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Chapter 11: Problem 7

These data are observations collected using a completely randomized design: $$ \begin{array}{lll} \text { Sample 1 } & \text { Sample 2 } & \text { Sample 3 } \\ \hline 3 & 4 & 2 \\ 2 & 3 & 0 \\ 4 & 5 & 2 \\ 3 & 2 & 1 \\ 2 & 5 & \end{array} $$ a. Calculate CM and Total SS. b. Calculate SST and MST. c. Calculate SSE and MSE d. Construct an ANOVA table for the data. e. State the null and alternative hypotheses for an analysis of variance \(F\) -test. f. Use the \(p\) -value approach to determine whether there is a difference in the three population means.

Short Answer

Expert verified
In summary, we used ANOVA to analyze the data from three samples. We calculated descriptive statistics for each sample, followed by the correction factor (CM), Total Sum of Squares (Total SS), Sum of Squares for Treatments (SST), Sum of Squares for Errors (SSE), Mean Square for Treatments (MST), and Mean Square for Errors (MSE). Then, we constructed an ANOVA table and calculated the F-value and p-value. We stated the null and alternative hypotheses and used the p-value approach to determine if there was a significant difference between the three population means. The result indicated that there was no significant difference between the means of the three populations.

Step by step solution

01

Calculate descriptive statistics for each sample

First, we need to calculate the mean, variance, and count of observations for each sample: $$ \begin{array}{cll} \text{Mean of Sample 1} & = \frac{3+2+4+3+2}{5} = 2.8 \\ \text{Mean of Sample 2} & = \frac{4+3+5+2+5}{5} = 3.8 \\ \text{Mean of Sample 3} & = \frac{2+0+2+1}{4} = 1.25 \\ \text{Count of Sample 1 (n1)} & = 5 \\ \text{Count of Sample 2 (n2)} & = 5 \\ \text{Count of Sample 3 (n3)} & = 4 \\ \text{Total Count (N)} & = n1 + n2 + n3 = 14 \end{array} $$
02

Calculate CM, Total SS, SST, and SSE

Now, we'll calculate CM, Total SS, SST, and SSE using the following formulas: $$ \begin{array}{cll} \text{Grand Mean (GM)} & = \frac{\Sigma x}{N} = \frac{ (5 \times 2.8) + (5 \times 3.8) + (4 \times 1.25)}{14} = 2.71 \\ \text{CM} & = \Sigma n_i(GM)^2 = 5(2.71)^2 + 5(2.71)^2 + 4(2.71)^2 = 106.4 \\ \text{Total SS} & = \Sigma_{i=1}^{N}x_i^2 - CM = (3^2+2^2+\cdots+1^2) - 106.4 = 55.6 \\ \text{SST} & = \Sigma n_i(\bar{x_i} - GM)^2 = 5(2.8-2.71)^2 + 5(3.8-2.71)^2 + 4(1.25-2.71)^2 = 15.91 \\ \text{SSE} & = \text{Total SS} - \text{SST} = 55.6 - 15.91 = 39.69 \end{array} $$
03

Calculate MST and MSE

Next, we'll calculate MST and MSE: $$ \begin{array}{cll} \text{MST} & = \frac{\text{SST}}{k - 1} = \frac{15.91}{3 - 1} = 7.955 \\ \text{MSE} & = \frac{\text{SSE}}{N - k} = \frac{39.69}{14 - 3} = 4.961 \\ \end{array} $$ Where k is the number of groups (3 in this case).
04

Calculate F-value and p-value

Now, we will calculate the F-value and p-value: $$ \text{F-value} = \frac{\text{MST}}{\text{MSE}} = \frac{7.955}{4.961} = 1.60 \\ \text{p-value} = P(F(2,11) > 1.60) = 0.244 $$
05

Construct ANOVA table

Using the values calculated above, we can construct the ANOVA table as follows: | Source | Sum of Squares | df | Mean Square | F-value | p-value | |:-------------:|:-------------:|:--:|:-----------:|:-------:|:-------:| | Treatments | 15.91 | 2 | 7.955 | 1.60 | 0.244 | | Errors | 39.69 | 11 | 4.961 | | | | Total | 55.60 | 13 | | | |
06

State the null and alternative hypotheses, and test for significance using the p-value approach

The null and alternative hypotheses for the ANOVA F-test are: H0: There is no significant difference between the means of the three populations. H1: There is a significant difference between the means of at least one pair of populations. The p-value (0.244) is greater than the significance level of 0.05, meaning we fail to reject the null hypothesis. This suggests there is no significant difference between the means of the three populations.

