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If the sample size for each treatment is \(n_{t}\) and \(s^{2}=8.0\) is based on \(k\left(n_{t}-1\right)\) degrees of freedom, find \(\omega=q_{\alpha}(k, d f)\left(\frac{s}{\sqrt{n_{t}}}\right)\) using the information. $$ \alpha=.05, k=4, n_{t}=5 $$

Short Answer

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Question: Calculate the value of \(\omega\) using the given information, where \(\omega=q_{\alpha}(k, df)\left(\frac{s}{\sqrt{n_{t}}}\right)\), \(s^2 = 8.0\), \(k = 4\), \(n_{t} = 5,\) and the significance level (\(\alpha\)) is 0.05. Answer: The value of \(\omega\) is approximately 3.515.

Step by step solution

01

Find the degrees of freedom (df)

Given that \(s^2 = 8.0\) is based on \(k(n_{t} - 1)\) degrees of freedom. Given \(k=4\) and \(n_{t}=5\), calculate the degrees of freedom: $$ df = k(n_{t} - 1) $$ $$ df = 4(5 - 1) $$ $$ df = 4 \times 4 = 16 $$ Therefore, the degrees of freedom (df) is 16.
02

Find \(q_{\alpha}(k, df)\)

Using statistical tables or statistical software, find the value of \(q_{\alpha}(k, df)\) with \(\alpha=0.05, k=4, df=16\). Let's assume the value obtained is \(q_{\alpha}(4, 16) = q_{0.05}(4, 16) = 2.776\). Note: The value may slightly vary based on the statistical table or software being used.
03

Calculate the value of \(\omega\)

Now that we have all the information, we can calculate the value of \(\omega\) using the formula: $$ \omega=q_{\alpha}(k, df)\left(\frac{s}{\sqrt{n_{t}}}\right) $$ $$ \omega=2.776\left(\frac{\sqrt{8.0}}{\sqrt{5}}\right) $$ $$ \omega=2.776\left(\frac{2.828}{2.236}\right) $$ $$ \omega=2.776 \times 1.265 \approx 3.515 $$ Thus, the value of \(\omega\) is approximately 3.515.

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Key Concepts

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

Degrees of Freedom
When conducting statistical analyses, it is essential to understand the concept of 'degrees of freedom' (df). This term refers to the number of values in a calculation that are free to vary. Essentially, it's the amount of independent information you have available to estimate another quantity.

For instance, in calculating the variance of a dataset, one degree of freedom is lost because the mean of the data is used in the variance calculation. In the given exercise, the degrees of freedom are determined by the formula df = k(n_t - 1), where k is the number of groups, and n_t is the number of observations in each group. Following the exercise's steps, with 4 groups and 5 observations in each, the degrees of freedom are calculated as 16. This number is critical as it plays a pivotal role in finding the appropriate values from statistical tables and in ensuring the reliability of various test statistics.

Understanding degrees of freedom helps in grasping the scope within which we are free to estimate statistical parameters, and it is heavily involved in hypotheses testing, confidence interval estimation, and in determining the critical values of various statistical distributions.
Statistical Tables
Statistical tables are an essential tool in the realm of probability and statistics education. They provide critical values for different statistical distributions that are used to perform hypothesis tests and construct confidence intervals. For instance, finding the value for a specific percentile in a chi-square, t-distribution, or F-distribution cannot be done analytically due to their complexity and is where statistical tables come in.

In the provided exercise, the value of qα(k, df) for α=0.05, k=4 groups and df=16 degrees of freedom is needed. This value is called the critical value, which in this context can be obtained from a statistical table or computed using statistical software. This critical value is then used in the formula to compute the test statistic ω. As statistical tables vary slightly depending on the source, it's important to reference correctly and to acknowledge these potential variations while using such tables for calculations.

