Translate the statement into a confidence interval for p. Approximate the level of confidence.

In a survey of 1000 U. S. adults, 19% are concerned that their taxes will be audited by the Internal Revenue Service. The survey's margin of error is plus or minus 3%. (Source: Rasmussen Reports)

Approximate the level of confidence.

The interval is approximately a % confidence interval.

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.

98 percent

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