Problem 5 - An Introduction to Mathematical Statistics and Its Applications

Question

The University of Missouri-St. Louis gave a validation test to entering students who had taken calculus in high school. The group of ninety-three students receiving no college credit had a mean score of \(4.17\) on the validation test with a sample standard deviation of \(3.70\). For the twenty- eight students who received credit from a high school dual-enrollment class, the mean score was \(4.61\) with a sample standard deviation of \(4.28\). Is there a significant difference in these means at the \(\alpha=0.01\) level? Assume the variances are equal.

Step-by-Step Solution

Answer

Based on the calculated t-value, we compare it to the critical value. If the absolute t-value is greater than the critical value, then there is a statistically significant difference between the two mean scores with a confidence level of 99%. If the value is smaller, there is not a significant difference.

Step 1: Define the Hypotheses

Our null hypothesis (H0) is that there is no significant difference in the mean scores of both groups, i.e., the mean score of Group 1 (students receiving no college credits) is equal to the mean score of Group 2 (students receiving credits from a high school dual-enrollment class). mathematically, this is \(H0: \mu1 = \mu2\). The alternative hypothesis (Ha), which is what we want to prove, is that there is a significant difference in the means, i.e., \(Ha: \mu1 \ne \mu2\).

Step 2: Calculate the Combined Standard Error

The combined standard error (SE) can be found with the formula: \(SE = \sqrt{(s1^2/n1) + (s2^2/n2)}\) where \(s1\) and \(s2\) are the standard deviations for the two groups, and \(n1\) and \(n2\) are the sample sizes of the two groups. We plug the given values into this formula: \(SE = \sqrt{(3.70^2/93) + (4.28^2/28)}\).

Step 3: Calculate the T-score

The t-score is calculated as: \(t = (\mu1 - \mu2) / SE\) where \(\mu1\) and \(\mu2\) are the mean scores of the two groups. We plug in the calculated SE and given values: \(t = (4.17 - 4.61) / SE.\)

Step 4: Find the Critical Value

We look for the critical t-value for a two-tailed test at the 0.01 significance level with degrees of freedom = \(n1 + n2 - 2 = 93 + 28 - 2 = 119\). Typically, we need to use a t-distribution table or statistical software for this. The critical value will be -2.615 in this case.

Step 5: Compare T-score with Critical Value

We compare the absolute value of the calculated t with the critical value. If the absolute t is greater than the critical value, we reject the null hypothesis. In other words, if there is sufficient evidence at the 0.01 significant level that the means are not equal, we will conclude that there is a significant difference.

Key Concepts

Null Hypothesis

When conducting a statistical hypothesis test, the null hypothesis is a foundational concept. It is essentially a statement that assumes no effect or no difference. In our exercise, the null hypothesis ( H0 ) implies that the mean scores of the two groups of students are the same.
The purpose of forming a null hypothesis is to set a baseline reference against which we can measure an effect or change. This hypothesis is subjected to scrutiny through a statistical test to determine if what we observe in the data can happen due to random chance, or if an actual effect is at play.
Notably, we start by assuming the null hypothesis is true. Our goal, however, is to gather sufficient evidence to challenge this assumption, rather than to prove it false directly.

• If the data significantly contradicts the null hypothesis, we may reject it in favor of an alternative explanation.
• Rejecting the null implies there's enough statistical backing to suggest a difference exists.

Alternative Hypothesis

Contrasting the null hypothesis (H0), the alternative hypothesis (Ha) proposes what you seek to prove through your test. It represents a new effect or difference between the groups being studied.
In the context of our exercise, the alternative hypothesis suggests there is a significant difference between the mean scores of the two student groups, represented mathematically as \(Ha: \mu1 e \mu2\).
This hypothesis comes into play if we find the null hypothesis too improbable given our data. In practice, we do not prove the alternative hypothesis; rather, we gather evidence that lends strong support to it.

• If you reject H0, you infer support for Ha.
• The acceptance of Haimplies that an effect exists beyond random variation.

T-test

A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's particularly useful for small sample sizes or when the population standard deviation is unknown.
In our exercise, we apply a T-test to assess whether the difference between the mean scores of the two student groups is significant.
By calculating the t-score, we quantify how far apart the sample means are relative to the variability in the samples.

• The larger the absolute t-score, the greater the difference between your sample means in comparison to the variation in your samples.
• The absolute value of the t-score is compared against a critical value to determine significance.

Significance Level

When performing hypothesis tests, the significance level (\( \alpha \)) helps you determine the threshold for rejecting the null hypothesis. It sets how extreme observed data must be to reject H0.
In practical terms, a common choice for \( \alpha \) is 0.05, but our exercise uses a 0.01 level, indicating a more stringent criterion for claiming a significant effect.
At \( \alpha = 0.01 \), only data crazily unlikely under H0 (less than 1% probable) leads us to reject it.

• A lower significance level means stronger evidence is needed to reject the null.
• A smaller \( \alpha \) reduces your chances of falsely finding a difference when there isn't one (Type I error).