Asurvey of 350 students is selected randomly on a large campus. they are asked if they use a laptop in class to take notes. suppose that based on the survey, 140 of the 350 students responded "yes."
a. what is the value of the sample proportion ^p?
b. what is the standard error of the sample proportation?
c. construct an approximate 95% confidence interval for the true proportion p by taking +- ses from the sample proportion.
A. The value of the sample proportion is 0.4
B. The standard error of the sample proportion is 0.02619
C. 0.3487 ≤ p ≤ 0.4513
The value of the sample proportion p' is calculated as:
Where x is the number of success in the sample or the number of students that use a laptop in class to take notes and n is the size of the sample or 350 students.
On the other hand, the standard error SE of the sample proportion is calculated as:
so, replacing the values, we get:
Finally, an approximate 95% confidence interval for the true proportion p is calculate as:
Where 1-α is equal to 95%, so is equal to 1.96. Then, replacing the values we get:
0.4 - 0.0513 ≤ p ≤ 0.4 + 0.0513
0.3487 ≤ p ≤ 0.4513
tickets for the harlem globetrotters
show at michigan state university in
2010 cost $16, $23, or, for vip seats, $40.
total revenue = 16x + 23y + 40z
if 9 times as many $16 tickets were sold
as vip tickets,
x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold
as vip tickets,
x = 9z => x – 9z = 0, sales of all three kinds of tickets would
16x + 23y + 40z = 46575, how many of each kind of ticket would
have been sold?
solve for x, y and z
x − 9z = 0
x − y − z = 55
16x + 23y + 40z = 46575
x = 1170
y = 985
z = 130
answer: f = 32.2m
step-by-step explanation: that is the answer