Math Problem Statement

A teacher in a university in México claims that group A performs better than group B. An analysis of last year’s exam scores for both groups shows:

  • The mean score for group A is 7.5, while the mean score for group B is 6.0 (on a scale of 10).
  • The standard deviation for both groups is 0.7 points.
  • Each group has 200 students.

Assuming the exam scores are normally distributed and there is no bias, determine if there is sufficient statistical evidence to support the teacher's claim that group A performs better than group B. Use a significance level of 0.05.

Solution

To determine if there is sufficient statistical evidence to support the teacher's claim, we can conduct a two-sample tt-test (or zz-test since the sample sizes are large). This will test whether the mean score of group A is significantly higher than that of group B. Here's the step-by-step solution:


1. Formulate the hypotheses

  • Null hypothesis (H0H_0): There is no difference in mean scores between the two groups, i.e., μA=μB\mu_A = \mu_B or μAμB=0\mu_A - \mu_B = 0.
  • Alternative hypothesis (HaH_a): The mean score for group A is higher than that for group B, i.e., μA>μB\mu_A > \mu_B.

This is a one-tailed test since we are testing for a specific direction (group A > group B).


2. Determine the test statistic

The test statistic is calculated using the formula: z=xˉAxˉBsA2nA+sB2nBz = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} where:

  • xˉA\bar{x}_A and xˉB\bar{x}_B are the sample means,
  • sAs_A and sBs_B are the standard deviations,
  • nAn_A and nBn_B are the sample sizes.

Substituting the given values:

  • xˉA=7.5\bar{x}_A = 7.5, xˉB=6.0\bar{x}_B = 6.0,
  • sA=sB=0.7s_A = s_B = 0.7,
  • nA=nB=200n_A = n_B = 200.
= \frac{1.5}{\sqrt{\frac{0.49}{200} + \frac{0.49}{200}}} = \frac{1.5}{\sqrt{0.0049 + 0.0049}} = \frac{1.5}{\sqrt{0.0098}} = \frac{1.5}{0.099} \approx 15.15$$ --- ### **3. Find the critical value** At a significance level ($$\alpha$$) of 0.05 for a one-tailed test, the critical $$z$$-value is: $$z_{\text{critical}} = 1.645$$ --- ### **4. Compare the test statistic with the critical value** The calculated $$z$$-value is $$15.15$$, which is much greater than $$z_{\text{critical}} = 1.645$$. --- ### **5. Conclusion** Since $$z > z_{\text{critical}}$$, we reject the null hypothesis. There is strong statistical evidence to support the teacher's claim that group A performs significantly better than group B. --- Would you like further details or additional examples? Here are 5 related questions: 1. How does a $$t$$-test differ from a $$z$$-test, and when should each be used? 2. What is the meaning of the $$p$$-value in hypothesis testing, and how can it be interpreted? 3. How do sample size and variability affect the power of a statistical test? 4. Can you explain why this test assumes normal distribution and how to handle non-normal data? 5. How can effect size be calculated and interpreted in this context? **Tip:** When performing hypothesis tests, always double-check the assumptions (e.g., normality, equal variances) to ensure the validity of the results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-test
Statistics
Normal Distribution

Formulas

z = (x̄₁ - x̄₂) / sqrt((s₁²/n₁) + (s₂²/n₂))
Critical z-value at α = 0.05 for a one-tailed test

Theorems

Central Limit Theorem
Properties of Normal Distribution

Suitable Grade Level

Undergraduate Statistics or Advanced High School