Question

In: Statistics and Probability

A 2003 Gallup poll asked 804 American adults who ate at fast food restaurants at least...

A 2003 Gallup poll asked 804 American adults who ate at fast food restaurants at least monthly, if they would choose a healthier menu option or stick with their current favorite item.

350 said they would choose a healthier menu option.

When the researchers report this result, what margin of error should they state (for 95% confidence)?

lower limit = ____________________

Upper limit = ____________________

Margin of error = ± __________________________ (express as a decimal to 4 decimal places)

Expert Solution

Sample proportion = 350 / 804 = 0.4353

95% confidence interval is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.4353 - 1.96 * sqrt( 0.4353 * 0.5647 / 804) < p < 0.4353 + 1.96 * sqrt( 0.4353 * 0.5647/ 804)

0.4010 < p < 0.4696

Lower limit = 0.4010

Upper limit = 0.4696

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

= 1.96 * sqrt( 0.4353 * 0.5647 / 804)

= 0.0311

orchestra answered 2 years ago

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