Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

Of 354 randomly selected medical students, 20 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

* Round to exactly 4 decimal places. Write in interval notation inside (). Leave one space after the comma. Include a leading 0. *

Answer 1

Solution :

Given that,

n = 354

x = 20

Point estimate = sample proportion = = x / n = 20/354=0.056

1 - = 1-0.056=0..944

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )   

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

= 1.96 (((0.056*0.944) / 354)

= 0.0240

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.056-0.0240 < p < 0.056'+0.0240

(0.0320< p < 0.080)

(0.0320 , 0.080)