Answer:
We are 99% confident that about (31.6 to 52.38) % of population would attend the Winter Formal.
Step-by-step explanation:
Given:
- The sample proportion p = 63 out of 150
- The sample size n = 150
Find:
- The required confidence interval and its interpretation.
Solution:
- To develop CI at 99%, we need Z-score value at significance level a = 1 - CI.
a = 1 - 0.99 = 0.01
CI: Z_a/2 = Z_0.005 = 2.58
-The formula for developing the confidence interval for any proportion p is given by:
![CI = p +/- Z_a/2 * \sqrt{\frac{p*(1-p)}{n} }](https://tex.z-dn.net/?f=CI%20%3D%20p%20%2B%2F-%20Z_a%2F2%20%2A%20%5Csqrt%7B%5Cfrac%7Bp%2A%281-p%29%7D%7Bn%7D%20%7D)
- We can now determine the 99% confidence interval for p using above expression.
p = 63 / 150 = 0.42
1-p = 0.58
![CI = 0.42 +/- 2.58 * \sqrt{\frac{0.42*(0.58)}{150} }\\\\CI = 0.42 +/- 0.10397\\\\0.316 < p < 0.5238](https://tex.z-dn.net/?f=CI%20%3D%200.42%20%2B%2F-%202.58%20%2A%20%5Csqrt%7B%5Cfrac%7B0.42%2A%280.58%29%7D%7B150%7D%20%7D%5C%5C%5C%5CCI%20%3D%200.42%20%2B%2F-%200.10397%5C%5C%5C%5C0.316%20%3C%20p%20%3C%200.5238)
- We are 99% confident that about (31.6 to 52.38) % of population would attend the Winter Formal.