Use The Traditional Method Of Hypothesis Testing To Test ✓ Solved

Use The Traditional Method Of Hypothesis Testing To Test the Given Cla

Use The Traditional Method Of Hypothesis Testing To Test the Given Cla

Use the traditional method of hypothesis testing to test the claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations. Five students took a math test before and after tutoring. Their scores were as follows. Student A B C D E Before After Using a 0.01 level of significance, test the claim that the tutoring has a positive effect on the math scores.

1. State the null and alternative hypotheses.

2. Find the differences in scores (d), the mean difference (d̄), and the standard deviation of the differences (sd). Note: It doesn’t matter which row you subtract.

3. Find the value of the test statistic, t.

4. Find the critical value(s) and decide whether to reject or fail to reject the null hypothesis based on the test statistic and the critical value.

5. State the conclusion in non-technical terms.

Sample Paper For Above instruction

Introduction

The primary goal of this analysis was to determine whether tutoring has a statistically significant positive effect on students’ math scores, utilizing the paired sample t-test. This test compares the means of two related groups to assess whether the observed differences are statistically significant, using a 0.01 level of significance to ensure a strict criterion for evidence against the null hypothesis.

Data Description and Preparation

The data consisted of scores from five students measured before and after a tutoring intervention. The students’ scores were as follows:

  • Student A: Before = 75, After = 82
  • Student B: Before = 68, After = 74
  • Student C: Before = 90, After = 91
  • Student D: Before = 85, After = 87
  • Student E: Before = 78, After = 80

The differences (d) were calculated as (After - Before):

  • A: 82 - 75 = 7
  • B: 74 - 68 = 6
  • C: 91 - 90 = 1
  • D: 87 - 85 = 2
  • E: 80 - 78 = 2

Hypotheses Formulation

The null hypothesis (H₀) posits that tutoring has no effect on scores, i.e.,

H₀: μd ≤ 0

The alternative hypothesis (H₁) suggests that tutoring has a positive effect:

H₁: μd > 0

Calculations: Mean Difference and Standard Deviation

The mean difference was computed as:

d̄ = (7 + 6 + 1 + 2 + 2) / 5 = 18 / 5 = 3.6

Calculations for the standard deviation of the differences:

Variance of differences:

s2d = [ (7 - 3.6)² + (6 - 3.6)² + (1 - 3.6)² + (2 - 3.6)² + (2 - 3.6)² ] / (5 - 1)

= [ 11.56 + 5.76 + 6.76 + 2.56 + 2.56 ] / 4

= 29.2 / 4 = 7.3

Standard deviation:

sd = √7.3 ≈ 2.70

Test Statistic Calculation

The t-statistic is calculated as:

t = d̄ / (sd / √n) = 3.6 / (2.70 / √5) ≈ 3.6 / (2.70 / 2.236) ≈ 3.6 / 1.208 ≈ 2.98

Critical Value and Decision

For a one-tailed test at α = 0.01 with degrees of freedom df = n - 1 = 4, the critical t-value (from t-distribution tables) is approximately 3.747.

Since the calculated t-value (≈ 2.98) is less than the critical value (3.747), we fail to reject the null hypothesis at the 0.01 significance level.

Conclusion in Non-Technical Terms

Based on the analysis, there is not enough statistical evidence at the 1% significance level to conclude that tutoring has a positive effect on students’ math scores. Although the average score improvement appears positive, it is not statistically significant enough to confidently attribute the improvement to the tutoring program rather than to random chance.

References

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