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A proportion test is commonly used to determine whether a sample proportion is equal to a specified population proportion. This helps us in situations where we want to find out if there is a significant difference between the observed data and what's expected. In the context of our exercise, we are trying to see if the proportion of driver fatalities involving intoxication in Kit Carson County is less than the national reported proportion. To do this, the proportion test compares the sample proportion (from the data we have) with the population proportion (the expected rate, 77% here). This test is particularly useful when working with dichotomous outcomes, like "intoxicated" or "not intoxicated.鈥� The test gives us a z-value that we compare against a critical value to determine significance. This comparison allows us to make a decision about our hypotheses.

In hypothesis testing, the significance level, denoted as \(\alpha\), is the threshold used to determine whether to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. A lower significance level means you demand stronger evidence to reject the null hypothesis. In our scenario, a significance level of 1% (0.01) is used. This indicates a very strict threshold, requiring strong evidence before we are willing to conclude that the proportion of intoxicated driver fatalities is less than 77%. The choice of significance level thus affects how we interpret the z-value calculated from the test statistic.

The z-value is a key element in hypothesis testing involving proportions. It tells us how many standard deviations away our sample proportion is from the hypothesized population proportion. Calculating the z-value involves subtracting the hypothesized population proportion from the sample proportion and dividing by the standard deviation. In our exercise, the formula used is:

• \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]

Plugging numbers into this formula helps us determine if the sample data is in line with the null hypothesis. The result of this calculation, the z-value, is compared to the critical z-value obtained from a standard normal distribution table. This comparison will tell us if the observed data provides enough evidence to reject the null hypothesis.

The null hypothesis \((H_0)\) serves as the starting point for hypothesis testing. It expresses the assumption that there is no effect or no difference. In our specific example, the null hypothesis posits that the proportion of intoxicated driver fatalities is equal to the national proportion (77%). It is a claim that there鈥檚 no significant difference between the sample data from Kit Carson County and the national standard. The purpose of the hypothesis test is to determine if there is enough statistical evidence to reject this assumption. By using the z-value and comparing it to the critical value, we assess whether the null hypothesis can be rejected, suggesting a difference exists.

The alternative hypothesis \((H_1)\) is what you want to prove in hypothesis testing. It stands in opposition to the null hypothesis and reflects the research hypothesis. For this exercise, it claims that the proportion of intoxicated driver fatalities in Kit Carson County is actually less than the national proportion of 77%. The alternative hypothesis is directional in this context because it specifies the direction of the effect ("less than").
Unlike the null hypothesis, the alternative hypothesis is not something you prove directly. Instead, you seek evidence to support it by showing that the null hypothesis can be rejected. The calculated z-value being in the rejection region of the critical value allows us to adopt this alternative hypothesis.