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    I have been confused by two separate questions, using different values for standard errors - i apologise for not knowing how to code, so pˆ is meant to be p-hat.

    The first: In a survey of 400 voters, 215 respond to vote for the incumbent, 185 for the challenger. Let p denote the fraction of all likely voters who preferred the incumbent at the time of the survey and pˆ be the fraction of survey respondents that prefer the incumbent.

    Now for the variance it is given by pˆ(1-pˆ )/n and when calculating the SE(pˆ ) we have to sq.rt. the variance to get 0.0249, and I am fine with this.

    The second question: In a given population 11% of voters are African American. A survey using a random sample of 600 landline telephone numbers finds 8% African Americans. Is there evidence the survey is biased?

    Now when calculating the t-statistic we use the null hypothesis with p=0.11, but it then states that se(pˆ )=pˆ (1-pˆ )/n

    Why do we no longer sq.rt. the value above to find the standard error? I imagine it must be to do with knowing the population variance?
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    (Original post by xander123456789)
    I have been confused by two separate questions, using different values for standard errors - i apologise for not knowing how to code, so pˆ is meant to be p-hat.

    The first: In a survey of 400 voters, 215 respond to vote for the incumbent, 185 for the challenger. Let p denote the fraction of all likely voters who preferred the incumbent at the time of the survey and pˆ be the fraction of survey respondents that prefer the incumbent.

    Now for the variance it is given by pˆ(1-pˆ )/n and when calculating the SE(pˆ ) we have to sq.rt. the variance to get 0.0249, and I am fine with this.

    The second question: In a given population 11% of voters are African American. A survey using a random sample of 600 landline telephone numbers finds 8% African Americans. Is there evidence the survey is biased?

    Now when calculating the t-statistic we use the null hypothesis with p=0.11, but it then states that se(pˆ )=pˆ (1-pˆ )/n

    Why do we no longer sq.rt. the value above to find the standard error? I imagine it must be to do with knowing the population variance?
    It depends on how you are defining standard error.... for the first time you are using the standard deviation, for the hypothesis test you have to test against what you'd expect the distribution to look like right? standard deviation won't tell you much about the bias of a sample no?
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    (Original post by natninja)
    It depends on how you are defining standard error.... for the first time you are using the standard deviation, for the hypothesis test you have to test against what you'd expect the distribution to look like right? standard deviation won't tell you much about the bias of a sample no?
    Thank you, I very much appreciate the help, though I don't quite understand what you are implying. I understand what a hypothesis test, but I'm not too sure what you're implying and why does it mean that you don't have to square root it in the second question?
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    (Original post by xander123456789)
    Thank you, I very much appreciate the help, though I don't quite understand what you are implying. I understand what a hypothesis test, but I'm not too sure what you're implying and why does it mean that you don't have to square root it in the second question?
    Depending on the problem/particular statistical analysis you may choose how to define your standard error, for a symmetric distribution standard deviation gives an ideal way of measuring standard error, however for a non-symmetric distribution the standard error may not be a constant across the distribution. In this case it has given you a formula for the standard error in the sample I think?
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    I'm still not quite sure what you are implying in terms of why this would mean that when we are using a sample we sq.rt. p(1-p)/n and when suddenly we have the value for the population value for p then we no longer have to sq.rt.
 
 
 
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