Question

translate the statement into a confidence interval. approximate the level of confidence.
in a survey of 900 adults in a country, 77% think teaching is one of the most important jobs in the country today.
the surveys margin of error is ±4%.
the confidence interval for the proportion is (0.73, 0.81)
(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: $0.77- 0.04=0.73$
Upper limit: $0.77 + 0.04=0.81$
So the confidence interval is $(0.73,0.81)$

Step3: Find the z - value

The formula for the margin of error for a proportion is $E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. We know that $n = 900$, $\hat{p}=0.77$, and $E=0.04$.
First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.77\times(1 - 0.77)}{900}}=\sqrt{\frac{0.77\times0.23}{900}}\approx\sqrt{\frac{0.1771}{900}}\approx\sqrt{0.0001968}\approx0.014$
Then, since $E = z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, we can solve for $z$: $z=\frac{E}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}}=\frac{0.04}{0.014}\approx2.86$

Step4: Find the confidence level

Using the standard normal distribution table, for $z = 2.86$, the area in the tails is $2\times(1-\Phi(2.86))$. From the standard normal table, $\Phi(2.86)\approx0.9979$. So the area in the tails is $2\times(1 - 0.9979)=2\times0.0021 = 0.0042$. The confidence level is $1-0.0042 = 0.9958\approx99.6\%$

Answer:

The interval is a $99.6\%$ confidence interval.