Question

I want to know what percent of these coffee shops sell espresso. Out of the 150 I sampled, only 40 offered espresso. Without the finite population correction factor, what is the margin of error for a 90% confidence interval around this proportion? Round to the third decimal point.

Answer 1

Solution :

Given that,

n = 150

x = 40

Point estimate = sample proportion = = x / n = 40/150=0.267

1 -   = 1- 0.267 =0.733

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.267*0.733) / 150)

E = 0.0594

Margin of error = E =0.059