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Key Concepts

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

Completely Randomized Design
In statistics, a Completely Randomized Design (CRD) is one of the simplest forms of experimental design. In a CRD, all experimental units are allocated at random to treatments. This randomness is used to control for the effects of variables not included in the experiment, which helps to ensure that the results obtained are due to the treatments and not some other factors.

In the context of the provided exercise, a CRD was used to collect observations for three different samples. This means that each sample had an equal chance of receiving any specific treatment or condition, ensuring the validity of the comparisons made between them. The CRD is particularly useful when the experimental units are homogeneous and the influence of external variables on the response is negligible.
Sum of Squares (SS)
The Sum of Squares (SS) is a statistical measure that quantifies the amount of variation within a set of data. In ANOVA (Analysis of Variance), it is used to measure the total variation, the variation due to treatments, and the variation due to error.

The Total SS represents the total variation from the overall mean. It's the sum of the squared differences between each observation and the grand mean. The Treatment SS (SST) captures the variation between the different group means and the grand mean, reflecting the effect of different treatments. The Error SS (SSE) measures the variation within the groups, representing the individual differences not accounted for by the treatments. The problem provides the formulas and calculations needed to find these values, giving insight into the sources of variation in the sample data.
Mean Square (MS)
The Mean Square (MS) in ANOVA is the average of the squared deviations (Sum of Squares) for the treatments and errors, used to analyze the variance of the groups' means. It is calculated by dividing the Sum of Squares by the respective degrees of freedom (df).

There are two types of mean squares in ANOVA: the Mean Square for Treatment (MST) and the Mean Square for Error (MSE). The MST represents the average variation due to the treatments, while the MSE is an estimate of the population variance based on sample data within groups. In the exercise, the MST and MSE were computed using calculated SST and SSE divided by their respective degrees of freedom, which were then used to calculate the F-value.
F-test
The F-test is a statistical test used in ANOVA to compare the variances of groups and determine if those variances are significantly different from each other. It is based on the F-distribution and the calculated F-value, which is the ratio of MST to MSE. A higher F-value indicates a greater degree of difference between group variances than expected by chance.

When performing an F-test, you're essentially testing the null hypothesis that the group means are equal against the alternative hypothesis that at least one group mean is different. In the exercise's ANOVA table, the F-value quantifies how much the observed group means deviate from the null hypothesis. The p-value associated with this F-value then informs us whether these deviations are statistically significant or not.
p-value
In hypothesis testing, the p-value measures the probability of obtaining results at least as extreme as the observed results, assuming that the null hypothesis is true. It is a fundamental concept in statistics used to determine the significance of the results.

A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so it is rejected, suggesting a statistically significant difference in means. In contrast, a larger p-value suggests insufficient evidence against the null hypothesis. In the textbook exercise, we compare the p-value with an alpha level of 0.05 to decide whether to reject the null hypothesis. As the calculated p-value of 0.244 is greater than 0.05, it suggests that the differences in sample means could be due to random chance rather than a significant effect of the treatments.

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Most popular questions from this chapter

An independent random sampling design was used to compare the means of six treatments based on samples of four observations per treatment. The pooled estimator of \(\sigma^{2}\) is \(9.12,\) and the sample means follow: \(\bar{x}_{1}=101.6 \quad \bar{x}_{2}=98.4 \quad \bar{x}_{3}=112.3\) \(\bar{x}_{4}=92.9 \quad \bar{x}_{5}=104.2 \quad \bar{x}_{6}=113.8\) a. Give the value of \(\omega\) that you would use to make pairwise comparisons of the treatment means for $$ \alpha=.05 . $$ b. Rank the treatment means using pairwise comparisons.

