Question

A computer software magazine compares the rates of malware infection forcomputers protected by security software A with the rates of infection for computersprotected by security software B.They found that out of 794 computers with security software A, 24 became infectedwith some type of malware after 1000 hours of internet interaction. For securitysoftware B, 47 out of 522 computers became infected after 1000 hours of internetinteraction.Assuming these to be random samples of infection rates for the two securitysoftware packages, construct a 95% confidence interval for the difference betweenthe proportions of infection for the two types of security software packages

Answer 1

Answer:

$(0.0302-0.090) - 1.96 \sqrt{\frac{0.0302(1-0.0302)}{794} +\frac{0.09(1-0.09)}{522}}=-0.087$  

$(0.0302-0.090) - 1.96 \sqrt{\frac{0.0302(1-0.0302)}{794} +\frac{0.09(1-0.09)}{522}}=-0.0325$  

And the 95% confidence interval would be given (-0.087;-0.0325).  

We are confident at 95% that the difference between the two proportions is between $-0.087 \leq p_A -p_B \leq -0.0325$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion of infected for A  

$\hat p_A =\frac{24}{794}=0.0302$ represent the estimated proportion infected for A

$n_A=794$ is the sample size required for Brand A

$p_B$ represent the real population proportion infected for B  

$\hat p_B =\frac{47}{522}=0.090$ represent the estimated proportion infected for B

$n_B=522$ is the sample size required infected for B

$z$ represent the critical value for the margin of error  

Solution to the problem

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.0302-0.090) - 1.96 \sqrt{\frac{0.0302(1-0.0302)}{794} +\frac{0.09(1-0.09)}{522}}=-0.087$  

$(0.0302-0.090) - 1.96 \sqrt{\frac{0.0302(1-0.0302)}{794} +\frac{0.09(1-0.09)}{522}}=-0.0325$  

And the 95% confidence interval would be given (-0.087;-0.0325).  

We are confident at 95% that the difference between the two proportions is between $-0.087 \leq p_A -p_B \leq -0.0325$