91Ó°ÊÓ

The t-statistic is a critical part of hypothesis testing when working with small sample sizes or when the population standard deviation is unknown. It is used to determine how far away your sample mean is from the population mean, in units of standard error.

The formula for the t-statistic is given by: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where:

• \( \bar{x} \) is the sample mean
• \( \mu \) is the population mean
• \( s \) is the sample standard deviation
• \( n \) is the sample size

In the given exercise, our sample size \( n \) is 31, with a sample mean \( \bar{x} \) of 8.1, and a sample standard deviation \( s \) of 1.9. The population mean for healthy adults is known to be \( \mu = 7.4 \). Plugging the values into the formula, we get a t-statistic of approximately 2.052.

This value helps us understand how significant our results are when compared to the hypothesized population mean.

The significance level, denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It serves as the threshold for determining whether our test statistic is extreme enough to reject the null hypothesis.

In this exercise, a 5% significance level, or \( \alpha = 0.05 \), was used. This means that we are willing to accept a 5% chance of incorrectly concluding that the drug changes the mean pH level when it does not.

In a two-tailed test, as in this example, the significance level is split between the two tails of the distribution, allowing us to test for changes in either direction. The critical values, therefore, are determined from the t-distribution, ensuring that the tails collectively cover 5% of the distribution. In this instance, the critical values around the population mean are approximately \( \pm 2.042 \). This defines our decision boundary. If the t-statistic lies beyond these critical values, we reject the null hypothesis.

The null hypothesis, often represented as \( H_0 \), posits that there is no effect or no difference. It serves as a starting point for statistical testing. The null hypothesis is assumed to be true until evidence suggests otherwise.

In the context of the given problem, the null hypothesis is that the drug does not change the mean pH level of arterial plasma. Formally, it is stated as: \[ H_0: \mu = 7.4 \] This hypothesis implies that any observed difference in the sample mean from 7.4 is due to random sampling variation and not the effect of the drug.

During hypothesis testing, we gather evidence from our sample data to determine whether it is statistically significant enough to reject the null hypothesis in favor of the alternative hypothesis. The decision is based on the calculated t-statistic and the predetermined significance level.

The alternative hypothesis, denoted as \( H_a \), reflects the opposite of the null hypothesis and is what you aim to provide evidence for in a statistical test. It suggests that there is a statistically significant effect or difference.

In this exercise, the alternative hypothesis is that the drug changes the mean pH level of the blood. It can be expressed as: \[ H_a: \mu eq 7.4 \] This indicates a two-tailed test scenario where any change, either increase or decrease in the mean pH, would support the alternative hypothesis.

The t-statistic calculated in the exercise ultimately helps determine if there is enough evidence to support the alternative hypothesis. If the absolute value of the t-statistic is greater than the critical value from the t-distribution at the set significance level, then the null hypothesis is rejected in favor of the alternative hypothesis, suggesting an effect caused by the drug.