To construct a 95% confidence interval for the mean waiting time, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown:
\[ CI = \bar{x} \pm t_{(n-1,\alpha/2)} \times \frac{s}{\sqrt{n}} \]
where:
- \(\bar{x}\) = sample mean
- \(s\) = sample standard deviation
- \(n\) = sample size
- \(t_{(n-1,\alpha/2)}\) = critical t-value for \(n-1\) degrees of freedom at the desired confidence level
Given:
- \(\bar{x} = 70.3\) minutes
- \(s = 28.2\) minutes
- \(n = 100\)
First, calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{28.2}{\sqrt{100}} = \frac{28.2}{10} = 2.82 \]
Next, find the critical t-value for a 95% confidence interval. With \(n - 1 = 100 - 1 = 99\) degrees of freedom and a two-tailed confidence level of 95%, we can use a t-table or calculator to find:
- \(t_{(99,0.025)}\), which is approximately 1.984 (this value can vary slightly depending on the table used).
Now we can construct the confidence interval:
\[ CI = 70.3 \pm 1.984 \times 2.82 \]
Calculating the margin of error:
\[ 1.984 \times 2.82 \approx 5.596 \]
Now, using this margin of error, we find the confidence interval:
\[ CI = [70.3 - 5.596, 70.3 + 5.596] = [64.704, 75.896] \]
Thus, the 95% confidence interval for the mean waiting time is approximately:
\[ \boxed{[64.7, 75.9]} \text{ minutes} \]