Respuesta :
Answer:
The 95% confidence interval would be given (0.174;0.226).
We are confident (95%) that true proportion of orders that were shipped late is between 0.174 and 0.226
Step-by-step explanation:
Data given and notation
n=900 represent the random sample taken
X=180 represent the orders that were shipped late
[tex]\hat p=\frac{180}{900}=0.2[/tex] estimated proportion of the orders that were shipped late
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)
p= population proportion of orders that were shipped late
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.2 - 1.96 \sqrt{\frac{0.2(1-0.2)}{900}}=0.174[/tex]
[tex]0.2 + 1.96 \sqrt{\frac{0.2(1-0.2)}{900}}=0.226[/tex]
And the 95% confidence interval would be given (0.174;0.226).
We are confident (95%) that true proportion of orders that were shipped late is between 0.174 and 0.226
