Question

A new radar device is being considered for a certain missile defense system. The system is checked by experimenting with aircraft in which a kill or a no kill is simulated. Two research group A and B participated the simulation, and it was found that 63 kills occur out of 100 trials by group A, and 59 kills occur out of 125 trials by group B. Is there a significant difference between the proportions of group A and B's simulation? Use P-value with significance level of 0.05 to draw a conclusion.

Answer 1

Answer:

There is significant difference between the proportions of group A and group B.

Step-by-step explanation:

In this question we need to perform an hypothesis test on the difference of proportions (Δπ) and determine if there is any significant difference between the two proportions.

The null and alternative hypothesis are:

$H_0: \Delta \pi=0\\\\H_1: \Delta \pi\neq0$

The proportion of group A is $p_1=63/100=0.630$.

The proportion of group B is $p_2=59/125=0.472$.

The weighted average of p (to calculate s) is:

$p=\frac{n_1*p_1+n_2*p_2}{n_1+n_2}=\frac{100*0.630+125*0.472}{100+125}=\frac{63+59}{100+125}=\frac{122}{225}=0.542$

The estimated standard deviation is:

$s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2} } =\sqrt{\frac{0.542(1-0.542)}{100}+\frac{0.542(1-0.542)}{125}} =\sqrt{\frac{0.248}{100} +\frac{0.248}{125} } \\\\s=\sqrt{0.00447} =0.067$

Then we can calculate z as

$z=\frac{p_1-p_2}{s}=\frac{0.630-0.472}{0.067}=\frac{0.158}{0.067}=2.36$

The P-value for z=2.36 is 0.009, which is less that the significance level of 0.05. The effect is significant and the null hypothesis is rejected.

There is significant difference between the proportions of group A and group B.