Question

A survey of 181 registered voters in one state reveals that 112 of them favor approval of a bill before the legislature. Construct a 98% confidence interval for the true proportion of all voters in the state who favor approval of the bill. Give your answers as decimals, rounded to 3 places after the decimal point (if necessary). 98% confidence interval for p: ( , )

Answer 1

Hence the approximate 98% confidence interval for the voters in favor of the approval of the bill is ( 0.541 , 0.683)

Explanation:

The sample proportion is = p = 112/181 = 0.61878= 0.612

q = 1-p = 1- 0.612= 0.388

The degree of confidence is 98 % so z₀.₀₂₅= 1.96 taking α = 5 at 95 %

The interval X~± 0.98 is a random variable because X does not have a particular value but takes different values in different samples.

In repeated samples of size 16 from a normal distribution with standard deviation 2 the interval X~± 0.98 will contain true unknown value of mean about 95 percent of the time .

p ± z ( base alpha by 2) √pq/n

Substituting the values

0.612 ± 1.96√0.612*0.388/181

Multiplying p and q

= 0.612 ± 1.96 √0.237456/181

Solving the square root

=0.612 ± 1.96( 0.03622)

Multiplying value of z with the value of square root

=0.612 ± 0.07099

Adding or subtracting will give 0.683, 0.541

Hence the approximate 98% confidence interval for the voters in favor of the approval of the bill is ( 0.541 , 0.683)