A random sample of n 1n1equals=139139 individuals results in x 1x1equals=3737 successes. An independent sample of n 2n2equals=147147 individuals results in x 2x2equals=5858 successes. Does this represent sufficient evidence to conclude that p 1 less than p 2p1

Answer:[tex]z=-2.32[/tex]  [tex]p_v =P(Z<-2.32)= 0.010[/tex]  If we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significant lower than the proportion 2 at 5% of significance.  Step-by-step explanation:1) Data given and notation  [tex]X_{1}=37[/tex] represent the number of people with characteristic 1[tex]X_{2}=58[/tex] represent the number of people with characteristic 2[tex]n_{1}=139[/tex] sample 1 selected[tex]n_{2}=147[/tex] sample 2 selected[tex]p_{1}=\frac{37}{139}=0.266[/tex] represent the proportion of people with characteristic 1[tex]p_{2}=\frac{58}{147}=0.395[/tex] represent the proportion of people with characteristic 2z would represent the statistic (variable of interest)  [tex]p_v[/tex] represent the value for the test (variable of interest)2) Concepts and formulas to use  We need to conduct a hypothesis in order to check if the proportion 1 is less than the proportion 2, the system of hypothesis would be:  Null hypothesis:[tex]p_{1} \geq p_{2}[/tex]  Alternative hypothesis:[tex]p_{1} < p_{2}[/tex]  We need to apply a z test to compare proportions, and the statistic is given by:  [tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{37+58}{139+147}=0.332[/tex]3) Calculate the statisticReplacing in formula (1) the values obtained we got this:  [tex]z=\frac{0.266-0.395}{\sqrt{0.332(1-0.332)(\frac{1}{139}+\frac{1}{147})}}=-2.32[/tex]  4) Statistical decisionFor this case we don't have a significance level provided [tex]\alpha[/tex] we can assuem it 0.05, and we can calculate the p value for this test.  Since is a one left tailed test the p value would be:  [tex]p_v =P(Z<-2.32)= 0.010[/tex]  So if we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significant lower than the proportion 2 at 5% of significance.