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In hypothesis testing, the null hypothesis is the starting point. It is a statement of no effect or no difference, and it reflects the assumption that any observed difference is due to random chance. In the context of the exercise, the null hypothesis (\(H_0\) ) posits that the true mean of daily tips is \$80 . This serves as a baseline that we test for statistical significance.

To reject or accept the null hypothesis, we gather data and perform a statistical test. If the evidence is strong enough (beyond a pre-set significance level), we reject \(H_0\) . Otherwise, we do not have sufficient evidence to discard it.

It's essential to remember that failing to reject the null hypothesis doesn't prove that \(H_0\) is true. It simply means there isn't enough evidence to prove it's false using the test conducted.

The test statistic, in hypothesis testing, is a standardized value that helps decide whether to reject the null hypothesis. It quantifies the difference between the observed sample data and what is expected under the null hypothesis.

In this exercise, we use a one-sample \(z\)-test test statistic. This is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where:

• \( \bar{x} = 84.85 \) is the sample mean
• \( \mu = 80 \) is the hypothesized population mean
• \( \sigma = 9.95 \) is the population standard deviation
• \( n = 35 \) is the sample size

The calculated \(z\) value helps determine how far the sample mean deviates from the hypothesized mean in terms of standard errors. If this deviation is sufficiently large compared to a critical value from the \(z\)-table , the null hypothesis can be rejected.

The \(p\)-value is a crucial concept in hypothesis testing. It represents the probability of observing a test statistic as extreme as the one computed, assuming the null hypothesis is true.

Within the exercise, the \(p\)-value is about 0.002, which implies a low probability of obtaining the observed data if the daily tips truly averaged \$80 .

A lower \(p\)-value , typically less than the significance level (here \(\alpha = 0.01\) ), suggests there is strong evidence against the null hypothesis. It essentially communicates the strength of the evidence to support the alternative hypothesis. When \(p\)-value is less than \(\alpha\), it indicates that there is a statistically significant difference, justifying the rejection of the null hypothesis.

Understanding the normal distribution is fundamental when working with hypothesis testing. It is a continuous probability distribution that is symmetric around the mean. Most of the observations cluster around the central peak, and probabilities for values taper off equally in both directions, providing a predictable pattern.

In this problem, the normal distribution is crucial as it allows the sample mean's distribution to be approximated by the \(z\)-distribution. Given a large enough sample (35 days, in this case), the sample mean follows a normal distribution due to the Central Limit Theorem.

The standard normal distribution has a mean of 0 and a standard deviation of 1. We use this property to find critical values and \(p\)-values for hypothesis testing. By converting observed sample means to \(z\)-scores, we determine how typical or unusual the sample mean is under the assumed distribution of the null hypothesis.