Answer:
0.372
Step-by-step explanation:
Total number of attacks = n = 591
Number of critical strikes = x = 251
Proportion of critical strikes = p = [tex]\frac{x}{n}=\frac{251}{591}[/tex]
Proportion of non-critical strikes = q = 1 - p = [tex]1-\frac{251}{591}=\frac{340}{591}[/tex]
Confidence Level = 99%
Z-score for this confidence level = 2.58
The Lower bound for the population proportion is given by:
[tex]p-z\sqrt{\frac{pq}{n}}[/tex]
Using the values, we get:
[tex]\frac{251}{591}-2.58\times\sqrt{\frac{\frac{251}{591}\times\frac{340}{591}}{591}} \\\\ =0.372[/tex]
The lower bound for the 99% confidence interval for the proportion of strikes that are critical strikes is 0.372
