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This problem is based on information regarding productivity in leading Silicon Valley companies (see reference in Problem 25). In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him- or herself talk. Let \(x_{1}\) be a random variable representing productive hours per week lost by peer employees of an intimidator. \(\begin{array}{llllllll}x_{1}: & 8 & 3 & 6 & 2 & 2 & 5 & 2\end{array}\) A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let \(x_{2}\) be a random variable representing productive hours per week lost by peer employees of a stressor. \(\begin{array}{lllllllll}x_{2}: & 3 & 3 & 10 & 7 & 6 & 2 & 5 & 8\end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=4.00, s_{1}=2.38, \bar{x}_{2}=5.5\), and \(s_{2}=2.78 .\) ii. Assuming the variables \(x_{1}\) and \(x_{2}\) are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a \(5 \%\) level of significance. (Assume the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetric.)

Step by step solution

01

Calculate Means and Standard Deviations

We first need to verify the calculated values of means and standard deviations for both variables, \(x_1\) and \(x_2\). Using the data: \(x_1: 8, 3, 6, 2, 2, 5, 2\) and \(x_2: 3, 3, 10, 7, 6, 2, 5, 8\), apply the formulas \(\bar{x} = \frac{\sum x}{n}\) for the mean and \(s = \sqrt{\frac{\sum (x - \bar{x})^2}{n-1}}\) for the standard deviation. Verify that \(\bar{x}_1 = 4.0, s_1 = 2.38\), \(\bar{x}_2 = 5.5, s_2 = 2.78\).

02

State the Hypotheses

Set up the null and alternative hypotheses. The null hypothesis \(H_0\): \(\mu_1 = \mu_2\) states that there is no difference in the population means. The alternative hypothesis \(H_a\): \(\mu_2 > \mu_1\) suggests that the mean time lost due to stressors is greater than that of intimidators.

03

Determine the Test Statistic

Given that both samples are independent and assuming normal distributions, we use a two-sample t-test for the difference of means. Compute the t-statistic using: \[ t = \frac{(\bar{x}_2 - \bar{x}_1)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(n_1 = 7\) and \(n_2 = 8\). Substitute the known values to calculate the t-statistic.

04

Find the Critical Value and Make the Decision

Using a significance level \(\alpha = 0.05\) and the degrees of freedom approximated using the smaller of \(n_1 - 1\) or \(n_2 - 1\), find the critical t-value from the t-distribution table. If the calculated t-statistic is greater than the critical value, reject \(H_0\). Otherwise, do not reject \(H_0\).

05

Conclusion

Based on the comparison of the test statistic and the critical value, conclude whether the data indicate that the mean time lost due to stressors is significantly greater than that lost due to intimidators.

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Key Concepts

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

Independent Samples t-test

The independent samples t-test is a statistical method used to compare the means of two independent groups. In the context of our exercise, we are interested in comparing the productive hours lost due to intimidators and stressors. The term 'independent' means that the samples from these two groups are separate and do not influence each other. For example, the data on intimidators does not affect the data on stressors.

This test helps determine if there is a statistically significant difference between the means of the two groups. The steps involve calculating the mean and standard deviation for each group. These values reflect the average and variability of lost hours, respectively. We then use these statistics to compute a t-statistic.

To conduct an independent samples t-test, it's important to assume that the data from both groups are normally distributed and have equal variances. However, some tests can still be robust even if these conditions are not strictly met. This allows for greater confidence in comparing data, especially when making decisions based on statistical evidence.

Population Mean Comparison

Population mean comparison focuses on assessing if two groups have significantly different central tendencies. In simple terms, we want to know if one group consistently has a higher average than the other.

For the problem we are tackling, we investigate whether stressors cause more lost productive time than intimidators. First, individual sample means from both groups are calculated. Then, through statistical testing, we estimate if these means differ significantly when it comes to the entire population of such data.

The null hypothesis we initially set states there is no difference between means. It acts like the default position. If our testing suggests that one mean is significantly larger than the other, we may reject the null hypothesis, suggesting one group indeed has a higher population mean in practice. It's about comparing averages to see apparent differences, leading to meaningful conclusions.

Statistical Significance

Statistical significance is a key concept for understanding if a result from data analysis reflects a real effect or if it could have occurred by chance. When we talk about significance in the context of hypothesis testing, we refer to how unlikely a result is, assuming the null hypothesis is true.

In our case, we are using a 5% significance level, which means we attribute significance to results only 5% likely to happen by random chance. During the hypothesis testing, if the t-statistic exceeds the critical value found in t-distribution tables, we reject the null hypothesis, affirming that any observed difference in means is statistically significant. This tells us that the population mean lost hours are genuinely greater for stressors than intimidators.

This does not equate to the conclusion being "important" or "large," but it highlights that the observed data strongly supports an actual difference in productivity loss between the two groups.