91Ó°ÊÓ

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is especially useful when the sample sizes are small and the population variance is unknown.

The t-test first calculates the t-statistic, which measures the size of the difference relative to the variation in the sample data. For a one-sample t-test, the formula for the t-statistic is:

\( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)

The variables in the formula represent:

• \(\bar{x}\): Sample mean
• \(\mu\): Population mean under the null hypothesis
• \(s\): Sample standard deviation
• \(n\): Sample size

In the given exercise, we calculated the t-statistic as follows:

\( t = \frac{18.3 - 20}{\frac{4.3}{\sqrt{18}}} \approx -1.68 \)
This value indicates how many standard deviations the sample mean is from the population mean assumed under the null hypothesis.

The null hypothesis, denoted as \(H_0\), is a statement that there is no effect or no difference in the population. It is the assumption that any observed difference in the sample is due to random variation.

In hypothesis testing, you start with the null hypothesis and then use sample data to determine whether there is enough evidence to reject it. The null hypothesis in this exercise is:
\(H_0: \mu = 20\)
The null hypothesis assumes that the population mean \(\mu\) is 20. By performing the t-test, we want to see if the sample provides strong evidence against this assumption.

Not rejecting the null hypothesis does not mean that the hypothesis is true. It simply means that the sample data does not provide strong enough evidence to conclude that the population mean is different from 20.

The p-value is a probability that measures the evidence against the null hypothesis. Specifically, it represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true.

In the exercise, the computed t-statistic was -1.68. The corresponding p-value is obtained from the t-distribution table or using statistical software. For 17 degrees of freedom (\(df = n - 1 = 18 - 1\)), the approximate p-value is 0.055.

A lower p-value indicates stronger evidence against the null hypothesis. The interpretation of the p-value is as follows:

• If the p-value is less than or equal to the significance level (\(\alpha\)), we reject the null hypothesis.
• If the p-value is greater than the significance level, we do not reject the null hypothesis.

In this exercise, since the p-value (0.055) is greater than the chosen significance level (0.05), we do not reject the null hypothesis.

The significance level, denoted as \(\alpha\), is a threshold set by the researcher to determine whether to reject the null hypothesis. It represents the probability of making a Type I error—that is, rejecting a true null hypothesis.

Common choices for \(\alpha\) are 0.05, 0.01, and 0.10. In this exercise, the significance level is set at 0.05.

The decision rule for hypothesis testing is based on the comparison of the p-value and the significance level:

• If the p-value \( \leq \alpha \), there is sufficient evidence to reject the null hypothesis.
• If the p-value \( > \alpha \), there is not enough evidence to reject the null hypothesis.

In the exercise, since the p-value of 0.055 is greater than the significance level of 0.05, the researcher does not reject the null hypothesis. This means there's not enough evidence to conclude that the population mean is less than 20 at the 0.05 significance level.