Answer:
The margin of error will be "0.65". A further explanation is provided below.
Step-by-step explanation:
The given values are:
n = 343
x = 110
At 99% confidence level,
[tex]\alpha = 1-99[/tex]%
[tex]=1-0.99[/tex]
[tex]=0.01[/tex]
then,
[tex]\frac{\alpha}{2} =\frac{0.01}{2}[/tex]
[tex]=0.005[/tex]
or,
[tex]Z_{\frac{\alpha}{2} }=Z_{0.005}[/tex]
[tex]=2.576[/tex]
Now,
The point estimate will be:
⇒ [tex]\hat{P}=\frac{x}{n}[/tex]
⇒ [tex]=\frac{110}{343}[/tex]
⇒ [tex]=0.321[/tex]
or,
⇒ [tex]1-\hat{P}=1-0.321[/tex]
⇒ [tex]=0.679[/tex]
The margin of error will be:
⇒ [tex]E=Z_{\frac{\alpha}{2} }\times \sqrt (\frac{(\hat{P}\times (1 - \hat{P})) }{n} )[/tex]
On substituting the above values, we get
⇒ [tex]=2.576\times \sqrt{\frac{0.321\times 0.679}{343} }[/tex]
⇒ [tex]=2.576\times \sqrt{\frac{0.217959}{343} }[/tex]
⇒ [tex]=0.065[/tex]
hence,
⇒ [tex]\hat{P}-E<p<\hat{P}+E[/tex]
⇒ [tex]0.321-0.065<p<0.321+0.065[/tex]
⇒ [tex]0.256,0.386[/tex]
