Suppose An Educator Is Interested In The Difference In Math

Suppose an educator is interested in the difference in math achievement of third

Suppose an educator is investigating the difference in math achievement between two groups of third-grade students subjected to different instructional environments. One group is taught in a highly structured setting with a fixed amount of time dedicated to teaching and learning math concepts, whereas the other group experiences a less structured environment where the same concepts are taught but without a set time limit for completion. The study involves randomly sampling students from each group at the end of a six-week period, and their math achievement is assessed through a standardized test.

The research question focuses on whether there is a significant difference in mean math achievement scores between these two groups, with the hypothesis being directional (specifically, testing if the highly structured environment leads to higher achievement). The significance level for testing is set at 0.05, and the educator requests guidance on conducting a statistical comparison, including calculating the critical value (cv), the test statistic, and interpreting the results based on the data provided.

Paper For Above instruction

Designing an appropriate statistical test to compare the two groups requires careful calculation of key parameters, including the critical value (cv) and the test statistic. Since the goal is to determine whether a difference exists in a specific direction—that is, whether the highly structured environment yields higher scores—a one-tailed t-test (specifically, a negative directional test) is suitable, assuming the data meets the assumptions of normality and homogeneity of variances.

Calculating the Critical Value (cv)

The significance level (α) is given as 0.05. For a one-tailed t-test, the critical value depends on the degrees of freedom (df). The degrees of freedom are calculated using the formula for independent samples:

df = (n₁ - 1) + (n₂ - 1)

where n₁ and n₂ are the sample sizes for groups 1 and 2, respectively. If specific sample sizes are provided, we substitute them into this formula. For example, if both groups have n₁ = n₂ = 30 students, then:

df = (30 - 1) + (30 - 1) = 58

Next, referring to a t-distribution table at α = 0.05 for a one-tailed test with 58 degrees of freedom, we find the critical t-value, which is approximately 1.671. Since the test is negative directional, we look for the negative critical value: -1.671.

Calculating the Test Statistic (t)

The test statistic for comparing two independent means is calculated as:

t = (Mean₁ - Mean₂) / SE

where SE (standard error) is:

SE = √( (s₁² / n₁) + (s₂² / n₂) )

Given the data points: means, sums, sums of squares, and variances, the standard deviations (s₁ and s₂) are computed as:

s_i = √(Variance_i)

Once the standard deviations and standard error are known, the t-statistic can be calculated. If specific values are provided—for example:

• Mean 1 = 85
• Mean 2 = 78
• n₁ = n₂ = 30
• Variance 1 = 100
• Variance 2 = 125

then:

s₁ = √(100) = 10

s₂ = √(125) ≈ 11.18

Standard error:

SE = √( (100 / 30) + (125 / 30) ) ≈ √(3.33 + 4.17) ≈ √7.5 ≈ 2.738

Test statistic:

t = (85 - 78) / 2.738 ≈ 7 / 2.738 ≈ 2.56

Interpreting the Results

The calculated t-value (approximately 2.56) is compared with the critical value (-1.671). Since 2.56 > 1.671, and considering the direction of the test (checking if group 1 > group 2), the test indicates a statistically significant difference. However, in a negative directional test, a t-value less than -1.671 would suggest significance in the opposite direction. Given our positive t-value, the evidence does not support the alternative hypothesis that the structured environment results in higher scores.

Alternatively, if the question hypothesizes that the less structured environment might lead to better outcomes, then a negative t-value exceeding the negative critical value would support that. The conclusion must be based on the exact hypothesis and the sign of the test statistic. In this example, since the t-value is positive, and the hypothesis was negative directional (testing if group 1

Overall, the results suggest that there is insufficient evidence at the 0.05 significance level to confirm that the type of environment significantly influences third-grade students’ math achievement scores in the tested direction. Further research with larger samples or additional variables may provide more conclusive insights.

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