Question

translate the statement into a confidence interval. approximate the level of confidence
in a survey of 1019 adults in a country, 73% said being able to speak the language is at the core of national identity.
the surveys margin of error is ±3.5%.

the confidence interval for the proportion is (0.695, 0.765)
(round to three decimal places as needed.)

the interval is a □% confidence interval
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The confidence interval for a proportion is given by $\hat{p}-E

Step2: Calculate the confidence interval

Substitute the values into the formula:
Lower limit: $0.73- 0.035=0.695$
Upper limit: $0.73 + 0.035=0.765$

Step3: Find the level of confidence

We know that for a proportion, the margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Here, $n = 1019$, $\hat{p}=0.73$, $E=0.035$. First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.73\times(1 - 0.73)}{1019}}=\sqrt{\frac{0.73\times0.27}{1019}}\approx\sqrt{\frac{0.1971}{1019}}\approx\sqrt{0.0001934}\approx0.0139$. Then, since $E = z_{\alpha/2}\times\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, we can find $z_{\alpha/2}=\frac{E}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}}=\frac{0.035}{0.0139}\approx2.52$. Looking up in the standard normal table, if $z_{\alpha/2}=2.52$, then $\alpha/2=1 - 0.9941=0.0059$ (using the standard - normal table: $P(Z<2.52)=0.9941$), and $\alpha=2\times0.0059 = 0.0118$. The level of confidence is $1-\alpha=1 - 0.0118=0.9882\approx98.8\%$

Answer:

The confidence interval for the proportion is $(0.695,0.765)$. The interval is a $98.8\%$ confidence interval.