Question

2. The New York Times and CBS News conducted a nationwide poll of 1048 randomly selected 13- to 17-year-olds. We can consider the sample to be a SRS. (a) Of these 1048 teenagers, 692 had a television in their room. Give a 95% confidence interval for the proportion of all people in this age group who had a TV in their room at the time of the poll

Answer 1

Answer:

The 95% confidence interval for the proportion of all people in this age group who had a TV in their room at the time of the poll is (0.6313, 0.6887).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

Of these 1048 teenagers, 692 had a television in their room. This means that $n = 1048, \pi = \frac{692}{1048} = 0.66$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.66 - 1.96\sqrt{\frac{0.66*0.34}{1048}} = 0.6313$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.66 + 1.96\sqrt{\frac{0.66*0.34}{1048}} = 0.6887$

The 95% confidence interval for the proportion of all people in this age group who had a TV in their room at the time of the poll is (0.6313, 0.6887).