The American Heart Association is about to conduct an anti-smoking campaign and wants to know the...

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 2222 who smoke.

Step 2 of 2 :

Suppose a sample of 1421 Americans over 22 is drawn. Of these people, 953 don't smoke. Using the data, construct the 90% confidence interval for the population proportion of Americans over 22 who smoke. Round your answers to three decimal places.

Solutions

Expert Solution

Solution :

Given that,

n = 1421

x = 1421 - 953 = 468

Point estimate = sample proportion = = x / n = 468 / 1421 = 0.329

1 - = 1 - 0.329 = 0.671

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 * $\sqrt$ (( * (1 - )) / n)

= 1.645 ( $\sqrt$ ((0.329 * 0.671) / 1421)

= 0.021

A 90% confidence interval for population proportion p is ,

± E

= 0.329 ± 0.021

= ( 0.308, 0.350 )

orchestra answered 1 year ago

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