Question

estimate the number of first-year students at an assembly meeting. He surveys 60 students and finds that 24 of them are first-year students. Identify the values needed to calculate a confidence interval at the 95% confidence level. Then find the confidence interval.

Answer 1

Answer:

The values needed to calculate a confidence interval at the 95% confidence level are $z = 1.96, n = 60, \pi = \frac{24}{60} = 0.4$

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).

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:

$n = 60, \pi = \frac{24}{60} = 0.4$

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 values needed to calculate a confidence interval at the 95% confidence level are $z = 1.96, n = 60, \pi = \frac{24}{60} = 0.4$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.4*0.6}{60}} = 0.276$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.96\sqrt{\frac{0.4*0.6}{60}} = 0.524$

The 95% confidence interval for the proportion of first-year students at an assembly meeting is (0.276, 0.524).