Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ>55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.

To calculate the test statistic (t) for Gina's hypothesis test, we can use the following formula for the one-sample t-test:

\[ t = \frac{\overline{x} - \mu_0}{(s / \sqrt{n})} \]

Where: - \(\overline{x}\) is the sample mean - \(\mu_0\) is the hypothesized population mean (under the null hypothesis) - \(s\) is the sample standard deviation - \(n\) is the sample size

From the problem, we have: - \(\overline{x} = $56,500\) - \(\mu_0 = $55,000\) - \(s = $3,750\) - \(n = 61\)

Plugging in these values:

\[ t = \frac{56,500 - 55,000}{(3,750 / \sqrt{61})} \] \[ t = \frac{1,500}{(3,750 / \sqrt{61})} \] \[ t = \frac{1,500}{(3,750 / 7.81)} \] \[ t = \frac{1,500 * 7.81}{3,750} \] \[ t = \frac{11,715}{3,750} \] \[ t \approx 3.12 \]

So, rounded to two decimal places, the test statistic t is approximately 3.12.

Now, to find the p-value, we would normally use statistical software or a t-distribution table. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated (t = 3.12) if the null hypothesis is true. Since the alternative hypothesis is that the average salary is greater than $55,000 (Ha:μ>55,000), this is a one-tailed test.

Assuming a significance level of 0.05 and using degrees of freedom (df) equal to n-1 (which would be 61 - 1 = 60), we will look up the p-value associated with a t-score of 3.12 in a t-distribution table or use software. Since we cannot look up the exact value in this case, let's use an approximation: a t-value of around 3.12 with 60 degrees of freedom would typically give us a p-value less than 0.005 (you would find this p-value to be even lower if using exact calculations with software).

Thus, the p-value is less than 0.005, and since we don't have the exact figure, we round the known significant digits, arriving at a p-value of approximately 0.002 or less.

Extra: In hypothesis testing, statisticians set up two contrasting hypotheses. The null hypothesis (\(H_0\)) represents a skeptical perspective or a claim to be tested, which in this case is the belief that the average salary is no more than $55,000. The alternative hypothesis (\(H_a\) or \(H_1\)) represents the new claim being tested, which Gina believes is that the average salary is higher than $55,000.

The test statistic (in this case, a t-value) helps us determine the likelihood of observing our sample data if the null hypothesis were true. If the test statistic falls into a critical region (or if the p-value is less than our chosen significance level, typically 0.05 or 5%), we reject the null hypothesis in favor of the alternative.

The p-value tells us about the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true. A low p-value indicates that such an extreme result is unlikely under the null hypothesis, leading us to reject the null hypothesis.

In most real-world situations, researchers use statistical software to calculate the exact p-value based on the t-distribution, since the actual p-value provides a more precise measure of how extreme the test statistic is. However, in educational settings, t-distribution tables serve as a useful tool for gaining an understanding of the principles behind hypothesis testing and how p-values can be estimated.