QUESTION 13 An online stock trading company makes part of their revenue from clients when the clients trade stocks therefore, it is important to the company to have an good idea of how many trades its clients are making in a given year. In a sample of 120 clients of an online stock trading company, the average number of trades per year was 82 with a standard deviation of 16 . If you were to test the hypothesis that the average number of trades per year is different than the previous year when the average number of trades was 85 (using the \( 5 \% \) level of significance), what is p-value for this test? (please round your answer to 4 decimal places)

To determine the p-value for testing whether the average number of trades per year has changed from the previous year, follow these steps: ### **Given:** - **Sample size (n):** 120 - **Sample mean (\(\bar{x}\)):** 82 - **Population mean under null hypothesis (\(\mu_0\)):** 85 - **Sample standard deviation (s):** 16 - **Significance level:** 5% (two-tailed test) ### **Steps:** 1. **Formulate Hypotheses:** - **Null Hypothesis (\(H_0\)):** \(\mu = 85\) - **Alternative Hypothesis (\(H_1\)):** \(\mu \neq 85\) 2. **Calculate the Test Statistic (z):** \[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{82 - 85}{16 / \sqrt{120}} = \frac{-3}{1.4606} \approx -2.054 \] 3. **Determine the p-value:** - Since this is a two-tailed test, the p-value is \(2 \times P(Z < -2.054)\). - Using standard normal distribution tables or a calculator: \[ P(Z < -2.054) \approx 0.0200 \] - Therefore, the p-value is: \[ 2 \times 0.0200 = 0.0400 \] ### **Conclusion:** The p-value for this test is **0.0400**. Since the p-value (0.0400) is less than the significance level (0.05), you would reject the null hypothesis, indicating that there is a statistically significant difference in the average number of trades per year compared to the previous year. **Answer:** 0.0400