A large city recently surveyed 1,354 streetlights, finding that 4.2% of them had burned out. Construct...

A large city recently surveyed 1,354 streetlights, finding that 4.2% of them had burned out. Construct a 99% confidence interval for the proportion of the city's streetlights that are burned out.

(no excel work)

Solutions

Expert Solution

Solution :

Given that,

n = 1354

Point estimate =4.2%=0.042

1 - = 1-0.042 =0.958

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 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.042*0.958) /1354 )

E = 0.014

A 99% confidence interval for proportion p is ,

- E < p < + E

0.042-0.014 < p < 0.042 + 0.014

0.028< p < 0.056

orchestra answered 2 years ago

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