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.

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].

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