91Ó°ÊÓ

In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable \(x_{1}\) measures a manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions: \(\begin{array}{llllllll}x_{1}: & 1 & 5 & 8 & 4 & 2 & 4 & 10\end{array}\) The variable \(x_{2}\) measures a manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable": \(\begin{array}{lllllllll}x_{2}: & 10 & 5 & 4 & 7 & 9 & 4 & 10 & 3\end{array}\) i. Use a calculator with sample mean and standard deviation keys to verify that \(\bar{x}_{1}=4.86, s_{1} \approx 3.18, \bar{x}_{2}=6.5\), and \(s_{2}=2.88\). ii. Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use \(\alpha=0.05\). Assume that the two lost-time population distributions are mound-shaped and symmetric.

Step by step solution

01

Calculate the sample means

First, we calculate the sample mean for each dataset. For \(x_1\): \(\bar{x}_1 = \frac{1 + 5 + 8 + 4 + 2 + 4 + 10}{7} = 4.86\). For \(x_2\): \(\bar{x}_2 = \frac{10 + 5 + 4 + 7 + 9 + 4 + 10 + 3}{8} = 6.5\).

02

Calculate the sample standard deviations

Using the standard deviation formula, calculate for each dataset. For \(x_1\), the standard deviation \(s_1\) is approximately \(3.18\), as given. For \(x_2\), \(s_2 = 2.88\), also as given in the problem statement.

03

Define the hypotheses

To determine if there's a significant difference in means, we set up the null hypothesis \(H_0: \mu_1 = \mu_2\), and the alternative hypothesis \(H_a: \mu_1 eq \mu_2\).

04

Calculate the test statistic

For two independent samples, we use the formula \(t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{4.86 - 6.5}{\sqrt{\frac{3.18^2}{7} + \frac{2.88^2}{8}}} \approx -1.32\).

05

Determine the degrees of freedom

Use the formula for degrees of freedom for two samples: \(df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}\). Substituting the values, we find \(df \approx 12.73\), which we round down to 12.

06

Compare the test statistic to the critical value

For \(\alpha = 0.05\) and \(df = 12\), the two-tailed critical t-value is approximately 2.179. Since \(|t| = 1.32\) is less than 2.179, we fail to reject the null hypothesis.

07

Conclusion

Since the test statistic does not exceed the critical value, there is not enough evidence to conclude that the mean time lost due to hot tempers is different from that lost due to disputes over superior attitudes.

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Key Concepts

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

Sample Mean

The sample mean is a simple yet foundational concept in statistics. It represents the average of a set of data points and gives a central value that summarizes the entire dataset. Calculating the sample mean, denoted as \( \bar{x} \), involves summing up all the values in the dataset and then dividing by the number of data points.

For example, if we have a dataset: \( \{1, 5, 8, 4, 2, 4, 10\} \), we find the sample mean by calculating:

• Add all numbers together: \( 1 + 5 + 8 + 4 + 2 + 4 + 10 = 34 \)
• Divide the sum by the number of values: \( \bar{x}_1 = \frac{34}{7} \approx 4.86 \)

This calculation indicates the average hours lost due to a specific issue in the dataset. The sample mean provides a quick summary of where the bulk of the data lies and is especially useful for comparing different datasets.

Standard Deviation

Standard deviation is a measure of how spread out the numbers in a dataset are. It tells us how much individual data points typically differ from the mean. A large standard deviation suggests that the data points are spread across a wide range, while a small standard deviation indicates they are closer to the mean.

To calculate the standard deviation for a set of data, we follow these steps:

• Calculate the mean of the dataset.
• Subtract the mean from each data point to find the deviation of each point.
• Square each deviation to avoid negatives and then average these squares to find the variance.
• Take the square root of the variance to find the standard deviation.

In our example, the standard deviation for dataset \( x_1 \) was approximately \( 3.18 \). This number tells us how much variation exists from the mean, providing insight into the consistency of the hours lost among the managers surveyed.

Two-Sample t-Test

The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. This test is particularly helpful when comparing two separate populations. In the context of our exercise, it helps test if the time lost to hot tempers differs significantly from time lost to technical disputes.

To perform a two-sample t-test:

• Calculate the mean and standard deviation for both samples.
• Use these values to compute the t-statistic with the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]
• This formula factors in the difference between sample means and the variation within those samples.

After computing, compare the t-statistic to a critical value from the t-distribution table. If the absolute value of the t-statistic is greater than the critical value, there is a significant difference; otherwise, there is not.

Null Hypothesis

When performing hypothesis testing, the null hypothesis (\(H_0\)) is a statement that there is no effect or no difference. It is typically the hypothesis that one attempts to disprove or reject. In our exercise, the null hypothesis is that the mean time lost due to hot tempers (\(\mu_1\)) is equal to the mean time lost due to technical disputes (\(\mu_2\)), or \( H_0: \mu_1 = \mu_2 \).

The null hypothesis serves as a baseline assumption for the test. If the test results are statistically significant, we may reject the null hypothesis, which suggests there is enough evidence to indicate a real difference between groups.

Failing to reject the null hypothesis implies that any observed variation could just be due to random sample variation, reinforcing that there's no substantial evidence to claim a difference in our context.

Alternative Hypothesis

The alternative hypothesis (\(H_a\)) is a statement that contradicts the null hypothesis. It posits that there is an effect, or a difference exists. In the context of our exercise, the alternative hypothesis suggests that the mean time lost due to hot tempers is not equal to the time lost due to technical disputes, expressed as \( H_a: \mu_1 eq \mu_2 \).

The alternative hypothesis is accepted if the evidence strongly contradicts the null hypothesis. This acceptance indicates that the observed data is unlikely under the assumption that the null hypothesis is true, at a specified level of significance (\(\alpha\)).

In our study, setting a significance level of \(0.05\), we would only accept \(H_a\) if the test statistic falls beyond the critical value, showing a less than 5% probability that the observed difference is due to random chance.