Question

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 1

Answer:

$z=-2.32$  

$p_v =P(Z$  

If we compare the p value and using any significance level for example $\alpha=0.05$ we see that $p_v$ 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  

$X_{1}=37$ represent the number of people with characteristic 1

$X_{2}=58$ represent the number of people with characteristic 2

$n_{1}=139$ sample 1 selected

$n_{2}=147$ sample 2 selected

$p_{1}=\frac{37}{139}=0.266$ represent the proportion of people with characteristic 1

$p_{2}=\frac{58}{147}=0.395$ represent the proportion of people with characteristic 2

z would represent the statistic (variable of interest)  

$p_v$ 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:$p_{1} \geq p_{2}$  

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

We need to apply a z test to compare proportions, and the statistic is given by:  

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

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{37+58}{139+147}=0.332$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.266-0.395}{\sqrt{0.332(1-0.332)(\frac{1}{139}+\frac{1}{147})}}=-2.32$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ 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:  

$p_v =P(Z$  

So if we compare the p value and using any significance level for example $\alpha=0.05$ we see that $p_v$ 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.