Respuesta :
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 [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex]
For this problem, we have that:
Seventy-two of the 225 students said they would attend the Spring Formal. So [tex]n = 225, \pi = \frac{72}{225} = 0.32[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.259[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 + 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.381[/tex]
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%.
