Question

In a simple random sample of size 50, taken from a population, 23 of the individuals met a specified criteria.

a) What is the margin of error for a 90% confidence interval for p, the population proportion?

Round your response to at least 3 decimal places.

   

b) What is the margin of error for a 95% confidence interval for p?

Round your response to at least 3 decimal places.

   

Answer 1

Solution :

Given that,

n = 50

x = 23

Point estimate = sample proportion = = x / n = 23/50=0.46

1 - = 1-0.46=0.54

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )

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

= 1.645 (((0.46*0.54) / 50)

Margin of error = E = 0.116

B.

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96} ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )   

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

= 1.96 (((0.46*0.54) / 50)

Margin of error = E    = 0.138

B.