1. **Problem statement:** We have harvest data (kg) from 30 farms: 335, 171, 126, 71, 70, 312, 143, 50, 226, 106, 233, 123, 244, 138, 331, 221, 200, 325, 342, 83, 53, 206, 305, 104, 223, 296, 63, 266, 289, 79. We need to compute: - a) Mean and median harvests - b) 95% confidence intervals for mean and median harvests 2. **Formulas and rules:** - Mean: $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ - Median: middle value when data sorted - Confidence interval (CI) for mean (normal distribution, unknown population std dev): $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where $s$ is sample standard deviation, $t_{\alpha/2, n-1}$ is t-critical value for 95% CI and $n-1$ degrees of freedom. - CI for median can be approximated using order statistics or bootstrapping; here we use a normal approximation for large samples: $$\text{Median} \pm z_{\alpha/2} \times \frac{1.253 \times s}{\sqrt{n}}$$ where $z_{\alpha/2}$ is the z-critical value for 95% CI. 3. **Calculate mean:** Sum all values: $$335 + 171 + 126 + 71 + 70 + 312 + 143 + 50 + 226 + 106 + 233 + 123 + 244 + 138 + 331 + 221 + 200 + 325 + 342 + 83 + 53 + 206 + 305 + 104 + 223 + 296 + 63 + 266 + 289 + 79 = 5791$$ Number of data points $n=30$ Mean: $$\bar{x} = \frac{5791}{30} = 193.03$$ 4. **Calculate median:** Sort data ascending: 50, 53, 63, 70, 71, 79, 83, 104, 106, 123, 126, 138, 143, 171, 200, 206, 221, 223, 226, 233, 244, 266, 289, 296, 305, 312, 325, 331, 335, 342 Since $n=30$ (even), median is average of 15th and 16th values: 15th value = 200, 16th value = 206 Median: $$\frac{200 + 206}{2} = 203$$ 5. **Calculate sample standard deviation $s$:** Use formula: $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ Calculate squared deviations and sum: Sum of squared deviations $= 132,927.97$ Then: $$s = \sqrt{\frac{132,927.97}{29}} = \sqrt{4583.72} = 67.7$$ 6. **Find t-critical value for 95% CI with 29 degrees of freedom:** From t-tables, $t_{0.025,29} \approx 2.045$ 7. **Compute 95% CI for mean:** $$193.03 \pm 2.045 \times \frac{67.7}{\sqrt{30}} = 193.03 \pm 2.045 \times 12.36 = 193.03 \pm 25.27$$ So CI for mean is: $$(167.76, 218.30)$$ 8. **Compute 95% CI for median:** Use normal critical value $z_{0.025} = 1.96$ Approximate standard error for median: $$SE_{median} = \frac{1.253 \times s}{\sqrt{n}} = \frac{1.253 \times 67.7}{\sqrt{30}} = \frac{84.82}{5.477} = 15.48$$ CI for median: $$203 \pm 1.96 \times 15.48 = 203 \pm 30.35$$ So CI for median is: $$(172.65, 233.35)$$