Question

In Mississippi, a random sample of 110 first graders revealed that 48 of them have been to a dentist at least once.

(a) Find a 95% confidence interval for the population proportion of first-graders in Mississippi who have been to a dentist at least once.
Enter the smaller number in the first box.

Confidence interval: ( , ).

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. A 9898% confidence interval is desired where the interval is no wider than 5 percentage points?

(a)    How many randomly selected students must be surveyed if we assume that there is no available information that could be used as an estimate of p^p^.

First Hint: note the question is limiting the "width" of the interval to 5 percentage points. Also, use at least five decimals places of accuracy with all values used in calculations. Your final answer however should be an integer.

Answer:

(b)    How many randomly selected students must be surveyed if we assume that another study indicated that 66% of college students carry weapons. Answer:

Answer 1

a)

p̂ = X / n = 48/110 = 0.436
p̂ ± Z(α/2) √( (p * q) / n)
0.436 ± Z(0.05/2) √( (0.4364 * 0.5636) / 110)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.436 - Z(0.05) √( (0.4364 * 0.5636) / 110) = 0.344
upper Limit = 0.436 + Z(0.05) √( (0.4364 * 0.5636) / 110) = 0.529
95% Confidence interval is ( 0.344 , 0.529 )

2)

a)

Margin of error E = Width / 2 = 0.05 / 2 = 0.025

When prior estimate for proportion is not given then p = 0.50

Sample size = Z2/2 * p ( 1- p) / E2

So,

n = 2.32632 * 0.5 * 0.5 / 0.0252

= 2164.66

n = 2165 (Rounded up to nearest integer)

b)

n = 2.32632 * 0.66 / ( 1 - 0.66) / 0.0252

= 1943.01

n = 1944 (Rounded up to nearest integer)