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denis23 [38]
2 years ago
13

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.
Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
6 0

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

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