Can someone please help me with this question?
2 answers:
As per the given graph we can observe the number of students which watched TV for certain intervals.
According to question, we have to find the common interval of minutes students spent watching TV~
That means : we have to find the time interval in which most of the students prefer to watch TV ~
Hence, the most common interval is :
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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:
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:
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).