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When conducting a hypothesis test, the null hypothesis ( H_0 ) serves as a statement of no effect or no difference. It sets the foundation for the statistical test by assuming that any observed effect is purely due to sampling variability. In our exercise, the null hypothesis was that the actual proportion, p , of patients experiencing headaches while on Diltiazem is 12%. This is written mathematically as H_0: p = 0.12 .

The alternative hypothesis ( H_a ), on the other hand, suggests that there is an effect or a difference. Here, we hypothesized that exercise might reduce headaches, setting H_a: p < 0.12 . This proposes that regular exercise lowers the headache proportion below 12%.

Testing the null hypothesis informs decision-making:

• If we reject it, we support the alternative hypothesis, suggesting a significant difference.
• If we fail to reject it, we continue to assume there's no significant change.

A z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. In hypothesis testing, the z-score helps determine how much an observed sample statistic deviates from what the null hypothesis predicts.

In our case, the z-score tells us whether the proportion of patients experiencing headaches differs from the 12% benchmark. We calculated the z-score using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]

where:

• \( \hat{p} \) is the sample proportion (here, 7.66%),
• \( p_0 \) is the population proportion under the null hypothesis (12%),
• \( n \) is the sample size (209 patients),

The z-score we found was approximately -1.93, indicating how far below the 12% the sample proportion falls in standard deviation units.

The significance level, denoted by \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It is the threshold for determining when an observed effect is statistically significant. Common levels of significance include 0.05 (5%) or 0.01 (1%).

In our scenario, a 1% significance level (\alpha = 0.01) was chosen. This lower threshold provides stricter criteria, reducing the risk of a false positive. It means we are willing to tolerate only a 1% risk of claiming that exercise reduces headaches when it does not.

A significance level allows us to set critical values for our test:

• If the test statistic falls in the critical region, we reject the null hypothesis.
• If it doesn't, we fail to reject it and conclude that the effect isn't statistically significant.

In this case, since the calculated z-score of -1.93 was not in the critical region set by a 1% significance level, the evidence was not strong enough to reject the null hypothesis.

The test statistic is a standardized value used in hypothesis testing. It helps determine whether to reject the null hypothesis. For a test concerning proportions, the z-score acts as the test statistic. It measures how far the sample proportion diverges from the proposed population proportion.To compute the test statistic, we use this equation:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\]where:

• \( \hat{p} \) is the observed sample proportion, which gives insight into the new study's observations (here, 0.0766).
• \( p_0 \) is the null hypothesis proportion of headaches (0.12).
• \( n \) is the number of subjects in the sample (209).

In this particular exercise, a test statistic of approximately -1.93 was calculated. This value measures the strength of evidence against the null hypothesis, indicating a smaller difference from the expected proportion than what might be considered statistically significant at a 1% level.