Question

A poll reported that only 593 out of a total of 1594 adults in a particular region said they had a​ "great deal of​ confidence" or​ "quite a lot of​ confidence" in the public school system. This was down 5 percentage points from the previous year. Assume the conditions for using the CLT are met. Complete parts​ (a) through​ (d) below.

a. Find a 95​% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the public​ schools, and interpret this interval.

The 95​% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the public schools is ?

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 1594

x = 593

Point estimate = sample proportion = = x / n = 593 / 1594 = 0.372

1 - = 1 - 0.372 = 0.628

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.96 (((0.372 * 0.628) / 1594)

= 0.024

A 95% confidence interval for population proportion p is ,

± E   

= 0.372  ± 0.024

= ( 0.348, 0.396 )

The 95​% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the public schools is between 0.348 and 0.396.