I believe the answer is 8
Answer
1st is yes
2nd is no
Step-by-step explanation:
Answer:
Step-by-step explanation:
8 I think
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}}](https://tex.z-dn.net/?f=CI%3Dp%5Cpm%20z_%7B%5Calpha%2F2%7D%5Csqrt%7B%5Cfrac%7Bp%281-%20p%29%7D%7Bn%7D%7D)
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}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7Bx%7D%7Bn%7D)
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:
thanks im not smart
Step-by-step explanation:
sn i knew that...