Question

A survey of 200 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. Suppose that based on the​ survey, 80 of the 200 students responded​ "yes."

​a) What is the value of the sample proportion ModifyingAbove p with caret​?

​b) What is the standard error of the sample​ proportion?

​c) Construct an approximate 95​% confidence interval for the true proportion p by taking plus or minus 2 SEs from the sample proportion.

Answer 1

Solution :

Given that,

n = 200

x = 80

(a) Point estimate = sample proportion = = x / n = 80 / 200 = 0.400

1 - = 1 - 0.400 = 0.600

(b) Standard error =  ((p * q )) / n)

= ((( 0.400 * 0.600)) / 200)

   = 0.035

The standard error of the sample​ proportion: 0.035

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

= 1.96 ((( 0.400 * 0.600)) / 200 )

= 0.068

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.400 - 0.068 < p < 0.400 + 0.068

0.332 < p < 0.468

( 0.332 , 0.468 )

The 95% confidence interval for the population proportion p is : ( 0.332 , 0.468 )