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.

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05[/tex] always [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 significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation

[tex]X_{1}=30[/tex] represent the number of people with a characteristic in 1

[tex]X_{2}=10[/tex] represent the number of people with a characteristic in 2

[tex]n_{1}=50[/tex] sample of 1 selected

[tex]n_{2}=50[/tex] sample of 2 selected

[tex]p_{1}=\frac{30}{50}=0.6[/tex] represent the proportion of people with a characteristic in 1

[tex]p_{2}=\frac{10}{50}=0.2[/tex] represent the proportion of people with a characteristic in 2

z 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 different from proportion 2 , the system of hypothesis would be:

Null hypothesis:[tex]p_{1} = p_{2}[/tex]

Alternative hypothesis:[tex]p_{1} \neq 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{30+10}{50+50}=0.4[/tex]

3) Calculate the statistic

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

[tex]z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082[/tex]

4) Statistical decision

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

Since is a two sided test the p value would be:

[tex]p_v =2*P(Z>4.082)=4.46x10^{-5}[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05[/tex] always [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 significantly different from proportion 2.