Question

Find the point estimate, the standard error, and the margin of error for the given confidence level and values of

x and n. x=45, n=97, confidence level 95%

(a) Find the point estimate.

(b) Find the standard error.

(c) Find the margin of error.

Answer 1

Solution :

Given that,

n = 97

x = 45

Point estimate = sample proportion = = x / n = 45/97=0.464

1 -   = 1- 0.464 =0.536

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 )

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

= 1.96 (((0.464*0.536) /97 )

= 0.0962

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.464-0.0962 < p <0.464+ 0.0962

0.3678< p < 0.5602

The 95% confidence interval for the population proportion p is : 0.3678,0.5602

Margin of error = E = 0.0962

Point estimate = sample proportion = =0.464

standard error =((( * (1 - )) / n) =(((0.464*0.536) /97 )=0.0506