Question

When n=343 college students are randomly selected and surveyed, it is found that x=110 own a car. Find a 99% confidence interval (CI) and margin of error (ME) for the true proportion of all college students who own a car.

Answer 1

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,

$\alpha = 1-99$%

  $=1-0.99$

  $=0.01$

then,

$\frac{\alpha}{2} =\frac{0.01}{2}$

  $=0.005$

or,

$Z_{\frac{\alpha}{2} }=Z_{0.005}$

     $=2.576$

Now,

The point estimate will be:

⇒  $\hat{P}=\frac{x}{n}$

⇒      $=\frac{110}{343}$

⇒      $=0.321$

or,

⇒  $1-\hat{P}=1-0.321$

⇒            $=0.679$

The margin of error will be:

⇒  $E=Z_{\frac{\alpha}{2} }\times \sqrt (\frac{(\hat{P}\times (1 - \hat{P})) }{n} )$

On substituting the above values, we get

⇒      $=2.576\times \sqrt{\frac{0.321\times 0.679}{343} }$

⇒      $=2.576\times \sqrt{\frac{0.217959}{343} }$

⇒      $=0.065$

hence,

⇒  $\hat{P}-E$

⇒  $0.321-0.065$

⇒  $0.256,0.386$