Question

A random sample of 245 students showed that 189 of them liked listening to music while studying. Find the 90% confidence interval for the proportion of students that like listening to music while studying.

What is the SE (standard error) for this sample?

Answer 1

Solution :

Given that,

n = 245

x = 189

Point estimate = sample proportion = = x / n = 189/245=0.771

1 - = 1-0.771=0.229

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

standard error= (( * (1 - )) / n) = (((0.771*0.229) /245 )=0.026844

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

= 1.645 (((0.771*0.229) /245 )

= 1.645*0.026844906

=0.044

A 90% confidence interval is ,

- E < p < + E

0.771-0.044 < p < 0.771+0.044

0.727<p<0.815