In a random sample of 1110 American adults aged 25 or older, it was found that $40 \%$ of them had a college degree. Construct a $95 \%$ confidence interval for the true proportion of all Americans aged 25 or older who have a college degree. Give your answers as decimals, rounded to 3 places after the decimal point (if necessary). 95\% confidence interval for $p$ :

Step 1 ：Given a random sample of 1110 American adults aged 25 or older, it was found that 40% of them had a college degree. We are asked to construct a 95% confidence interval for the true proportion of all Americans aged 25 or older who have a college degree.

Step 2 ：The first step to solve this problem is to calculate the standard error of the proportion. The formula for the standard error of a proportion is \(\sqrt{\frac{p(1-p)}{n}}\), where p is the sample proportion and n is the sample size. In this case, p = 0.40 and n = 1110.

Step 3 ：Calculate the standard error (SE): \(SE = \sqrt{\frac{0.4(1-0.4)}{1110}} = 0.0147\)

Step 4 ：We can find the 95% confidence interval by using the formula p ± Z*SE, where Z is the Z-score for a 95% confidence interval, which is approximately 1.96.

Step 5 ：Calculate the lower and upper bounds of the confidence interval: \(CI_{lower} = p - Z*SE = 0.4 - 1.96*0.0147 = 0.371\) and \(CI_{upper} = p + Z*SE = 0.4 + 1.96*0.0147 = 0.429\)

Step 6 ：Final Answer: The 95% confidence interval for the true proportion of all Americans aged 25 or older who have a college degree is \(\boxed{[0.371, 0.429]}\).