Answer:
95% confidence interval for the true proportion of university students is [0.284 , 0.416].
Step-by-step explanation:
We are given that a survey of 200 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. The result of the survey is that 70 of the 200 students responded "yes".
Let p = true proportion of university students who use laptop in class to take notes
So, the pivotal quantity for 95% confidence interval for true proportion is given by;
P.Q. = ~ N(0,1)
where, = proportion of university students who use laptop in class to take notes in a survey of 200 students = = 0.35
n = sample of students = 200
So, 95% confidence interval for the true proportion, p is;
P(-1.96 < N(0,1) < 1.96) = 0.95
P(-1.96 < < 1.96) = 0.95
P( < < ) = 0.95
P( < p < ) = 0.95
<u><em>95% confidence interval for p</em></u> = [ , ]
= [ , ]
= [0.284 , 0.416]
Therefore, 95% confidence interval for the true proportion of university students who use laptop in class to take notes is [0.284 , 0.416].