A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked, "In the last 20 years the proportion of the world population living in extreme poverty has...", and three choices were provided: 1.)"increased" 2.) "remained more or less the same" and 3.) "decreased". Of 373 university degree holders, 45 responded with the correct answer: decreased; of 639 non-degree respondents, 57 responded with the correct answer1. We would like to test if the percent of correct answers is significantly different between degree holders and non-degree holders. Let group 1 be the degree holders and let group 2 be the non-degree holders.

If we compare the p 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 FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

Step-by-step explanation:

Data given and notation

[tex]X_{1}=45[/tex] represent the number of correct answers for university degree holders

[tex]X_{2}=57[/tex] represent the number of correct answers for university non-degree holders

[tex]n_{1}=373[/tex] sample 1 selected

[tex]n_{2}=639[/tex] sample 2 selected

[tex]p_{1}=\frac{45}{373}=0.121[/tex] represent the proportion of correct answers for university degree holders

[tex]p_{2}=\frac{57}{639}=0.0892[/tex] represent the proportion of correct answers for university non-degree holders

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportions are different between the two groups, 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{45+57}{373+639}=0.101[/tex]

Calculate the statistic

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

[tex]z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620[/tex]

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 one side test the p value would be:

[tex]p_v =2*P(Z>1.620)=0.105[/tex]

If we compare the p 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 FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance