Answer:
(0.6231 , 0.6749)
Step-by-step explanation:
With the information we have, it is impossible to solve the exercise, therefore I was looking for information to complete it and we have to:
the sample proportion is 64.9%, or 0.649 plus the sample size is 1300 (n)
Now, we have that the standard error is given by:
SE = (p * (1 - p) / n) ^ (1/2)
replacing
SE = (0.649 * (1 - 0.649) / 1300) ^ (1/2)
SE = 0.0132
Now we have that confidence level is 95%, hence α = 1 - 0.95 = 0.05
α / 2 = 0.05 / 2 = 0.025, Zc = Z (α / 2) = 1.96
With this we can calculate margin of error like so:
ME = z * SE
ME = 1.96 * 0.0132
ME = 0.0259
Finally the interval would be:
CI = (p - ME, p + ME)
CI = (0.649 - 0.0259, 0.649 + 0.0259)
CI = (0.6231, 0.6749)
Answer:
mean is equal to 5.12
Step-by-step explanation:
<em>We are given that 32% of college students work fulltime. We have to find the mean for the number of student s who are working full time in a sample of 16</em>
success rate, p = 32% = 0.32
Sample size is denoted as n = 16
The forumla of mean is given as
mean = sample size × success rate
mean = n × p
= 16 × 0.32
= 5.12
X(1/2) = 12
x(1/2)(2) = 12(2)
x(1) = 24
x = 24
hope this helps
The first one doesn't make sense but
1/3 is less than 1/2
1.75 is i think greater than 1.5