A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked, “In the last

Question

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.

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Piper 2 weeks 2021-09-13T10:13:59+00:00 1 Answer 0

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    2021-09-13T10:15:39+00:00

    Answer:

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

    p_v =2*P(Z>1.620)=0.105  

    If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha 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  

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

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

    n_{1}=373 sample 1 selected

    n_{2}=639 sample 2 selected

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

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

    z would represent the statistic (variable of interest)  

    p_v 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: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{45+57}{373+639}=0.101

    Calculate the statistic

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

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

    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 side test the p value would be:  

    p_v =2*P(Z>1.620)=0.105  

    If we compare the p value and using any significance level for example \alpha=0.05 always p_v>\alpha 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

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45:7+7-4:2-5:5*4+35:2 =? ( )