Question

Let n1equals50​, Upper X 1equals30​, n2equals50​, and Upper X 2equals10. Complete parts​ (a) and​ (b) below. a. At the 0.05 level of​ significance, is there evidence of a significant difference between the two population​ proportions? Determine the null and alternative hypotheses. Choose the correct answer below. A. Upper H 0 : pi 1 equals pi 2 Upper H 1 : pi 1 not equals pi 2 Your answer is correct.B. Upper H 0 : pi 1 greater than or equals pi 2 Upper H 1 : pi 1 less than pi 2 C. Upper H 0 : pi 1 less than or equals pi 2 Upper H 1 : pi 1 greater than pi 2 D. Upper H 0 : pi 1 not equals pi 2 Upper H 1 : pi 1 equals pi 2 Calculate the test​ statistic, Upper Z Subscript STAT​, based on the difference p1minusp2.

Answer 1

Answer:

Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$    

$p_v =2*P(Z>4.082)=4.46x10^{-5}$  

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation  

$X_{1}=30$ represent the number of people with a characteristic in 1

$X_{2}=10$ represent the number of people with a characteristic in 2

$n_{1}=50$ sample of 1 selected  

$n_{2}=50$ sample of 2 selected  

$p_{1}=\frac{30}{50}=0.6$ represent the proportion of people with a characteristic in 1

$p_{2}=\frac{10}{50}=0.2$ represent the proportion of people with a characteristic in 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 different from proportion 2 , the system of hypothesis would be:  

Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{30+10}{50+50}=0.4$  

3) Calculate the statistic  

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

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$    

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 two sided test the p value would be:  

$p_v =2*P(Z>4.082)=4.46x10^{-5}$  

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.