Question

suppose that 320 out of a random sample of 400 letters mailed in the united states were delivered the day after they were mailed. based on this, compute a 99% confidence interval for the proportion of all letters mailed in the united states that were delivered the day after they were mailed. then find the lower limit and upper limit of the 99% confidence interval.
carry your intermediate computations to at least three decimal places. round your answers to two decimal places. (if necessary, consult a list of formulas.)
lower limit:
upper limit:

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{320}{400} = 0.8$.

Step2: Find critical value

For a 99% confidence interval, the significance level $\alpha=1 - 0.99=0.01$, and $\alpha/2=0.005$. The critical value $z_{\alpha/2}$ from the standard normal distribution table is $z_{0.005} = 2.576$.

Step3: Calculate standard error

The standard error $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.8\times(1 - 0.8)}{400}}=\sqrt{\frac{0.8\times0.2}{400}}=\sqrt{\frac{0.16}{400}}=\sqrt{0.0004}=0.02$.

Step4: Calculate lower and upper limits

The lower limit of the confidence interval is $\hat{p}-z_{\alpha/2}\times SE=0.8-2.576\times0.02=0.8 - 0.05152 = 0.75$ (rounded to two decimal places).
The upper limit of the confidence interval is $\hat{p}+z_{\alpha/2}\times SE=0.8 + 2.576\times0.02=0.8+0.05152 = 0.85$ (rounded to two decimal places).

Answer:

Lower limit: $0.75$
Upper limit: $0.85$