Question

A random sample of 332 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 = 332

x = 176

Point estimate = sample proportion = = x / n = 176/332=0.530

1 - = 1-0.530=0.470

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 (((0.530*0.470) /332 )

E = 0.064

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.530-0.064 < p < 0.530+ 0.064

0.466< p < 0.594

The 98% confidence interval for the population proportion p is :lower limit 0.466, upper limit=0.594