Question

In a 2-sample z-test for two proportions, you find the following: X1 = 24 n1 = 200 X2 = 17 n2 = 150 You decide to run a test for which the alternative hypothesis is H1: p1 > p2. Find the appropriate test statistic for the test

Answer 1

Answer:

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{24+17}{200+150}=0.117$  

Replacing the info given we got:

$z=\frac{0.12-0.113}{\sqrt{0.117(1-0.117)(\frac{1}{200}+\frac{1}{150})}}=0.202$    

We can calculate the p value with this probability:

$p_v =P(Z>0.202)= 0.420$    

Since the p value is a very higher value we don't have enough evidence to conclude that the true proportion for population 1 is higher than the trrue proportion for population 2

Step-by-step explanation:

Information given

$X_{1}=24$ represent the number of people with the characteristic 1  

$X_{2}=17$ represent the number of people with the characteristic 2  $n_{1}=200$ sample 1 selected  

$n_{2}=150$ sample 2 selected  

$p_{1}=\frac{24}{200}=0.12$ represent the proportion estimated for the sample 1  

$p_{2}=\frac{17}{150}=0.113$ represent the proportion estimated for the sample 2  

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)    

$p_v$ represent the value for the test

Hypothesis to test

We want to check if the proportion for population 1 is higher than the proportion for population 2, the system of hypothesis would be:    

Null hypothesis:$p_{1} \leq p_{2}$    

Alternative hypothesis:$p_{1} > p_{2}$    

The statistic would be given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{24+17}{200+150}=0.117$  

Replacing the info given we got:

$z=\frac{0.12-0.113}{\sqrt{0.117(1-0.117)(\frac{1}{200}+\frac{1}{150})}}=0.202$    

We can calculate the p value with this probability:

$p_v =P(Z>0.202)= 0.420$    

Since the p value is a very higher value we don't have enough evidence to conclude that the true proportion for population 1 is higher than the trrue proportion for population 2