Question

An App developing company recently developed a new game that can be played on the iPhones. The company wants to run a survey on the satisfaction rate on the game. They randomly send out a survey to 600 iPhone users who have downloaded the app and 498 of them are happy with the game.

What is a 95% Confidence interval for the true proportion that are satisfied with the app companies game?

(.79, .87)

(.80, .86)

(.76, .89)

(.83, .88)

Answer 1

Solution :

Given that,

n = 600

x = 498

Point estimate = sample proportion = = x / n = 498/600=0.83

1 -   = 1- 0.83 =.17

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z / 2    * ((( * (1 - )) / n)

= 1.96 (((0.83*0.17) /600 )

E= 0.03006

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.83-0.03006 < p <0.83+ 0.03006

0.80< p < 0.86

The 95% confidence interval for the population proportion p is : 0.80, 0.86