Question

For service improvement purposes, the manager at a large restaurant would like to conduct a survey to estimate the proportion of customers who were unhappy with some aspect of their dining experience. What (minimum) sample size is needed to form a 95% confidence interval to estimate the true proportion of unhappy customers to within ± 0.06?

Do not round intermediate calculations. Round up your final answer to the next whole number.

Sample size =

customers

Answer #1

Solution :

Given that,

= 0.5 ( assume 0.5)

1 - = 1 - 0.5 = 0.5

margin of error = E = ± 0.06

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 / 0.06)2 * 0.5 * 0.5

= 266.78

Sample size =267

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