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In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were \(n_{1}=n_{2}=30\) children in each group. The Gates- MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was \(\bar{x}_{1}=344.5\), with sample standard deviation \(s_{1}=49.1\). For the control group, the mean score on the same test was \(\bar{x}_{2}=354.2\), with sample standard deviation \(s_{2}=50.9\). Use a \(5 \%\) level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began.

Step by step solution

01

Define Hypotheses

We start by defining our null and alternative hypotheses. The null hypothesis \(H_0\) states that there is no difference in the mean vocabulary scores between the two groups, i.e., \(\mu_1 = \mu_2\). The alternative hypothesis \(H_a\) is that there is a difference, i.e., \(\mu_1 eq \mu_2\).

02

Determine the Test Statistic

To test the hypotheses, we use a two-sample t-test since we compare means from two independent samples. The formula for the t-statistic in this case is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values: \(n_1 = 30\), \(n_2 = 30\), \(\bar{x}_1 = 344.5\), \(\bar{x}_2 = 354.2\), \(s_1 = 49.1\), and \(s_2 = 50.9\).

03

Calculate the Test Statistic

First, calculate the combined standard error:\[ SE = \sqrt{\frac{49.1^2}{30} + \frac{50.9^2}{30}} = \sqrt{\frac{2409.21}{30} + \frac{2590.81}{30}} = \sqrt{80.307 + 86.360} = \sqrt{166.667} \approx 12.91 \]Next, find the t-statistic:\[ t = \frac{344.5 - 354.2}{12.91} = \frac{-9.7}{12.91} \approx -0.75 \]

04

Determine the Critical Value and Decision Rule

For a two-tailed test at a 5% significance level with \(n_1 + n_2 - 2 = 58\) degrees of freedom, the critical t-value is approximately \(\pm 2.002\). The decision rule is: if \(|t| > 2.002\), reject \(H_0\); otherwise, fail to reject \(H_0\).

05

Make a Decision and Conclusion

The calculated t-statistic is \(-0.75\), which is less than the critical value of 2.002. Therefore, we fail to reject the null hypothesis \(H_0\). This means there is not enough evidence to suggest a significant difference in vocabulary scores between the experimental and control groups before instruction.

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Key Concepts

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

Two-Sample t-Test

The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. This test is highly useful in experimental designs where two groups, such as an experimental group and a control group, are compared. In this scenario, children were split into two groups, and their vocabulary scores were analyzed to see if peer tutoring affected their performance.
The formula for calculating the t-statistic is crucial as it enables you to measure whether the difference in means is statistically significant. For this, you calculate the standard error which considers the standard deviations and sample sizes of both groups. By comparing the calculated t-value with a critical value, you can determine the significance of the observed difference. This step is vital for making informed conclusions.

Null and Alternative Hypotheses

In hypothesis testing, the first step is defining the null hypothesis (H_0) and the alternative hypothesis (H_a). These hypotheses act as a guide for your statistical test. The null hypothesis (H_0) represents a statement of no effect or no difference — in this case suggesting that the mean vocabulary scores of the two groups are equal. It posits that any observed differences are due to random chance.
On the other hand, the alternative hypothesis (H_a) suggests that there is a real effect, meaning the mean scores differ. Formulating these hypotheses clearly is critical, as they provide the basis for making decisions after conducting the test.

• Null Hypothesis (H_0): There is no difference in the mean scores (\(\mu_1 = \mu_2\)).
• Alternative Hypothesis (H_a): There is a difference in means (\(\mu_1 eq \mu_2\)).

Statistical Significance

Statistical significance is crucial in hypothesis testing as it helps you decide if the differences you've observed are meaningful or if they could have occurred by random chance. In this context, it involves comparing the calculated t-statistic with a critical value based on a predetermined significance level, often set to 0.05 or 5%.
This level implies that you allow a 5% chance of making an error in stating that there is a difference when there isn't one (Type I error). If the calculated t-value exceeds the critical value, the difference is considered statistically significant, suggesting that the effect seen is unlikely due to chance alone.
However, if the t-value does not exceed the threshold, as in this exercise, the results are not statistically significant, indicating insufficient evidence to claim the groups differ meaningfully.

Critical Value

Critical values play a key role in statistical tests. They provide a threshold that the calculated test statistic must exceed for the results to be deemed significant. In a two-sample t-test, the critical value depends on the significance level and the degrees of freedom, calculated as the total sample size minus the number of groups (here, 2).
For example, with a 5% significance level and 58 degrees of freedom, the critical t-value is approximately \(\pm 2.002\). The test is two-tailed because it assesses for differences in either direction. By understanding and referring to this critical value, you can make a well-grounded decision on whether to reject or fail to reject the null hypothesis, guiding your conclusions about the experimental effects.

Experimental and Control Groups

In experimental research, distinguishing between experimental and control groups is essential to draw valid conclusions. The experimental group receives the treatment—in this case, peer tutoring along with regular instruction, aiming to boost their learning outcomes.
The control group, on the other hand, follows standard learning methods without additional interventions. This setup allows researchers to isolate the effect of the experimental treatment—in this case, tutoring—and attribute any differences in performance specifically to this intervention.
By comparing the results from both groups using statistical analysis, you gain insights into the effectiveness of the treatment, helping to answer important research questions about educational methods and interventions.