Question 12, 6.3.31-T HW score: $87.7 \%, 10.52$ of 12 Part 2 of 2 points Points: 0 of 1 Save Translate the statement into a confidence interval. Approximate the level of confidence. In a survey of 800 adults in a country, $68 \%$ think teaching is one of the most important jobs in the country today. The survey's margin of error is $\pm 4 \%$. The confidence interval for the proportion is $(0.64,0.72)$. (Round to three decimal places as needed.) The interval is a $\square \%$ confidence interval. (Round to one decimal place as needed.)

Step 1 ：Given that the proportion (p) is 0.68 and the sample size (n) is 800, we can calculate the standard deviation (σ) using the formula for the standard deviation of a proportion: \(σ = \sqrt{(p(1 - p)) / n}\).

Step 2 ：Substituting the given values into the formula, we get \(σ = \sqrt{(0.68(1 - 0.68)) / 800} = 0.0165\).

Step 3 ：We know the confidence interval is (0.64, 0.72). The mean (μ) is the point estimate, which is the middle of the confidence interval.

Step 4 ：We can calculate the Z-score for the lower and upper bounds of the interval using the formula: \(Z = (X - μ) / σ\).

Step 5 ：For the lower bound (0.64), the Z-score is \(Z_{lower} = (0.64 - 0.68) / 0.0165 = -2.43\).

Step 6 ：For the upper bound (0.72), the Z-score is \(Z_{upper} = (0.72 - 0.68) / 0.0165 = 2.43\).

Step 7 ：We can use a Z-table to find the corresponding probabilities for these Z-scores. The probability for \(Z_{lower} = -2.43\) is 0.0076 and for \(Z_{upper} = 2.43\) is 0.9923.

Step 8 ：The confidence level is the difference between the probabilities for the upper and lower bounds, which is \(0.9923 - 0.0076 = 0.9847\) or 98.47%.

Step 9 ：Rounding to one decimal place, the confidence level is 98.5%.

Step 10 ：Final Answer: The interval is a \(\boxed{98.5 \%}\) confidence interval.