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 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 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 \(\bar{x}_{1} \approx 4.86, s_{1} \approx 3.18, \bar{x}_{2}=6.5\), and \(s_{2} \approx 2.88\). 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 Mean of \(x_1\)

To find the mean of \(x_1\), sum all the values of \(x_1\) and divide by the total number of values. Given values: \(x_1 = [1, 5, 8, 4, 2, 4, 10]\)Sum: \(1 + 5 + 8 + 4 + 2 + 4 + 10 = 34\)Number of values: 7Mean: \(\bar{x}_{1} = \frac{34}{7} \approx 4.86\)

02

Calculate the Standard Deviation of \(x_1\)

First, determine the square of the difference between each value and the mean, then take their average, and finally, compute the square root. \(s_{1} = \sqrt{\frac{\sum (x_{1i} - \bar{x}_{1})^2}{n_1 - 1}}\)The variance for \(x_1\): \((1-4.86)^2 + (5-4.86)^2 + (8-4.86)^2 + (4-4.86)^2 + (2-4.86)^2 + (4-4.86)^2 + (10-4.86)^2 \approx 60.64\)Divide by 6 (\(n_1-1\)): \(60.64/6 \approx 10.107\)Standard deviation: \(s_{1} \approx \sqrt{10.107} \approx 3.18\)

03

Verify Mean and Standard Deviation for \(x_2\)

Sum all \(x_2\) values and divide by the total number for the mean:Given values: \(x_2 = [10, 5, 4, 7, 9, 4, 10, 3]\)Sum: \(10 + 5 + 4 + 7 + 9 + 4 + 10 + 3 = 52\)Number of values: 8Mean: \(\bar{x}_{2} = \frac{52}{8} = 6.5\)Already given that the standard deviation is \(s_{2} \approx 2.88\).

04

State Hypotheses for Mean Comparison

The null hypothesis \(H_0\) and alternative hypothesis \(H_a\) for comparing population means are:\(H_0: \mu_1 - \mu_2 = 0\) (no difference)\(H_a: \mu_1 - \mu_2 eq 0\) (difference exists)

05

Compute Test Statistic

Use the formula for the test statistic \(t\) for comparing means:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]With means and standard deviations known:\[t = \frac{4.86 - 6.5}{\sqrt{\frac{3.18^2}{7} + \frac{2.88^2}{8}}}\]\[t = \frac{-1.64}{\sqrt{1.4453 + 1.0368}}\]\[t \approx \frac{-1.64}{1.667} \approx -0.983\]

06

Determine Critical t-value

Find the critical t-value for a two-tailed test with degrees of freedom approximately calculated using smaller sample size, here available from tables at \(\alpha = 0.05\). With vine approximation for \(\text{df} \approx 13\), critical t-value is approximately \(\pm 2.160\).

07

Decision Making

Compare the test statistic with critical values. \(-0.983\) is between \(-2.160\) and \(2.160\).Thus, we fail to reject the null hypothesis \(H_0\).

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Key Concepts

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

Population Mean Comparison

When comparing the mean of two different populations, such as the hours lost by a manager due to two distinct causes, it is crucial to ensure that we correctly calculate and interpret their averages. Here, the two populations are represented by the variables $x_1$ and $x_2$. To find the mean for each dataset, add all the numbers in a dataset and divide by the total number of entries. For example, to compute the mean of $x_1$, add up all observations: 1, 5, 8, 4, 2, 4, and 10. The result is 34. Divide 34 by 7 (the number of observations), resulting in an average of approximately 4.86 hours.
Similarly, for $x_2$, adding up all numbers results in 52, and dividing by 8 gives a mean of 6.5 hours.
This comparison aims to determine if there is a significant difference in the average time lost on different issues. If the means are close, the issues might impact the manager's time equivalently. If they are far apart, one may be a more significant time loss factor than the other.

Standard Deviation Calculation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most numbers are close to the average. A high standard deviation means the numbers are spread out.
For \(x_1\), the standard deviation \(s_1\) is calculated using the formula:\( s_1 = \sqrt{\frac{\sum (x_{1i} - \bar{x}_1)^2}{n_1 - 1}}\)This involves several steps:

• Find the mean \( \bar{x}_1 \).
• Subtract the mean from each data point and square the result.
• Sum these squared differences.
• Divide the sum by \( n_1 - 1 \), where \( n_1 \) is the number of observations.
• Take the square root of the result, giving \( s_1 \approx \) 3.18 hours.

Similarly, perform these steps for \(x_2\), resulting in a standard deviation \(s_2\) of approximately 2.88 hours. This shows how scattered or concentrated the values are around the mean.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions based on data from populations. It involves making an initial assumption (the null hypothesis) and determining whether evidence collected from data significantly contradicts this assumption.
For this problem, the null hypothesis \( H_0 \) states that there is no difference in means: \( \mu_1 - \mu_2 = 0 \). The alternative hypothesis \( H_a \) suggests a difference: \( \mu_1 - \mu_2 eq 0 \).
To test these hypotheses, calculate the test statistic \( t \) using:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]With known means and standard deviations, this evaluates approximately to \( t \approx -0.983 \). With a significance level \( \alpha = 0.05 \), and comparing the value of t to critical values, we can determine if the null hypothesis should be rejected. Here, since \( t \) falls between critical values \(-2.160\) and \(2.160\), we fail to reject \( H_0 \). Thus, there is insufficient evidence of a difference in population means.

Independent Random Variables

In statistical analysis, independent random variables are those whose outcomes do not affect each other.
This concept is essential when comparing data such as $x_1$ and $x_2$ in the exercise. Each variable represents different issues affecting managerial time but independently measures distinct sources of lost time. Knowing the value of one variable does not provide any information about the other.
This independence is crucial for statistical methods applied, like hypothesis testing and standard deviation calculations, as it justifies assuming that the outcomes of the variables do not correlate. Therefore, when treated as independent subsets of data, each can be analyzed separately, allowing for unbiased comparisons and robust statistical conclusions.