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: ( , )

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

Step-by-step 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)

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00