } & \text { Control } \\ \hline \text { Sample size } & 86 & 95 & 53 \\ \text { Sample mean score } & 29.90 & 30.94 & 29.32 \\ \begin{array}{l} \text { Sample standard } \\ \text { deviation } \end{array} & 3.73 & 5.14 & 3.78 \\ \hline \end{array}$$ In addition, \(\overline{\bar{x}}=30.19\). Suppose that it is reasonable to regard these three samples as random samples from the three student populations of interest. Is there sufficient evidence to conclude that the mean Hopkins score is not the same for the three student populations? Use \(\alpha=.05\).

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