1. **State the problem:** We want to find the 95% confidence interval for the average daily sales of a café based on a sample of 20 days, where the sample mean is $4800$ and the sample standard deviation is $400$. 2. **Identify the known values:** - Sample mean $\bar{x} = 4800$ - Sample standard deviation $s = 400$ - Sample size $n = 20$ - Confidence level = 95% 3. **Determine the appropriate distribution:** Since the sample size is less than 30 and population standard deviation is unknown, we use the $t$-distribution. 4. **Find the critical $t$ value:** Degrees of freedom $df = n - 1 = 19$. Using a $t$-table or calculator for 95% confidence, two-tailed, $t_{0.025,19} \approx 2.093$. 5. **Calculate the margin of error (ME):** $$ ME = t \times \frac{s}{\sqrt{n}} = 2.093 \times \frac{400}{\sqrt{20}} $$ Calculate $\sqrt{20} \approx 4.472$: $$ ME = 2.093 \times \frac{400}{4.472} = 2.093 \times 89.44 \approx 187.11 $$ 6. **Determine the confidence interval:** $$ \text{Lower bound} = 4800 - 187.11 = 4612.89 $$ $$ \text{Upper bound} = 4800 + 187.11 = 4987.11 $$ 7. **Interpretation:** We are 95% confident that the true mean daily sales lie between $4612.89$ and $4987.11$. **Final answer:** The 95% confidence interval is $$\left(4612.89, 4987.11\right)$$.