Question

A random sample of 330 medical doctors showed that 176 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.)

(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 = 330

x = 176

Point estimate = sample proportion = = x / n = 176/330=0.533

1 - = 1-0.533=0.467

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_/2 = Z_0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.533*0.467) /330 )

= 0.064

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.533-0.064< p <0.533+0.064

(0.469 ,0.597)

lower limit 0.469

upper limit 0.597

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

= 2.326 (((0.533*0.467) /330 )

= 0.064