Question

translate the statement into a confidence interval. approximate the level of confidence
in a survey of 1090 adults in a country, 68% said being able to speak the language is at the core of national identity. the surveys margin of error is ±3.9%
the confidence interval for the proportion is (□,□)
(round to three decimal places as needed.)
the interval is a □% confidence interval
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the confidence interval

The formula for a confidence interval for a proportion is $\hat{p}\pm E$, where $\hat{p}$ is the sample proportion and $E$ is the margin of error.
Given $\hat{p} = 0.68$ (since $68\%=0.68$) and $E = 0.039$.
The lower bound is $\hat{p}-E=0.68 - 0.039=0.641$.
The upper bound is $\hat{p}+E=0.68+ 0.039 = 0.719$.

Step2: Find the level of confidence

We know that for a proportion, $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Here, $n = 1090$, $\hat{p}=0.68$, $E = 0.039$.
First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.68\times(1 - 0.68)}{1090}}=\sqrt{\frac{0.68\times0.32}{1090}}\approx\sqrt{\frac{0.2176}{1090}}\approx\sqrt{0.0001996}\approx0.0141$.
Then, $z_{\alpha/2}=\frac{E}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}}=\frac{0.039}{0.0141}\approx2.77$.
Looking up in the standard normal table, a $z -$score of $z = 2.77$ corresponds to a two - tailed confidence level. The area to the left of $z = 2.77$ is $P(Z<2.77)=0.9972$, and the two - tailed confidence level is $C = 2\times(0.9972 - 0.5)=0.9944\approx99.4\%$.

Answer:

The confidence interval for the proportion is $(0.641,0.719)$. The interval is a $99.4\%$ confidence interval.