/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Twenty third graders were random... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\alpha=.05\).

Short Answer

Expert verified
Answer: No, there is not enough evidence at a 0.05 significance level to indicate a difference in the average scores for the four teaching methods.

Step by step solution

01

Determine the type of design

For this experiment, as twenty third graders were randomly separated into four equal groups and each group was taught a mathematical concept using a different teaching method, it is safe to say that the design is a completely randomized design.
02

Compute necessary statistics

We need to calculate the number of samples (n), group means, overall mean, and sum of squares for the four teaching methods. n total = 19 students (excluding the missing score) n per group = 5, 5, 4, 5 students Group means: \(\bar{x}_1 = \frac{112+92+124+89+97}{5} = 102.8\) \(\bar{x}_2 = \frac{111+129+102+136+99}{5} = 115.4\) \(\bar{x}_3 = \frac{140+121+130+106}{4} = 124.25\) \(\bar{x}_4 = \frac{101+116+105+126+119}{5} = 113.4\) Overall mean: \(\bar{x} = \frac{\sum{\text{all group scores}}}{19} = \frac{1486}{19} = 78.2105\) Sum of squares: \(SS_{within} = \sum{(x-\bar{x}_i)^2} = 4560.60\) \(SS_{between} = 5(\bar{x}_1-\bar{x})^2 + 5(\bar{x}_2-\bar{x})^2 + 4(\bar{x}_3-\bar{x})^2 + 5(\bar{x}_4-\bar{x})^2 = 1220.2496\) \(SS_{total} = \sum{(x-\bar{x})^2} = 5781.8272\)
03

Create ANOVA table

The ANOVA table will consist of components "source" (between, within, total), "degrees of freedom" (df), "sum of squares" (SS), "mean squares" (MS), "F-statistic" (F), and "p-value" (P). ANOVA table: | Source | df | SS | MS | F | P | |-----------|----|-----------|-----------|------|------------| | Between | 3 | 1220.2496 | 406.74987 | 2.92 | 0.06206685 | | Within | 15 | 4560.60 | 304.04 | | | | Total | 18 | 5781.8272| | | | To compute the F value and P value, we compare the between-group variance (MS_between) with the within-group variance (MS_within) using the F-distribution: \(F_{3, 15} = \frac{MS_{between}}{MS_{within}} = \frac{406.74987}{304.04} = 2.92\) The p-value = P(F > 2.92) = 0.06206685
04

Hypothesis test and conclusion

We will conduct the hypothesis test using a significance level of 0.05. Hypothesis: \(H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4\) (No difference in average scores among the groups) \(H_a:\) At least one group's average score is significantly different. Since our p-value (0.06206685) is greater than the significance level (0.05), we fail to reject the null hypothesis. In conclusion, there is not enough evidence at a 0.05 significance level to indicate a difference in the average scores for the four teaching methods.

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Key Concepts

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

Completely Randomized Design
In educational experiments, such as the one involving third graders and teaching methods, a completely randomized design is often used to ensure that each participant or subject is randomly assigned to an experimental group. This approach helps to minimize bias and ensure that each group is comparable before the intervention begins. In our specific case, this type of design ensures that personal characteristics or prior knowledge do not influence the outcome of the experiment, allowing for a fair comparison of teaching methods.

In completely randomized designs, the focus is on comparing several treatment conditions or groups that are formed entirely by chance. It is one of the simplest and most straightforward experimental designs used in hypothesis testing and is particularly beneficial when experiments involve a relatively small number of treatment groups and subjects. For the third graders in the problem, the use of such a design helps in determining whether the observed differences in test scores are due to the teaching methods or just random chance.
ANOVA Table
The ANOVA table is a keystone in understanding the analysis of variance. ANOVA, which stands for Analysis of Variance, is a statistical technique used to compare the means of three or more samples by analyzing variances within the samples and between the samples. The table organizes all the necessary calculations and presents them in an efficient, easy-to-read format.

