1. **State the problem:** We want to test if the mean length of novels written by the club members is greater than 50,000 words at a significance level $\alpha = 0.10$. 2. **Set hypotheses:** - Null hypothesis $H_0$: $\mu = 50000$ (mean length is 50,000 words) - Alternative hypothesis $H_a$: $\mu > 50000$ (mean length is greater than 50,000 words) 3. **Data given:** Lengths: 48972, 50100, 51560, 49800, 50020, 49900, 52193 Sample size $n=7$. 4. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{48972 + 50100 + 51560 + 49800 + 50020 + 49900 + 52193}{7} = \frac{352545}{7} = 50363.57$$ 5. **Calculate sample standard deviation $s$:** First find squared deviations: $(48972 - 50363.57)^2 = 193,927,716.5$ $(50100 - 50363.57)^2 = 6,995,306.5$ $(51560 - 50363.57)^2 = 144,388,306.5$ $(49800 - 50363.57)^2 = 31,995,306.5$ $(50020 - 50363.57)^2 = 11,877,306.5$ $(49900 - 50363.57)^2 = 21,995,306.5$ $(52193 - 50363.57)^2 = 33,195,306.5$ Sum of squared deviations $= 444,374,556$ Sample variance $s^2 = \frac{444,374,556}{7-1} = 74,062,426$ Sample standard deviation $s = \sqrt{74,062,426} = 8603.07$ 6. **Calculate test statistic $t$:** $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{50363.57 - 50000}{8603.07/\sqrt{7}} = \frac{363.57}{3252.68} = 0.112$$ 7. **Determine critical value:** Degrees of freedom $df = n-1 = 6$. At $\alpha=0.10$ for a one-tailed test, critical $t_{0.10,6} = 1.440$ (from t-distribution table). 8. **Decision:** Since $t = 0.112 < 1.440$, we fail to reject the null hypothesis. 9. **Conclusion:** There is insufficient evidence at the 0.10 significance level to conclude that the mean length of novels is greater than 50,000 words.