Question

Faculty members at Splitty Town High School want to determine whether there are enough students to have a Spring Formal. Seventy-two of the 225 students said they would attend the Spring Formal. Construct and interpret a 95% confidence interval for p.A. The 95% confidence interval is (0.2591, 0.3810). There is a 95% chance that a randomly selected student who will attend the Spring Formal lies between 25.91% and 38.10%.B. The 95% confidence interval is (0.2591, 0.3810). Ninety-five percent of all samples of this size will yield a confidence interval of (0.2591, 0.3810).C. The 95% confidence interval is (0.2591, 0.3810). We are 95% confident that the true proportion of students attending the Spring Formal is between 25.91% and 38.10%.D. The 95% confidence interval is (0.6190, 0.7410). Ninety-five percent of all samples of this size will yield a confidence interval of (0.6190, 0.7410).E. The 95% confidence interval is (0.6190, 0.7410). We are 95% confident that the true proportion of students attending the Spring Formal is between 61.90% and 74.10%.

Answer 1

Answer:

C. The 95% confidence interval is (0.2591, 0.3810). We are 95% confident that the true proportion of students attending the Spring Formal is between 25.91% and 38.10%.

Step-by-step explanation:

The interpretation of a confidence interval at a x% confidence level if that we are x% sure that the true proportion(mean) of the population is in this interval.

Confidence interval

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, 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:

Seventy-two of the 225 students said they would attend the Spring Formal. So $n = 225, \pi = \frac{72}{225} = 0.32$

95% confidence interval

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.259$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 + 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.381$

The correct answer is:

C. The 95% confidence interval is (0.2591, 0.3810). We are 95% confident that the true proportion of students attending the Spring Formal is between 25.91% and 38.10%.