Question

WOW: 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 244 critical strikes out of 593 attacks. Part 1 of 3 (a) Construct an 80% confidence interval for the proportion of strikes that are critical strikes. Round the answer to at least three decimal places. An 80% confidence interval for the proportion of strikes that are critical strikes is .385 < p < 1.437). Part: 1/3 Part 2 of 3 (b) Construct a 90% confidence interval for the proportion of strikes that are critical strikes. Round the answer to at least three decimal places. A 90% confidence interval for the proportion of strikes that are critical strikes is

Answer 1

Solution

given : r = 244, n = 593

244 Sample proportion (f)=-= = 0.4115

(a) for 80% confidence level. za = 1.282

download link for excel to calculate za

https://drive.google.com/open?id=1miDykB956rSHGgAYD93YaJKxmD9a_oZ1

formula : Confidence interval for the population proportion, p

$=\hat p \pm z_{\frac{\alpha }{2}}*\left \{\sqrt{ \frac{\hat p*(1-\hat p)}{n}} \right \}$

= 0.4115+1.282 * 0.4115 *(1-0.4115) 593

= 0.41150.0259

Lower Bound = 0.4115 -0.0259 = 0.3856 0.386

Upper Bound = 0.4115 +0.0259 = 0.4374 0.437

...0.386 <p < 0.437

------------------------------------------------------------------------------------------------------------------

(b) for 90% confidence level, ze = 1.645

download link for excel to calculate za

https://drive.google.com/open?id=1miDykB956rSHGgAYD93YaJKxmD9a_oZ1

formula : Confidence interval for the population proportion, p

$=\hat p \pm z_{\frac{\alpha }{2}}*\left \{\sqrt{ \frac{\hat p*(1-\hat p)}{n}} \right \}$

= 0.41151.645米。 0.4115 *(1-0.4115) 593

= 0.41150.0332

Lower Bound = 0.4115 -0.0332 = 0.3782 -0.378

Upper Bound = 0.4115 +0.3782 = 0.44470.445

...0.378 <p < 0.445