Question

18#Suppose that 303 out of a random sample of 375 letters mailed in the United States were delivered the day after they were mailed. Based on this, compute a 90% 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 90% 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:

Answer 1

Lower limit: 0.77

Upper limit: 0.84

Explanation:

Given:

x = 303

n = 375

We calculate the value of the proportion in the following way:

`\begin{gathered} p=(x)/(n)=(303)/(375) \\ \\ p=0.808 \end{gathered}`

For a 90% confidence interval we have the following:

`\begin{gathered} \alpha=100\%-90\%=10\%=0.1 \\ \\ \alpha\text{/2}=0.1=0.05 \\ \\ \text{ For the normal table this corresponds to:} \\ \\ Z_{\alpha\text{/2}}=1.645 \end{gathered}`

We calculate the limits of the 90% confidence interval using the following formula:

`\begin{gathered} \text{ Lower limit: }p-Z_{\alpha\text{/2}}\cdot\sqrt{(p\cdot(1-p))/(n)}=\:0.808-1.645\cdot\sqrt{(0.808\cdot\left(1-0.808\right))/(375)}\:=0.77 \\ \\ \:\text{Upper limit: }p-Z_{\alpha\text{/2}}\cdot\sqrt{(p\cdot\left(1-p\right))/(n)}\:=0.808+1.645\cdot\sqrt{(0.808\cdot\left(1-0.808\right))/(375)}=0.84 \end{gathered}`