Question

A random sample of 324 medical doctors showed that 178 had a solo practice.

(b) Find a 98% confidence interval for p. (Use 3 decimal places.)

lower limit
upper limit

What is the margin of error based on a 98% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that,

n = 324

x = 178

Point estimate = sample proportion = = x / n = 178/324=0.549

1 -   = 1- 0.549 =0.451

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.549*0.451) /324 )

= 0.064

A 98% confidence interval is ,

- E < p < + E

0.549 - 0.064 < p < 0.549 +0.064

0.485< p < 0.613

lower limit 0.485
upper limit    0.613

b.

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

= 2.326 (((0.549*0.451) /324 )

= 0.064