Opinion Polls (Cambridge Stats Notes) - Clarification Required












0












$begingroup$


Looking at these again.



Page 18 describes opinion polls. I understand everything up to where they say this:



$$mathbb{P} left( hat{p} - 0.03 leq p leq hat{p} + 0.03 right) = mathbb{P} left( -frac{0.03}{sqrt{p(1-p)/n}} leq frac{hat{p}-p}{sqrt{p(1-p)/n}} leq frac{0.03}{sqrt{p(1-p)/n}} right)$$



I don't understand how they jump from the lhs to the rhs. Any tips, please?



Also how do they then go to the next line? (the approximate one). Thank you.



EDIT:



I was about to write that I get the following, to obtain $p$ in the middle:



$$mathbb{P} left( hat{p} - eta sqrt{p(p-1)/n} leq p leq hat{p} - xi sqrt{p(1-p)n} right)$$



is this actually useful, hmm










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:43












  • $begingroup$
    ahhhhhhhhh. I see thanks...
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:47










  • $begingroup$
    very neat. I like it.
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:48










  • $begingroup$
    You are welcome.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:49












  • $begingroup$
    um the numerator is then $p-hat{p}$
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:53
















0












$begingroup$


Looking at these again.



Page 18 describes opinion polls. I understand everything up to where they say this:



$$mathbb{P} left( hat{p} - 0.03 leq p leq hat{p} + 0.03 right) = mathbb{P} left( -frac{0.03}{sqrt{p(1-p)/n}} leq frac{hat{p}-p}{sqrt{p(1-p)/n}} leq frac{0.03}{sqrt{p(1-p)/n}} right)$$



I don't understand how they jump from the lhs to the rhs. Any tips, please?



Also how do they then go to the next line? (the approximate one). Thank you.



EDIT:



I was about to write that I get the following, to obtain $p$ in the middle:



$$mathbb{P} left( hat{p} - eta sqrt{p(p-1)/n} leq p leq hat{p} - xi sqrt{p(1-p)n} right)$$



is this actually useful, hmm










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:43












  • $begingroup$
    ahhhhhhhhh. I see thanks...
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:47










  • $begingroup$
    very neat. I like it.
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:48










  • $begingroup$
    You are welcome.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:49












  • $begingroup$
    um the numerator is then $p-hat{p}$
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:53














0












0








0





$begingroup$


Looking at these again.



Page 18 describes opinion polls. I understand everything up to where they say this:



$$mathbb{P} left( hat{p} - 0.03 leq p leq hat{p} + 0.03 right) = mathbb{P} left( -frac{0.03}{sqrt{p(1-p)/n}} leq frac{hat{p}-p}{sqrt{p(1-p)/n}} leq frac{0.03}{sqrt{p(1-p)/n}} right)$$



I don't understand how they jump from the lhs to the rhs. Any tips, please?



Also how do they then go to the next line? (the approximate one). Thank you.



EDIT:



I was about to write that I get the following, to obtain $p$ in the middle:



$$mathbb{P} left( hat{p} - eta sqrt{p(p-1)/n} leq p leq hat{p} - xi sqrt{p(1-p)n} right)$$



is this actually useful, hmm










share|cite|improve this question











$endgroup$




Looking at these again.



Page 18 describes opinion polls. I understand everything up to where they say this:



$$mathbb{P} left( hat{p} - 0.03 leq p leq hat{p} + 0.03 right) = mathbb{P} left( -frac{0.03}{sqrt{p(1-p)/n}} leq frac{hat{p}-p}{sqrt{p(1-p)/n}} leq frac{0.03}{sqrt{p(1-p)/n}} right)$$



I don't understand how they jump from the lhs to the rhs. Any tips, please?



Also how do they then go to the next line? (the approximate one). Thank you.



EDIT:



I was about to write that I get the following, to obtain $p$ in the middle:



$$mathbb{P} left( hat{p} - eta sqrt{p(p-1)/n} leq p leq hat{p} - xi sqrt{p(1-p)n} right)$$



is this actually useful, hmm







statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 17:47







i squared - Keep it Real

















asked Dec 20 '18 at 17:32









i squared - Keep it Reali squared - Keep it Real

1,5871927




1,5871927








  • 3




    $begingroup$
    Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:43












  • $begingroup$
    ahhhhhhhhh. I see thanks...
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:47










  • $begingroup$
    very neat. I like it.
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:48










  • $begingroup$
    You are welcome.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:49












  • $begingroup$
    um the numerator is then $p-hat{p}$
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:53














  • 3




    $begingroup$
    Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:43












  • $begingroup$
    ahhhhhhhhh. I see thanks...
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:47










  • $begingroup$
    very neat. I like it.
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:48










  • $begingroup$
    You are welcome.
    $endgroup$
    – APC89
    Dec 20 '18 at 17:49












  • $begingroup$
    um the numerator is then $p-hat{p}$
    $endgroup$
    – i squared - Keep it Real
    Dec 20 '18 at 17:53








3




3




$begingroup$
Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
$endgroup$
– APC89
Dec 20 '18 at 17:43






$begingroup$
Subtract $hat{p}$ from both sides and divide by $sqrt{p(1-p)/n}$.
$endgroup$
– APC89
Dec 20 '18 at 17:43














$begingroup$
ahhhhhhhhh. I see thanks...
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:47




$begingroup$
ahhhhhhhhh. I see thanks...
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:47












$begingroup$
very neat. I like it.
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:48




$begingroup$
very neat. I like it.
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:48












$begingroup$
You are welcome.
$endgroup$
– APC89
Dec 20 '18 at 17:49






$begingroup$
You are welcome.
$endgroup$
– APC89
Dec 20 '18 at 17:49














$begingroup$
um the numerator is then $p-hat{p}$
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:53




$begingroup$
um the numerator is then $p-hat{p}$
$endgroup$
– i squared - Keep it Real
Dec 20 '18 at 17:53










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047786%2fopinion-polls-cambridge-stats-notes-clarification-required%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047786%2fopinion-polls-cambridge-stats-notes-clarification-required%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Wiesbaden

Marschland

Dieringhausen