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Virty [35]
3 years ago
8

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.
Mathematics
2 answers:
Lina20 [59]3 years ago
7 0

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 - <em>α</em>)% confidence interval for population proportion <em>P</em> is:

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

The information provided is:

<em>x</em> = number of students who responded as"yes" = 70

<em>n</em> = 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).

Gemiola [76]3 years ago
7 0

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

<u><em>95% confidence interval for p</em></u> = [  \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].

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Answer:

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Step-by-step explanation:

Data given and notation    

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s=8.6 represent the population standard deviation    

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\alpha=0.01 represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

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The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

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We can replace in formula (1) the info given like this:    

t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73    

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Hello!!

\rule{300}{5}

\huge\blue{{\fbox{\tt{ANSWER:-\:}}}}

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