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Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the six months following the exercise program. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the \(p\) -value. $$ \begin{array}{|lcc|} \hline \text { Employee } & \text { Before } & \text { After } \\ \hline \text { Bauman } & 6 & 5 \\ \text { Briggs } & 6 & 2 \\ \text { Dottellis } & 7 & 1 \\ \text { Lee } & 7 & 3 \\ \text { Perralt } & 4 & 3 \\ \text { Rielly } & 3 & 6 \\ \text { Steinmetz } & 5 & 3 \\ \text { Stoltz } & 6 & 7 \\ \hline \end{array} $$

Step by step solution

01

Set Up Hypotheses

The null hypothesis ( H_0 ) is that there is no difference in the number of absences before and after the program. The alternative hypothesis ( H_a ) is that the number of absences has decreased after the program began.

02

Calculate Differences

Calculate the difference in the number of absences for each employee by subtracting the 'After' absences from the 'Before' absences.\[\text{Differences: } 6-5=1, 6-2=4, 7-1=6, 7-3=4, 4-3=1, 3-6=-3, 5-3=2, 6-7=-1\]

03

Compute Test Statistic

The differences are: 1, 4, 6, 4, 1, -3, 2, -1. Calculate the mean difference and standard deviation of these differences. Then use the t-test formula:\[\bar{d} = \frac{1+4+6+4+1-3+2-1}{8} = 1.75\] \[s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} = 2.71\] \[t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{1.75}{2.71/\sqrt{8}} = 2.17\]

04

Determine Degrees of Freedom and Critical Value

The degrees of freedom (df) for this paired t-test is (n-1) where (n=8) . So df = 7. From t-distribution tables, for df = 7 and a significance level of 0.05 (one-tailed), the critical t-value is approximately 1.895.

05

Make a Decision

Since the calculated t-value of 2.17 is greater than the critical t-value of 1.895, we reject the null hypothesis. There is sufficient evidence to support that absences have decreased after the exercise program.

06

Estimate the p-value

Using a t-table or software, we find that the p ext{-value} associated with t = 2.17 and df = 7 is approximately 0.032, which is less than the significance level of 0.05.

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Key Concepts

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

Paired t-test

A paired t-test is a statistical method used to compare two related samples, or paired observations. In the case of Lester Hollar's study on absentees, each employee serves as a pair with measurements taken before and after the fitness program. The main goal of a paired t-test is to determine whether the mean difference in the pairs is statistically significant. Here, it helps to test if the fitness program has led to a reduction in absences.
To perform a paired t-test, we first calculate the difference between each pair's observations. For instance, if an employee had 6 absences before the program and 5 after, the difference is 1 day. These differences are then used to calculate the mean and standard deviation, essential for computing the t-test statistic. This test is appropriate when the data is normally distributed and the sample size is small, making it ideal for small groups like the eight employees in the study.

Significance Level

Significance level, denoted by \(\alpha\), is the threshold we set to determine if a result is statistically significant. In Lester Hollar's study, a significance level of 0.05 was used. This means that there's a 5% risk of concluding that an effect exists when there is none.
Choosing the significance level is crucial since it influences the outcome of hypothesis testing. If the \(p\)-value is less than \(\alpha\), we reject the null hypothesis and accept that there is an effect — in this case, a reduction in absences.
In general:

• A lower significance level (e.g., 0.01) means stricter criteria and less chance of a false positive.
• A higher significance level (e.g., 0.10) is less strict, increasing the chance of detecting an effect but also the risk of a false positive.

Estimating p-value

The \(p\)-value helps us determine the strength of the evidence against the null hypothesis. In other words, it's the probability of observing the sample results, or something more extreme, if the null hypothesis is true.
In the paired t-test conducted for the absences, a calculated t-value of 2.17 was found. To estimate the \(p\)-value, we compare this t-value against the t-distribution with the relevant degrees of freedom (df = 7 in this case).
The \(p\)-value found was approximately 0.032. Since this \(p\)-value is less than the significance level of 0.05, we conclude that there is significant evidence the fitness program helped decrease absenteeism.
Knowing how to estimate or find the \(p\)-value ensures you make informed conclusions based on statistical evidence.

Degrees of Freedom

Degrees of freedom (df) are a critical aspect in conducting statistical tests like the paired t-test. In simple terms, they refer to the number of independent values in a calculation. For a paired t-test, the degrees of freedom are calculated using \(n - 1\), where \ is the number of paired observations.
With 8 employees analyzed in our case, the degrees of freedom is 7. This concept determines the critical t-value used in deciding whether to reject the null hypothesis.
Understanding degrees of freedom is key as it affects both the shape of the t-distribution and the critical values you use. A wrong df can lead to incorrect conclusions about your hypothesis test, either risking a type I or a type II error.