Let's solve **Problems 4.11 and 4.12** step by step using the data from the table. --- # Problem 4.11 **Task:** Estimate the range of values for which you would expect 95% of all possible measured values for each column (Set 1, Set 2, Set 3) to fall. Discuss in terms of finite statistics. ### **Step 1: Calculate Mean (\(\bar{x}\)) and Standard Deviation (\(s\))** #### **Set 1 Values:** 51.9, 51., 50.3, 49.6, 51., 50., 48.9, 50.5, 50.9, 52.4, 51.3, 507, 52., 49.4, 49.7, 50.5, 50.7, 49.4, 49.9, 49.2 - **Mean (\(\bar{x}\))**: \[ \bar{x}_1 = \frac{\sum x_i}{n} = \frac{101.3}{20} = 50.515 \] - **Standard deviation (\(s\))**: \[ s_1 \approx 1.016 \] #### **Set 2 Values:** 51.9, 48.7, 51.1, 51.7, 49.9, 48.8, 52.5, 51.7, 51.3, 52.6, 49.4, 50.3, 50.3, 50.2, 50.9, 52.1, 49.3, 50.7, 50.5, 49.7 - **Mean (\(\bar{x}\))**: \[ \bar{x}_2 = \frac{1015.7}{20} = 50.785 \] - **Standard deviation (\(s\))**: \[ s_2 \approx 1.255 \] #### **Set 3 Values:** 51.1, 50.1, 51.4, 50.5, 49.7, 51.6, 51., 49.5, 52.4, 49.5, 51.6, 49.4, 51.6, 49.4, 50.8, 50.8, 50.2, 50.1, 52.3, 48.9, 50.4, 51.5 There are 21 values; let's use the first 20 for consistency. - **Mean (\(\bar{x}\))**: \[ \bar{x}_3 = \frac{101.2}{20} = 50.51 \] - **Standard deviation (\(s\))**: \[ s_3 \approx 1.012 \] --- ### **Step 2: 95% Range for Measured Values** For a normal distribution, 95% of values fall within \(\bar{x} \pm 2s\) (more precisely, \(\bar{x} \pm 1.96s\)). - **Set 1:** Range: \(50.515 \pm 1.96 \times 1.016 = (48.53, 52.50)\) - **Set 2:** Range: \(50.785 \pm 1.96 \times 1.255 = (48.33, 53.24)\) - **Set 3:** Range: \(50.51 \pm 1.96 \times 1.012 = (48.53, 52.49)\) --- ### **Discussion (Finite Statistics)** - The range estimates vary because each set is a finite sample of a larger population. - Sampling variation means each set's statistics differ, even when measuring the same thing. - With more data, the estimates would converge and ranges would stabilize. --- # Problem 4.12 **Task:** Estimate the mean value at a 95% probability level (confidence interval for the mean) for each set. Compare to previous answers. ### **Step 1: 95% Confidence Interval for the Mean** The 95% confidence interval for the mean: \[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \] For \(n=20\), degrees of freedom = 19. \(t^* \approx 2.093\). #### **Set 1:** \[ \text{CI} = 50.515 \pm 2.093 \times \frac{1.016}{\sqrt{20}} = 50.515 \pm .475 \] **Interval:** (50.04, 50.99) #### **Set 2:** \[ \text{CI} = 50.785 \pm 2.093 \times \frac{1.255}{\sqrt{20}} = 50.785 \pm .587 \] **Interval:** (50.20, 51.37) #### **Set 3:** \[ \text{CI} = 50.51 \pm 2.093 \times \frac{1.012}{\sqrt{20}} = 50.51 \pm .474 \] **Interval:** (50.04, 50.98) --- ### **Discussion** - Confidence intervals for the mean are much narrower than the 95% data ranges. - The interval reflects uncertainty in the estimate of the mean, not the spread of individual values. - Differences between sets reflect finite sample statistics—real variability in small samples. - With increasing sample size, intervals would become narrower, and means would converge. --- ## **Summary Table** | Set | Mean (\(\bar{x}\)) | Std Dev (\(s\)) | 95% Range for Measured Values | 95% CI for Mean | |--------|---------------------|-----------------|------------------------------|-------------------| | 1 | 50.515 | 1.016 | (48.53, 52.50) | (50.04, 50.99) | | 2 | 50.785 | 1.255 | (48.33, 53.24) | (50.20, 51.37) | | 3 | 50.510 | 1.012 | (48.53, 52.49) | (50.04, 50.98) | --- ### **Key Concept** - **95% range for measured values**: Where most individual measurements are expected to fall. - **95% confidence interval for mean**: Where the true mean of all possible measurements is likely to be. Let me know if you need further breakdowns or calculations!