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

1,62511027




1,62511027








  • 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










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