Question

A poll is taken in which 265 out of 1000 randomly selected voters indicated their preference for a certain candidate. Find a 90% confidence interval. Remember σpˆ = ?pˆ(1−pˆ)

(a)

[4 pts] Find Zα/2:

(b)

[4 pts] Find the confidence interval:

(c)

[4 pts] Find the margin of error:

(d)

[4 pts] Without doing any calculations, indicate whether the margin of error is larger or smaller or the same for a 95% confidence interval. Why you think so?

Answer 1

Solution :

Given that,

n = 1000

x = 265

Point estimate = sample proportion = = x / n = 265/1000=0.265

1 -   = 1- 0.265 =0.735

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

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

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

= 1.645 *((0.265*0.735) /1000 )

E = 0.0230

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.265-0.0230 < p <0.265+ 0.0230

0.2420< p < 0.2880

The 90% confidence interval for the population proportion p is : 0.2420,0.2880

answer is margin of error is larger for a 95% confidence interval