Aresearcher wants to construct a 99% confidence interval for the proportion of elementary school students in seward county who receive free or reduced-price school lunches. what sample size is needed so that the confidence interval will have a margin of error of at most 0.07?

The sample size can be calculated using the formula:

n = p (1-p) (Zc / m)^2

where:

p = proportion of students who received lunches= 0.5

Zc = taken from standard tables at 99% CI = 2.575

m = margin of error =  0.07

Substituting known values:

n = 0.5 (1-0.5) (2.575/0.07)^2

n = 338.3

n = 339 samples

b or 2, just plug these equations into your graphing calculator then check the graph to see the image. if it matches up then that's your answer!

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.