Opinion Polls (Cambridge Stats Notes) - Clarification Required
$begingroup$
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
statistics
$endgroup$
|
show 4 more comments
$begingroup$
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
statistics
$endgroup$
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
|
show 4 more comments
$begingroup$
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
statistics
$endgroup$
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
statistics
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
|
show 4 more comments
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
|
show 4 more comments
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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