Let's break down the problem and solve it step by step: **Given:** - Null hypothesis: \( H_0 : \mu = \$37,600 \) - Alternative hypothesis: \( H_1 : \mu > \$37,600 \) (right-tailed test) - Level of significance: \( \alpha = 0.01 \) - Sample size: \( n = 26 \) - Sample mean: \( \bar{x} = \$40,150 \) - Population standard deviation: \( \sigma = \$8,650 \) - The data is normally distributed. --- ### **a. Compute the value of the test statistic and its corresponding p-value** #### **Step 1: Calculate the test statistic (z)** \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Plug in the values: \[ z = \frac{40,150 - 37,600}{\frac{8,650}{\sqrt{26}}} \] First, calculate the denominator: \[ \frac{8,650}{\sqrt{26}} \approx \frac{8,650}{5.099} \approx 1,696.58 \] Now, calculate the numerator: \[ 40,150 - 37,600 = 2,550 \] So, \[ z = \frac{2,550}{1,696.58} \approx 1.50 \] **z = 1.50 (rounded to 2 decimal places)** --- #### **Step 2: Find the p-value** For a right-tailed test, we find \( P(Z > 1.50) \). From the Z-table, \( P(Z < 1.50) = 0.9332 \). So, \[ p\text{-value} = 1 - 0.9332 = 0.0668 \] **p-value = 0.0668 (rounded to 4 decimal places)** --- ### **b. Conclusion (Reject or Fail to Reject \( H_0 \))** - **If \( p \leq \alpha \), reject \( H_0 \).** - **If \( p > \alpha \), fail to reject \( H_0 \).** Here, \( p = 0.0668 > 0.01 \). **Conclusion:** Fail to reject \( H_0 \). --- ### **c. State the decision error type that is possible** Since we failed to reject \( H_0 \), the possible error is a **Type II Error** (failing to reject a false null hypothesis). --- ## **Summary Table** | Step | Value/Conclusion | |---------------------|-------------------------------| | Test statistic (z) | 1.50 | | p-value | 0.0668 | | Conclusion | Fail to reject \( H_0 \) | | Possible error type | Type II Error | Let me know if you need further clarification or calculations!