Question

A random sample of n 1 equals 135n1=135 individuals results in x 1 equals 40x1=40 successes. An independent sample of n 2 equals 150n2=150 individuals results in x 2 equals 60x2=60 successes. Does this represent sufficient evidence to conclude that p 1 less than p 2p1

Answer 1

Answer:

$p_v =P(Z$  

If we compare the p value with any significance level for example $\alpha=0.05,0.1$ 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 true proportion 1 is less than the true proportion 2, at 5% or 10% of significance .  

Step-by-step explanation:

1) Data given and notation  

$X_{1}=40$ represent the number of successes for 1

$X_{2}=60$ represent the number of successes for 2

$n_{1}=135$ sample of 1 selected

$n_{2}=150$ sample of 2 selected

$\hat p_{1}=\frac{40}{135}=0.296$ represent the sample proportion for 1

$\hat p_{2}=\frac{60}{150}=0.40$ represent the sample proportion 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:  

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

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{40+60}{135+150}=0.3509$

3) Calculate the statistic

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

$z=\frac{0.296-0.40}{\sqrt{0.3509(1-0.3509)(\frac{1}{135}+\frac{1}{150})}}=-1.837$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a one left tailed test the p value would be:  

$p_v =P(Z$  

If we compare the p value with any significance level for example $\alpha=0.05,0.1$ 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 true proportion 1 is less than the true proportion 2, at 5% or 10% of significance .