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:
[tex]CI = p +/- Z_a/2 * \sqrt{\frac{p*(1-p)}{n} }[/tex]
- We can now determine the 99% confidence interval for p using above expression.
p = 63 / 150 = 0.42
1-p = 0.58
[tex]CI = 0.42 +/- 2.58 * \sqrt{\frac{0.42*(0.58)}{150} }\\\\CI = 0.42 +/- 0.10397\\\\0.316 < p < 0.5238[/tex]
- We are 99% confident that about (31.6 to 52.38) % of population would attend the Winter Formal.
