In: Statistics and Probability
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?
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