Answer:
The interval is approximately a 98% confidence interval
Step-by-step explanation:
From the question : Error, E= 3% = 0.03, Total population, n=1000, number of people that are concerned that their taxes will be audited, p = 19% = 0.19
E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}
0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}
z_{\alpha/2}^{2}=5.848
z_{\alpha/2}=2.418
Area right to 2.418 is 0.0078. So
\alpha/2=0.0078
Therefore \alpha=0.0156\approx 0.02
Thus, the interval is approximately a 98% confidence interval.
