Answer: [2.07%; 12.93%]

## Explanation Since the population variance is unknown and the population is normally distributed, and the sample size is less than 30, we use a t-statistic. The t-statistic for a 95% confidence interval and 27 degrees of freedom (df = n-1 = 28-1 = 27) is 2.052. ### Step 1: Calculate the Standard Error \[ \text{Standard Error} = \frac{\text{Sample Standard Deviation}}{\sqrt{\text{Sample Size}}} = \frac{14}{\sqrt{28}} = \frac{14}{5.2915} \approx 2.646 \] ### Step 2: Calculate the Margin of Error \[ \text{Margin of Error} = t_{\alpha/2, df} \times \text{Standard Error} = 2.052 \times 2.646 \approx 5.43 \] ### Step 3: Calculate the Confidence Interval \[ \text{Lower Bound} = \text{Sample Mean} - \text{Margin of Error} = 7.5 - 5.43 = 2.07\% \] \[ \text{Upper Bound} = \text{Sample Mean} + \text{Margin of Error} = 7.5 + 5.43 = 12.93\% \] Therefore, the 95% confidence interval is **[2.07%; 12.93%]**. ### Key Points: - **t-distribution** is used when population variance is unknown and sample size is small (n ≤ 30) - **Degrees of freedom** = n - 1 = 27 - **t-critical value** for 95% confidence with 27 df is 2.052 - Using z-statistic instead would give incorrect results for small samples with unknown population variance

Author: Tanishq Prabhu