1. **State the problem:** We want to test if the average time spent on the company's website per session is significantly different from the industry average of 5 minutes. 2. **Set up hypotheses:** - Null hypothesis $H_0$: $\mu = 5$ (mean time equals industry average) - Alternative hypothesis $H_a$: $\mu \neq 5$ (mean time is different) 3. **Formula for one-sample t-test:** $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ where $\bar{x}$ is sample mean, $\mu_0$ is hypothesized mean, $s$ is sample standard deviation, and $n$ is sample size. 4. **Plug in values:** - $\bar{x} = 4.5$ - $\mu_0 = 5$ - $s = 1.2$ - $n = 50$ Calculate standard error: $$ SE = \frac{1.2}{\sqrt{50}} = \frac{1.2}{7.071} \approx 0.1697 $$ Calculate t-statistic: $$ t = \frac{4.5 - 5}{0.1697} = \frac{-0.5}{0.1697} \approx -2.947 $$ 5. **Interpretation:** With $n-1=49$ degrees of freedom, compare $t = -2.947$ to critical t-values or find p-value to decide if the difference is significant. **Final answer:** The calculated t-statistic is approximately $-2.947$, which can be used to test if the average time differs significantly from 5 minutes.