Water samples were taken at four different locations in a river to determine whether the quantity of dissolved oxygen, a measure of water pollution, varied from one location to another. Locations 1 and 2 were selected above an industrial plant, one near the shore and the other in midstream; location 3 was adjacent to the industrial water discharge for the plant; and location 4 was slightly downriver in midstream. Five water specimens were randomly selected at each location, but one specimen, corresponding to location \(4,\) was lost in the laboratory. The data and a MINITAB analysis of variance computer printout are provided here (the greater the pollution, the lower the dissolved oxygen readings). $$ \begin{array}{llllll} \text { Location } && {\text { Mean Dissolved }} {\text { Oxygen Content }} \\\ \hline 1 &&& 5.9 & 6.1 & 6.3 & 6.1 & 6.0 \\ 2 &&& 6.3 & 6.6 & 6.4 & 6.4 & 6.5 \\ 3 &&& 4.8 & 4.3 & 5.0 & 4.7 & 5.1 \\ 4 &&& 6.0 & 6.2 & 6.1 & 5.8 & \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in the mean dissolved oxygen contents for the four locations? b. Compare the mean dissolved oxygen content in midstream above the plant with the mean content adjacent to the plant (location 2 versus location 3 ). Use a \(95 \%\) confidence interval.

Suppose you were to conduct a two-factor factorial experiment, factor \(\mathrm{A}\) at four levels and factor \(\mathrm{B}\) at two levels, with \(r\) replications per treatment. a. How many treatments are involved in the experiment? b. How many observations are involved? c. List the sources of variation and their respective degrees of freedom.

Refer to Exercise \(11.63 .\) The means of all observations, at the factor A levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\) are \(\bar{x}_{1}=3.7\) and \(\bar{x}_{2}=1.4,\) respectively. Find a \(95 \%\) confidence interval for the difference in mean response for factor levels \(\mathrm{A}_{1}\) and \(\mathrm{A}_{2}\)

Swampy Sites An ecological study was conducted to compare the rates of growth of vegetation at four swampy undeveloped sites and to determine the cause of any differences that might be observed. Part of the study involved measuring the leaf lengths of a particular plant species on a preselected date in May. Six plants were randomly selected at each of the four sites to be used in the comparison. The data in the table are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. The MINITAB analysis of variance computer printout for these data is also provided. $$ \begin{array}{lllllll} \text { Location } & {\text { Mean Leaf Length (cm) }} \\ \hline 1 && 5.7 & 6.3 & 6.1 & 6.0 & 5.8 & 6.2 \\ 2 && 6.2 & 5.3 & 5.7 & 6.0 & 5.2 & 5.5 \\ 3 && 5.4 & 5.0 & 6.0 & 5.6 & 4.9 & 5.2 \\ 4 && 3.7 & 3.2 & 3.9 & 4.0 & 3.5 & 3.6 \end{array} $$ a. You will recall that the test and estimation procedures for an analysis of variance require that the observations be selected from normally distributed (at least, roughly so) populations. Why might you feel reasonably confident that your data satisfy this assumption? b. Do the data provide sufficient evidence to indicate a difference in mean leaf length among the four locations? What is the \(p\) -value for the test? c. Suppose, prior to seeing the data, you decided to compare the mean leaf lengths of locations 1 and \(4 .\) Test the null hypothesis \(\mu_{1}=\mu_{4}\) against the alternative \(\mu_{1} \neq \mu_{4}\) d. Refer to part c. Construct a \(99 \%\) confidence interval for \(\left(\mu_{1}-\mu_{4}\right)\) e. Rather than use an analysis of variance \(F\) -test, it would seem simpler to examine one's data, select the two locations that have the smallest and largest sample mean lengths, and then compare these two means using a Student's \(t\) -test. If there is evidence to indicate a difference in these means, there is clearly evidence of a difference among the four. (If you were to use this logic, there would be no need for the analysis of variance \(F\) -test.) Explain why this procedure is invalid.
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