Answer: 19.06

## Explanation To calculate the test statistic for this hypothesis test, we use the formula for a one-sample t-test (or z-test when population standard deviation is known, but here we use sample standard deviation): **Step 1: Calculate the standard error of the sample mean** \[\text{Standard Error} = \frac{s}{\sqrt{n}}\] Where: - \(s = \$4,500\) (sample standard deviation) - \(n = 115\) (sample size) \[\text{Standard Error} = \frac{4500}{\sqrt{115}} = \frac{4500}{10.7238} \approx 419.6272\] **Step 2: Calculate the test statistic** \[t = \frac{\bar{x} - \mu_0}{\text{Standard Error}}\] Where: - \(\bar{x} = \$65,000\) (sample mean) - \(\mu_0 = \$57,000\) (hypothesized population mean) \[t = \frac{65,000 - 57,000}{419.6272} = \frac{8,000}{419.6272} \approx 19.06\] **Why this is correct:** 1. The test statistic measures how many standard errors the sample mean is away from the hypothesized population mean 2. A value of 19.06 indicates the sample mean is 19.06 standard errors above the hypothesized mean of $57,000 3. This is a very large test statistic, suggesting strong evidence against the null hypothesis **Note:** Although we're using sample standard deviation, with a sample size of 115, the t-distribution is very close to the normal distribution, so this test statistic would be highly significant.

Author: Nikitesh Somanthe