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In hypothesis testing, the null hypothesis is a statement that indicates there is no effect or no difference being tested. When you conduct a two-sample t-test, like in our exercise with students from Florida State University and the University of Florida, the null hypothesis assumes that the average age of students at both universities is the same. In mathematical terms, this is expressed as

• \( H_{0}: \mu_{1} = \mu_{2} \)

where \( \mu_{1} \) and \( \mu_{2} \) represent the population means of the student ages from the two universities. Establishing a null hypothesis is crucial as it lays the foundation for testing if there is a statistically significant difference between two sample groups. Then, using statistical methods, we attempt to prove or disprove this hypothesis, thus opening the possibility to accept the alternative hypothesis.

The alternative hypothesis is opposite to the null hypothesis and indicates the effect or difference that you want to test for. For this exercise, the alternative hypothesis is that the average age of the students at the two universities is different. It is expressed mathematically as follows:

• \( H_{1}: \mu_{1} eq \mu_{2} \)

This hypothesis represents the conclusion if the null hypothesis is not supported by the data. It aims to prove that there is a statistically significant difference in the average ages between the two groups being studied. In hypothesis testing, rejecting the null hypothesis leads to accepting the alternative hypothesis, suggesting the observed data might show a real effect or difference rather than random chance.

The critical value is a key element in hypothesis testing, used to determine the threshold at which the null hypothesis can be rejected. For a two-sample t-test, the critical value depends on the level of significance (often set at 0.05 for a 95% confidence interval) and the degrees of freedom.
In this exercise, a two-tailed test is used at a 0.05 level of significance. The degrees of freedom are calculated as the smaller of the sample sizes minus one, i.e.,

• df = min(\(n_{1}-1\), \(n_{2}-1\))

For our samples, this becomes df = min(119, 99). The critical value is obtained from a t-distribution table or calculator, resulting in \( \pm 1.96 \). If the calculated test statistic falls beyond these critical values, the null hypothesis is rejected, suggesting there is a statistically significant difference between the groups.

The test statistic is a calculated value from the sample data that is compared against the critical value to make a decision about the null hypothesis. In a two-sample t-test, the test statistic measures how far apart the sample means are from each other, accounting for variation within the samples. In our exercise, the test statistic is computed using this formula:\[ t = \frac{(\overline{X}_{1} - \overline{X}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{S^2_{1}}{n_{1}} + \frac{S^2_{2}}{n_{2}}}} \]where:

• \(\overline{X}_{1}\) and \(\overline{X}_{2}\) are the sample means
• \(S_{1}\) and \(S_{2}\) are the standard deviations
• \(n_{1}\) and \(n_{2}\) are the sample sizes

Plugging in the given values,\( t \approx -4.59 \).This calculated test statistic is compared with the critical value. A test statistic that falls outside the range defined by the critical value supports rejecting the null hypothesis, showing substantial evidence of a significant age difference between the two student groups.