Melody

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Melody  11 feb 2022
 #1
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Bertie, (or anyone with statistics knowledge) can you help please ?

Pretty please with sprinkles on top :)))

 

                                          

 

I do not know if this is correct.

I don't know what to do with the 0.04 either  

I am copying Bertie's answer from here

 

 

$$\\SE=\sqrt{\dfrac{p*q}{n}}\\\\
p=\frac{62}{70},\qquad q=\frac{8}{70},\qquad and \qquad n=70\\\\
SE=\sqrt{\dfrac{p*q}{n}}\\\\
SE=\sqrt{\dfrac{\frac{62}{70}*\frac{8}{70}}{70}}\\\\
SE=0.0380\qquad $4 dec places)$$

 

$${\sqrt{{\frac{\left(\left({\frac{{\mathtt{62}}}{{\mathtt{70}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{70}}}}\right)\right)}{{\mathtt{70}}}}}} = {\mathtt{0.038\: \!027\: \!150\: \!037\: \!068\: \!1}}$$

 

$${\frac{{\mathtt{62}}}{{\mathtt{70}}}} = {\frac{{\mathtt{31}}}{{\mathtt{35}}}} = {\mathtt{0.885\: \!714\: \!285\: \!714\: \!285\: \!7}}$$

 

$$0.8857$$    is the mean of this distribution.

 

A 95% confidence interval is within 2 standard errors.   Mmm

So the 95% conficence limits will be

$$\\\displaystyle 0.8857 \pm 2\times 0.0380 = 0.8857\pm 0.076\\\\
$So the lower limit is $0.8097 \\
$and the upper limit is $0.9617\\$$

 

Now I am guessing EVEN more.  :)

NO I  am not going to bother because I really have no idea what I am doing  :((

Maybe Bertie can help :/

18 abr 2015