It is believed that the majority of Americans do not eat any lunch. A 95% confidence...

It is believed that the majority of Americans do not eat any lunch. A 95% confidence interval for the population proportion of Americans who do not eat any lunch is between .475 and .535. What was the sample size?

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Expert Solution

Solution :

Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = (0.475+0.535) / 2=0.505

Margin of error = E = Upper confidence interval - = 0.535-0.505=0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.03)2 * 0.505 * 0.495

= 1067.00

Sample size =1067

orchestra answered 1 year ago

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