Answer:
We are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%
Step-by-step explanation:
-From the given information,
.
-We calculate the confidence interval using this value at 95% confidence level:
![CI=\hat p\pm z \sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\\\=0.65\pm 1.96\times \sqrt{\frac{0.65\times 0.35}{12000}}\\\\\\=0.65\pm 0.0085\\\\\\=[0.6415,0.6585]](https://tex.z-dn.net/?f=CI%3D%5Chat%20p%5Cpm%20z%20%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D%5C%5C%5C%5C%5C%5C%3D0.65%5Cpm%201.96%5Ctimes%20%5Csqrt%7B%5Cfrac%7B0.65%5Ctimes%200.35%7D%7B12000%7D%7D%5C%5C%5C%5C%5C%5C%3D0.65%5Cpm%200.0085%5C%5C%5C%5C%5C%5C%3D%5B0.6415%2C0.6585%5D)
So, the 95% confidence interval is (0.6515,0.6585).
Hence, we are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%.
Answer:
c. We are 95% confident that, among the population of students that play basketball recreationally, the proportion who live on campus is between 0.12 and 0.18.
Step-by-step explanation:
According to the question, it is mentioned that there is 95% confidence interval and it is calculated (0.12,0.18)
So the correct interpretation is that they are 95% confident that the students who play basketballs the proportion should lies between 0.12 and 0.12
So, the option C is correct
The rest of the options are wrong
Answer:
Step-by-step explanation: idk
N= 7.01 :D I’m pretty sure that’s the answer :)