1. **State the problem:** We want to test if the average effect level of change in foreign exchange rates is greater than 0.4 using the given sample data and a 5% significance level. 2. **Identify the hypotheses:** - Null hypothesis $H_0$: $\mu = 0.4$ (the average effect level is 0.4) - Alternative hypothesis $H_a$: $\mu > 0.4$ (the average effect level is greater than 0.4) 3. **Given data:** Sample values: $0.593, 0.42, 0.329, 0.69, 0.231, 0.893, 0.512, 0.392, 0.418, 0.575, 0.413, 0.419, 0.617, 0.375, 0.482$ Sample size $n = 15$ Significance level $\alpha = 0.05$ 4. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{\sum x_i}{n} = \frac{0.593 + 0.42 + 0.329 + 0.69 + 0.231 + 0.893 + 0.512 + 0.392 + 0.418 + 0.575 + 0.413 + 0.419 + 0.617 + 0.375 + 0.482}{15}$$ $$\bar{x} = \frac{7.859}{15} = 0.5239$$ 5. **Calculate the sample standard deviation $s$:** First find each squared deviation $(x_i - \bar{x})^2$, sum them, then divide by $n-1$ and take the square root. Sum of squared deviations $= 0.593^2 + 0.42^2 + ...$ (calculated as 0.0937 approx) More precisely, $$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = 0.158$$ (calculated value) 6. **Calculate the test statistic $t$:** $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{0.5239 - 0.4}{0.158/\sqrt{15}} = \frac{0.1239}{0.0408} = 3.04$$ 7. **Determine the critical value:** For a one-tailed test with $\alpha=0.05$ and $df = n-1 = 14$, the critical $t$ value is approximately $1.761$. 8. **Decision:** Since $t = 3.04 > 1.761$, we reject the null hypothesis. 9. **Conclusion:** There is sufficient evidence at the 5% significance level to conclude that the average effect level of change in foreign exchange rates is greater than 0.4.