Respuesta :
Answer:
The 95% confidence interval for the true proportion of university students who use laptop in class to take notes is (0.2839, 0.4161).
Step-by-step explanation:
The (1 - α)% confidence interval for population proportion P is:
[tex]CI=p\pm z_{\alpha/2}\sqrt{\frac{p(1- p)}{n}}[/tex]
The information provided is:
x = number of students who responded as"yes" = 70
n = sample size = 200
Confidence level = 95%
The formula to compute the sample proportion is:
[tex]p=\frac{x}{n}[/tex]
The R codes for the construction of the 95% confidence interval is:
> x=70
> n=200
> p=x/n
> p
[1] 0.35
> s=sqrt((p*(1-p))/n)
> s
[1] 0.03372684
> E=qnorm(0.975)*s
> lower=p-E
> upper=p+E
> lower
[1] 0.2838966
> upper
[1] 0.4161034
Thus, the 95% confidence interval for the true proportion of university students who use laptop in class to take notes is (0.2839, 0.4161).
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. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = proportion of university students who use laptop in class to take notes in a survey of 200 students = [tex]\frac{70}{200}[/tex] = 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 < [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95
P([tex]-1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < [tex]{\hat p -p}[/tex] < [tex]1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.95
P( [tex]\hat p -1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < p < [tex]\hat p +1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.95
95% confidence interval for p = [ [tex]\hat p -1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] , [tex]\hat p +1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ]
= [ [tex]0.35 -1.96 \times \sqrt{\frac{0.35(1-0.35)}{200} }[/tex] , [tex]0.35 +1.96 \times \sqrt{\frac{0.35(1-0.35)}{200} }[/tex] ]
= [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].