Question

In the computer game World of Warcraft, some of the strikes are critical strikes, which do more damage. Assume that the probability of a critical strike is the same for every attack, and that attacks are independent. During a particular fight, a character has 251 critical strikes out of 591 attacks. What is the lower bound for the 99% confidence interval for the proportion of strikes that are critical strikes.

Answer 1

0.372

Explanation:

Total number of attacks = n = 591

Number of critical strikes = x = 251

Proportion of critical strikes = p =
`(x)/(n)=(251)/(591)`

Proportion of non-critical strikes = q = 1 - p =
`1-(251)/(591)=(340)/(591)`

Confidence Level = 99%

Z-score for this confidence level = 2.58

The Lower bound for the population proportion is given by:

`p-z\sqrt{(pq)/(n)}`

Using the values, we get:

`(251)/(591)-2.58*\sqrt{((251)/(591)*(340)/(591))/(591)} \\\\ =0.372`

The lower bound for the 99% confidence interval for the proportion of strikes that are critical strikes is 0.372