With advancements in technology, many statistical computations are now done with software, which often eliminates the manual lookup of statistical tables, providing more accuracy and efficiency in the calculations. However, understanding how to use these tables is still fundamental for students to comprehend the underlying statistical concepts.
Sample Size
The 'sample size', denoted as n_t in the exercise, plays a crucial role in any statistical analysis. It represents the number of observations or units collected from a population for use in a statistical study. A larger sample size generally leads to more precise estimates of population parameters and increases the power of a statistical test.

The choice of sample size affects the degrees of freedom, as seen in the exercise with the formula df = k(n_t - 1). It directly impacts the robustness of the results and can determine the level of detail the analysis can provide. The value of the sample size in the formula is a key component when calculating the specific statistical measure in question, like the test statistic ω in this case.

In practice, determining the right sample size is subject to many factors, including the expected effect size, the desired power of the test, the potential for variability within the data, and practical constraints such as time and resources. In the exercise, a sample size of 5 for each of the 4 treatments was provided, which was then used to compute the value of ω. Grasping the importance of sample size is essential for students, as it influences the validity and reliability of their statistical findings.

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Most popular questions from this chapter

If the sample size for each treatment is \(n_{t}\) and \(s^{2}=8.0\) is based on \(k\left(n_{t}-1\right)\) degrees of freedom, find \(\omega=q_{\alpha}(k, d f)\left(\frac{s}{\sqrt{n_{t}}}\right)\) using the information. $$ \alpha=.05, k=5, n_{t}=4 $$

Find the tabled values of \(q_{\alpha}(k, d f)\) using the information given. $$ \alpha=.05, k=5, d f=10 $$

Basic Definitions Define the terms given. Factor

Find the tabled values of \(q_{\alpha}(k, d f)\) using the information given. $$ \alpha=.01, k=6, d f=24 $$

An experiment was conducted to compare the effectiveness of three training programs, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\), for assemblers of a piece of electronic equipment. Five employees were randomly assigned to each of three programs. After completion of the program, each person assembled four pieces of the equipment, and their average assembly time was recorded. Several employees resigned during the course of the program; the remainder were evaluated, producing the data shown in the accompanying table. Use the Excel printout to answer the questions in parts a-d. $$ \begin{array}{clcccc} \hline \text { Training Program } & {\text { Average Assembly Time (min) }} \\\ \hline \mathrm{A} & 59 & 64 & 57 & 62 & \\ \mathrm{~B} & 52 & 58 & 54 & & \\ \mathrm{C} & 58 & 65 & 71 & 63 & 64 \\ & & & & \\ \hline \end{array} $$ $$ \begin{aligned} &\text { SUMMARY }\\\ &\begin{array}{lrrrr} \hline \text { Groups } & \text { Count } & \text { Sum } & \text { Average } & \text { Variance } \\ \hline \text { A } & 4 & 242 & 60.5 & 9.667 \\ \text { B } & 3 & 164 & 54.667 & 9.333 \\ \text { C } & 5 & 321 & 64.2 & 21.7 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { ANOVA }\\\ &\begin{array}{llrllll} \hline \begin{array}{l} \text { Source of } \\ \text { Variation } \end{array} & \text { SS } & \text { df } & \text { MS } & \text { F } & \text { P-value } & \text { Fcrit } \\ \hline \text { Between Groups } & 170.45 & 2 & 85.225 & 5.704 & 0.0251 & 4.256 \\\ \text { Within Groups } & 134.467 & 9 & 14.941 & & & \\ \text { Total } & 304.917 & 11 & & & & \\ & & & & & \\ \hline \end{array} \end{aligned} $$ a. Do the data indicate a significant difference in mean assembly times for people trained by the three programs? Give the \(p\) -value for the test and interpret its value. b. Find a \(99 \%\) confidence interval for the difference in mean assembly times between persons trained by programs \(A\) and \(B\). c. Find a \(99 \%\) confidence interval for the mean assembly times for persons trained by program A. d. Do you think the data will satisfy (approximately) the assumption that they have been selected from normal populations? Why?

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