Question

A programming team is in the process of testing a new software module. As part of the effort, they need to estimate the success rate of the module when used with a particular operating system. To do this, they plan to run the module on a randomly selected set of computers, record how many individual runs execute properly, and use that result to calculate the sample success rate (p-hat, the number of successes divided by the total number of tests). Assuming a confidence level of 99%, calculate n, the number of computers they need to use for the test in order to ensure a 0.03 margin of error in the success rate. Calculate n for the following two cases: (1) no assumption is made about the value of the sample success rate, and (2) in a recent test of a similar software module, that module ran successfully in 94% of the tests. Round your answers upward to the next higher integer.

(1) If no assumptions are made about the sample success rate, the sample size required to ensure a margin of error of 0.03 is n = .

(2) If it is assumed that the new module will run successfully roughly in 94% of the tests, the required sample size required to ensure a margin of error of 0.03 is n =

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.03

At 99% confidence level

= 1 - 99%

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 /0.03 )2 * 0.5 *0.5

= 1843.27

sample size = n = 1844

b) = 0.94

1 - = 1 - 0.94 = 0.06

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 /0.03 )2 * 0.94 *0.06

= 415.84

sample size = n = 416