Question

In: Statistics and Probability

Five hundred seventeen (517) homes in a certain southern California community are randomly surveyed to determine...

Five hundred seventeen (517) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-one (171) of the homes surveyed met the minimum recommendations for earthquake preparedness and 346 did not.

Find the confidence interval at the 90% confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. (Round your answers to three decimal places.)

( _____ , _______ )

Solutions

Expert Solution

Solution :

Given that,

n = 517

x = 171

Point estimate = sample proportion = = x / n = 171 / 517 = 0.331

1 - = 1 - 0.331 = 0.669

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.331 * 0.669) / 517 )

= 0.034

A 90% confidence interval for population proportion p is ,

± E   

= 0.331  ± 0.034

= ( 0.297, 0.365 )


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