1. **State the problem:** We want to test if the average time teenagers spend on social media is 3 hours per day based on a sample of 100 teenagers. 2. **State the hypotheses:** - Null hypothesis ($H_0$): $\mu = 3$ hours (the average time is 3 hours) - Alternative hypothesis ($H_a$): $\mu \neq 3$ hours (the average time is not 3 hours) 3. **Formula for the Z-test statistic:** $$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$$ where: - $\bar{X} = 2.84$ (sample mean) - $\mu = 3$ (claimed population mean) - $\sigma = 0.8$ (sample standard deviation) - $n = 100$ (sample size) 4. **Calculate the test statistic:** $$Z = \frac{2.84 - 3}{0.8 / \sqrt{100}} = \frac{-0.16}{0.08} = -2$$ 5. **Decision rule:** - Critical Z values at significance level 0.05 (two-tailed) are $\pm 1.96$. - If $Z < -1.96$ or $Z > 1.96$, reject $H_0$. 6. **Conclusion:** - Calculated $Z = -2$ is less than $-1.96$, so we reject the null hypothesis. - There is enough evidence at the 0.05 significance level to conclude the average time spent on social media by teenagers is different from 3 hours. **Final answer:** The hypothesis test shows sufficient evidence to reject the claim that teenagers spend 3 hours per day on social media.