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Chapter 11: Problem 4

Suppose you wish to compare the means of four populations based on independent random samples, each of which contains six observations. Insert, in an ANOVA table, the sources of variation and their respective degrees of freedom.

Short Answer

Expert verified
The degrees of freedom for each source of variation in the ANOVA table are as follows: - Between Groups: 3 degrees of freedom - Within Groups: 20 degrees of freedom

Step by step solution

01

Identify the Sources of Variation

In an ANOVA table, there are typically two sources of variation: between groups and within groups. The "between groups" variation captures the differences among the group means, while the "within groups" variation captures the differences within each group.
02

Calculate the Degrees of Freedom for Between Groups

The degrees of freedom for the "between groups" variation is calculated by subtracting 1 from the number of groups/populations. In this case, we have 4 populations, so the degrees of freedom for the between groups variation is 4 - 1 = 3.
03

Calculate the Degrees of Freedom for Within Groups

The degrees of freedom for the "within groups" variation is calculated by multiplying the number of populations by the degrees of freedom within each population. Since there are 6 observations within each population, there are 6 - 1 = 5 degrees of freedom within each population. So, the total degrees of freedom for the within groups variation is 4 * 5 = 20.
04

Set Up the ANOVA Table

Now that we have identified the sources of variation and calculated their respective degrees of freedom, we can set up the ANOVA table. | Source of Variation | Degrees of Freedom | |---------------------|--------------------| | Between Groups | 3 | | Within Groups | 20 | | Total | 23 | In summary, the ANOVA table contains two sources of variation: between groups with 3 degrees of freedom and within groups with 20 degrees of freedom.

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Most popular questions from this chapter

Suppose you wish to compare the means of six populations based on independent random samples, each of which contains 10 observations. Insert, in an ANOVA table, the sources of variation and their respective degrees of freedom.

Twenty third graders were randomly separated into four equal groups, and each group was taught a mathematical concept using a different teaching method. At the end of the teaching period, progress was measured by a unit test. The scores are shown below (one child in group 3 was absent on the day that the test was administered). $$ \begin{array}{rrrr} && {\text { Group }} \\ \hline \\ 1 & 2 & 3 & 4 \\ \hline 112 & 111 & 140 & 101 \\ 92 & 129 & 121 & 116 \\ 124 & 102 & 130 & 105 \\ 89 & 136 & 106 & 126 \\ 97 & 99 & & 119 \end{array} $$ a. What type of design has been used in this experiment? b. Construct an ANOVA table for the experiment. c. Do the data present sufficient evidence to indicate a difference in the average scores for the four teaching methods? Test using \(\alpha=.05\).

The sample means corresponding to populations 1 and 2 in Exercise 11.4 are \(\bar{x}_{1}=88.0\) and \(\bar{x}_{2}=83.9\) a. Find a \(90 \%\) confidence interval for \(\mu_{1}\). b. Find a \(90 \%\) confidence interval for the difference \(\left(\mu_{1}-\mu_{2}\right)\)

An experiment was conducted to compare the effectiveness of three training programs, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) in training assemblers of a piece of electronic equipment. Fifteen employees were randomly assigned, five each, to the three programs. After completion of the courses, each person was required to assemble four pieces of the equipment, and the average length of time required to complete the assembly was recorded. Several of the employees resigned during the course of the program; the remainder were evaluated, producing the data shown in the accompanying table. Use the MINITAB printout to answer the questions. $$ \begin{array}{lllll} \text { Training Program } & {\text { Average Assembly Time (min) }} \\ \hline \text { A } && 59 & 64 & 57 & 62 \\ \text { B } && 52 & 58 & 54 & \\ \text { C } && 58 & 65 & 71 & 63 & 64 \end{array} $$ a. Do the data provide sufficient evidence to indicate a 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 \(\mathrm{A}\) and \(\mathrm{B}\) c. Find a \(99 \%\) confidence interval for the mean assembly times for persons trained in program A. d. Do you think the data will satisfy (approximately) the assumption that they have been selected from normal populations? Why?

Exercise 10.40 examined an advertisement for Albertsons, a supermarket chain in the western United States. The advertiser claims that Albertsons has consistently had lower prices than four other full-service supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total based on the prices of approximately 95 items is given for five different supermarket chains recorded during 4 consecutive weeks. $$ \begin{array}{llrlll} & \text { Albertsons } & \text { Ralphs } & \text { Vons } & \text { Alpha Beta } & \text { Lucky } \\ \hline \text { Week 1 } & \$ 254.26 & \$ 256.03 & \$ 267.92 & \$ 260.71 & \$ 258.84 \\ \text { Week 2 } & 240.62 & 255.65 & 251.55 & 251.80 & 242.14 \\ \text { Week 3 } & 231.90 & 255.12 & 245.89 & 246.77 & 246.80 \\ \text { Week 4 } & 234.13 & 261.18 & 254.12 & 249.45 & 248.99 \end{array} $$ a. What type of design has been used in this experiment? b. Conduct an analysis of variance for the data. c. Is there sufficient evidence to indicate that there is a difference in the average weekly totals for the five supermarkets? Use \(\alpha=.05\) d. Use Tukey's method for paired comparisons to determine which of the means are significantly different from each other. Use \(\alpha=.05 .\)
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