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In hypothesis testing, determining whether a test is one-tailed or two-tailed is crucial. For a one-tailed test, we're examining the data in only one direction. This means we check if a sample statistic is significantly greater than or less than a specified value, not both.

In the given exercise, the hypotheses are stated as follows:

• Null Hypothesis (\( H_0 \)): \( \mu \ge 220 \)
• Alternative Hypothesis (\( H_1 \)): \( \mu < 220 \)

The alternative hypothesis \( ( H_1: \mu < 220 ) \) suggests that we are interested in whether the mean is less than 220. Hence, it's a lower-tailed, or simply, a one-tailed test.

This focus on one direction makes the test more "powerful" for detecting an effect in that specific direction.

The p-value in hypothesis testing is a measure of the evidence against the null hypothesis. It reflects the probability of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true.

In this exercise, the null hypothesis is that \( \mu \geq 220 \) and the alternative hypothesis is \( \mu < 220 \). After calculating the z-value to be -2.67, we look at standard normal distribution tables to find the p-value. It comes out as approximately 0.0037.

A p-value below the significance level (0.03 in this case) indicates strong evidence against the null hypothesis, thus suggesting it should be rejected. Here, since 0.0037 < 0.03, there is strong evidence that the population mean is indeed less than 220.

A decision rule in hypothesis testing helps determine whether to reject or not reject the null hypothesis. It involves comparing the calculated test statistic with a critical value.

For a one-tailed test at a 0.03 significance level, the critical z-value is approximately -1.88. Our decision rule is:

• Reject \( H_0 \) if the calculated z-value is less than -1.88.
• Otherwise, do not reject \( H_0 \).

These rules provide a clear boundary. If our test statistic falls in the "rejection region" (below -1.88), it suggests that the data are unlikely under the null hypothesis.

Given that the calculated z-value is -2.67, we fall into the rejection zone. Therefore, we reject the null hypothesis based on our decision rule.

The test statistic is a standard numerical measurement that indicates how far the sample mean deviates from the null hypothesis mean. This calculation standardizes the sample mean difference so it can be compared against a probability distribution.

In this exercise, the test statistic is calculated using the formula:\[ z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}} \]where:

• \( \bar{x}\)
  is the sample mean (215)
• \( \mu_{0}\)
  is the population mean under the null hypothesis (220)
• \( \sigma\)
  is the population standard deviation (15)
• \( n\)
  is the sample size (64)

By plugging in these values, we get:\[ z = \frac{215 - 220}{\frac{15}{\sqrt{64}}} = -2.67 \]This result of -2.67 shows the number of standard deviations the sample mean is from the null hypothesis mean.