Question

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". Using RStudio, construct and interpret a 95% confidence interval for the true proportion of university students who use laptop in class to take notes.

Answer 1

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:

$CI=p\pm z_{\alpha/2}\sqrt{\frac{p(1- p)}{n}}$

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:

$p=\frac{x}{n}$

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 2

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. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of university students who use laptop in class to take notes in a survey of 200 students = $\frac{70}{200}$ = 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 < $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P($-1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p -p}$ < $1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p -1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p +1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

95% confidence interval for p = [  $\hat p -1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ ,  $\hat p +1.96 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

                                 = [ $0.35 -1.96 \times \sqrt{\frac{0.35(1-0.35)}{200} }$ , $0.35 +1.96 \times \sqrt{\frac{0.35(1-0.35)}{200} }$ ]

                                 = [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].