Question

In: Statistics and Probability

A survey of 280 homeless persons showed that 63 were veterans. Construct a 99% confidence interval...

A survey of 280 homeless persons showed that 63 were veterans. Construct a 99% confidence interval for the proportion of homeless persons who are veterans

Expert Solution

Solution :

Given that,

n =280

x = 63

= x / n =63 /280 = 0.225

1 - = 1 - 0.225 = 0.775

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 2.576 * ( $\sqrt$ ((0.225* 0.775) / 280) = 0.0643

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.225 - 0.0643 < p < 0.225+ 0.0643

0.1607 < p < 0.2893

The 99% confidence interval for the population proportion p is : ( 0.1607 , 0.2893)

orchestra answered 2 years ago

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