In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
water drinking habits:
mean = 62 ounces
standard deviation = 5.2 ounces
Step 02:
normally distribution:
95% ===> 2 SD
(62 + 5.2 + 5.2) ounces = 72.4 ounces ==> + 2 SD
(62 - 5.2 - 5.2) ounces = 51.6 ounces ==> - 2 SD
The answer is:
51.6 ounces - 72.4 ounces
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).