An ANOVA table usually includes several components such as the source of variation (between-group or within-group), degrees of freedom (df), sum of squares (SS), mean squares (MS), the F-statistic (F), and the p-value (P). The 'between-group' variation measures how much the sample means differ from the overall mean, while the 'within-group' variation indicates the variability within each sample. These calculations allow us to test the null hypothesis that all groups have the same population mean, by comparing the F-statistic with the critical value of F or by examining the p-value. In the present experiment, the calculated F-statistic and p-value are used to determine the likelihood that any observed differences in group means are due to random chance.
Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistical analysis in experiments, including educational experiments. It involves making an assumption, or 'null hypothesis', about a population parameter, and then determining whether the observed data provide enough evidence to reject this hypothesis in favor of an 'alternative hypothesis'.

In our textbook example, the null hypothesis (\(H_0\)) assumes that there is no difference in average scores between the four teaching methods for third graders. The alternative hypothesis (\(H_a\)) indicates that at least one group's average score is significantly different from the others. We use the p-value obtained from the ANOVA analysis to make our decision. If the p-value is less than our chosen significance level (commonly \( \alpha = 0.05 \)), we reject the null hypothesis, suggesting that at least one teaching method has a significantly different effect. Since our p-value is greater than the significance level in this case, we fail to reject the null hypothesis, indicating that the differences in teaching method do not result in statistically significant differences in student scores.

One critical aspect of hypothesis testing is setting a significance level that is appropriate for the context of the experiment. This significance level represents the threshold for what we consider statistically significant. It is chosen before analyzing the data and is based on the degree of certainty we require to reject the null hypothesis. In educational and social sciences, \( \alpha = 0.05 \) is commonly used, but researchers may choose a more stringent level, like 0.01, for more sensitive experiments.

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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.

An experiment was conducted to compare the effects of four different chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) in producing water resistance in textiles. A strip of material, randomly selected from a bolt, was cut into four pieces, and the four pieces were randomly assigned to receive one of the four chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) or \(\mathrm{D} .\) This process was replicated three times, thus producing a randomized block design. The design, with moistureresistance measurements, is as shown in the figure (low readings indicate low moisture penetration). Analyze the experiment using a method appropriate for this randomized block design. Identify the blocks and treatments, and investigate any possible differences in treatment means. If any differences exist, use an appropriate method to specifically identify where the differences lie. What are the practical implications for the chemical producers? Has blocking been effective in this experiment? Present your results in the form of a report. $$$ \begin{aligned} &\text { Blocks (bolt samples) }\\\ &\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline \mathrm{C} & \mathrm{D} & \mathrm{B} \\ 9.9 & 13.4 & 12.7 \\ \mathrm{A} & \mathrm{B} & \mathrm{D} \\ 10.1 & 12.9 & 12.9 \\ \mathrm{B} & \mathrm{A} & \mathrm{C} \\ 11.4 & 12.2 & 11.4 \\ \mathrm{D}_{2} & \mathrm{C} & \mathrm{A} \\ 12.1 & 12.3 & 11.9 \end{array} \end{aligned} $$

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.

If the sample size for each treatment is \(n_{t}\) and if \(s^{2}\) is based on \(12 d f\), find \(\omega\) in these cases: a. \(\alpha=.05, k=4, n_{t}=5\) b. \(\alpha=.01, k=6, n_{t}=8\)

The calcium content of a powdered mineral substance was analyzed five times by each of three methods, with similar standard deviations: $$ \begin{array}{llllll} \text { Method } & {\text { Percent Calcium }} \\ \hline 1 && .0279 & .0276 & .0270 & .0275 & .0281 \\ 2 && .0268 & .0274 & .0267 & .0263 & .0267 \\ 3 && .0280 & .0279 & .0282 & .0278 & .0283 \end{array} $$ Use an appropriate test to compare the three methods of measurement. Comment on the validity of any assumptions you need to make.

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