A wedding website states that the average cost of a wedding is \( \$ 26,987 \). One concerned bride hopes that the average is less than reported. To see if her hope is correct, she surveys 40 recently married couples and finds that the average cost of weddings in the sample was \( \$ 24,998 \). Assuming that the population standard deviation is \( \$ 6039 \), is there sufficient evidence to support the bride's hope at the 0.05 level of significance? Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

To compute the value of the test statistic, we use the formula for the z-test when the population standard deviation is known: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where: - \(\bar{x}\) = sample mean = \$24,998 - \(\mu\) = population mean (claimed average) = \$26,987 - \(\sigma\) = population standard deviation = \$6,039 - \(n\) = sample size = 40 First, we calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{6039}{\sqrt{40}} \approx \frac{6039}{6.3246} \approx 954.18 \] Next, we substitute the values into the z-test formula: \[ z = \frac{24998 - 26987}{954.18} \approx \frac{-1989}{954.18} \approx -2.09 \] Rounded to two decimal places, the value of the test statistic is: \[ \mathbf{-2.09} \]