A computer software magazine compares the rates of malware infection for computers protected by security software A with the rates of infection for computers protected by security software B.
They found that out of 787 computers with security software A, 72 became infected with some type of Malware after 1000 hours of internet interaction. For security software B, 37 out of 834 computers Became infected after 1000 hours of internet interaction.
Assuming these to be random samples of infection rates for the two security software packages, construct a 98% confidence interval for the difference between the proportions of infection for the two types of
Security software packages. Does the confidence interval contradict the claim that the proportion of
Infections is the same for the two types of security software?

Question

Quizplus · Accepted Answer

We can use the formula for the difference between two proportions to construct the confidence interval:

p1 - p2 ± z*sqrt((p1*q1/n1) + (p2*q2/n2))

where:
- p1 and p2 are the sample proportions of infection for the two security software packages
- q1 and q2 are the complements of p1 and p2 respectively (1 - p1 and 1 - p2)
- n1 and n2 are the sample sizes for the two packages
- z is the z-score for the desired confidence level (98% in this case)

Using the given values, we get:

p1 = 72/787 = 0.0915
q1 = 1 - p1 = 0.9085
n1 = 787
p2 = 37/834 = 0.0444
q2 = 1 - p2 = 0.9556
n2 = 834
z = 2.33 (since we want a 98% confidence interval)

Plugging in these values into the formula, we get:

0.0915 - 0.0444 ± 2.33*sqrt((0.0915*0.9085/787) + (0.0444*0.9556/834))
= 0.0471 ± 0.0271

Therefore, the 98% confidence interval for the difference between the proportions of infection is (0.0200, 0.0742). Since this interval does not include 0 (which would indicate no difference between the two packages), we can conclude that the proportions are different with 98% confidence.

A Computer Software Magazine Compares the Rates of Malware Infection