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When conducting hypothesis testing, the P-value helps us understand the strength of the evidence against the null hypothesis. The P-value represents the probability of obtaining a sample statistic as extreme, or more extreme, than the observed one, assuming the null hypothesis is true.
This means that the smaller the P-value, the stronger the evidence against the null hypothesis. In practical terms:

• If the P-value is less than or equal to the significance level (often 0.05), 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, the P-value is 0.18, which is greater than 0.05. Thus, there is not enough statistical evidence to reject the null hypothesis. Therefore, it seems plausible that the observed sample proportion could have occurred by random chance when the true proportion is 0.50.

The null hypothesis, often symbolized as \( H_0 \), is a statement of no effect or no difference. It serves as a starting point for statistical testing. In the context of this exercise, the null hypothesis states that 50% of the residents support the proposed tax increase.
The null hypothesis forms the basis against which the alternative hypothesis is tested. It assumes that any observed effect in the sample is purely due to random sampling variability rather than any actual difference. For this exercise, the null hypothesis is:

• \( H_0: p = 0.50 \)

This hypothesis suggests that half of the city's residents support the tax increase, and the data we collected from the sample is not significantly different from this assumption.

The alternative hypothesis, denoted \( H_1 \) or \( H_a \), is the statement you want to test against the null hypothesis. It represents what we suspect or are trying to demonstrate with statistics. In this scenario, the alternative hypothesis proposes that more than 50% of the city's residents support the tax increase.
This hypothesis is important because the outcome of the hypothesis test hinges on whether the data provides enough evidence to reject the null hypothesis in favor of the alternative. For this problem, the alternative hypothesis is:

• \( H_1: p > 0.50 \)

When we calculate the P-value, we are specifically testing against this alternative hypothesis to see if the data strongly supports that more than half of the residents are in favor.

The sample proportion is a statistic representing the fraction of the sample that has a specific characteristic. In the current exercise, it is calculated as the number of individuals in the sample who support the tax increase divided by the total number of individuals in the sample.
The formula used is \( \hat{p} = \frac{158}{300} = 0.527 \).
This value informs us that 52.7% of the sample supports the tax increase. The sample proportion is crucial because it is used to estimate the true population proportion and to compare against the hypothesized population proportion defined in the null hypothesis. In statistical testing, sample proportions help determine how likely the observed outcomes would occur under the null hypothesis. This process allows us to assess whether an observed effect is statistically significant.