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Hypothesis testing begins with defining a null hypothesis, often denoted as \( H_0 \). This is a statement that gives us a starting point for any statistical test. It generally posits that there is no effect or no difference, implying that any observed variation is due to sampling or experimental error.
In the context of our problem, the null hypothesis suggests that the mean annual automobile insurance rate in Nevada is not greater than \( \\(1388 \). Mathematically, it can be expressed as \( \mu \leq 1388 \). This hypothesis assumes that the insurance rate remains at \( \\)1388 \) or less, reflecting the existing belief or status quo.

The alternative hypothesis, noted as \( H_1 \), challenges the null hypothesis. It proposes a new theory or claim and is what we are aiming to find evidence for in the study.
For our exercise, the insurance broker suspects that the current average annual rate exceeds \( \$1388 \). Thus, the alternative hypothesis is \( \mu > 1388 \). This claim suggests that there has been an increase in the insurance rate, indicating a shift from the previous average documented.

The Z-test is a statistical test used to determine if there is a significant difference between sample and population means. It is particularly useful when dealing with large sample sizes or when the population variance is known.
To perform a Z-test, we calculate the Z-value using the formula: \[ z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}}\]Here, \( \bar{x} \) is the sample mean, \( \mu_{0} \) is the population mean under the null hypothesis, \( \sigma \) is the standard deviation, and \( n \) is the sample size. For the exercise, the sample mean is \( \\(1413 \), and the population mean is \( \\)1388 \). By substituting these values in, we derive the Z-value to help compare against the critical value.

The significance level, denoted as \( \alpha \), is a threshold set before the data analysis. It represents the risk of rejecting the null hypothesis when it is, in fact, true, often set at \( 5\% \), \( 1\% \), or \( 0.1\% \). For this problem, we have a \( 1\% \) significance level, implying a high standard, requiring strong evidence to reject \( H_0 \).
Using a significance level of \( 1\% \), our critical value for a one-sided test can be found from the standard normal distribution, which is approximately \( 2.33 \). We compare our calculated Z-value to this critical value to decide whether to reject the null hypothesis. If the Z-value exceeds \( 2.33 \), it marks the difference as statistically significant, leading to the rejection of the null hypothesis. This means the mean insurance rate is deemed higher than \( \$1388 \) based on the sample data.