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In a sample of 70 stores, 62 violated a scanner accuracy standard. It has been demonstrated that the conditions for a valid large-sample confidence interval for the true proportion of the stores that violate the standard were not met. Determine the number of stores that must be sampled in order to estimate the true proportion to within .04 with 95% confidence using the large-sample method

Guest Apr 17, 2015

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 #2
avatar+1035 
+5

....

 

$$\Displaystyle \Text$ {This question is asking what size sample is required to estimate the mean \mathrm{\mu} \Text$ {to within a given margin of error (E) for a \mathrm(Z_{1 -\alpha}) \Text$ {confidence interval.}\\
\Displaystyle \Text{Specifically, what sample size (N) is required for a confidence interval of 95\% with an error of 4\% or less.} \\\\
\\ \Displaystyle E= Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{N}} \\\\
\Text$ {Solve for (N) to find minimum. Sample size} \\\
\\ N= \left( {\frac {Z_\alpha /_2}{E} \right)^{2} \hat{p}\left(1-\hat{p}\right) \\
\noindent N= \left( {\frac {1.96}{0.04} \right)^{2} \ * \ 0.886\left(0.114 \right) \; \ = \; 242.51 \Leftarrow \Text$ {Round up to next integer}\\
\Text {Minimum sample size 243}$$

Nauseated  Apr 19, 2015
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 #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 :/

Melody  Apr 18, 2015
 #2
avatar+1035 
+5
Best Answer

....

 

$$\Displaystyle \Text$ {This question is asking what size sample is required to estimate the mean \mathrm{\mu} \Text$ {to within a given margin of error (E) for a \mathrm(Z_{1 -\alpha}) \Text$ {confidence interval.}\\
\Displaystyle \Text{Specifically, what sample size (N) is required for a confidence interval of 95\% with an error of 4\% or less.} \\\\
\\ \Displaystyle E= Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{N}} \\\\
\Text$ {Solve for (N) to find minimum. Sample size} \\\
\\ N= \left( {\frac {Z_\alpha /_2}{E} \right)^{2} \hat{p}\left(1-\hat{p}\right) \\
\noindent N= \left( {\frac {1.96}{0.04} \right)^{2} \ * \ 0.886\left(0.114 \right) \; \ = \; 242.51 \Leftarrow \Text$ {Round up to next integer}\\
\Text {Minimum sample size 243}$$

Nauseated  Apr 19, 2015
 #3
avatar+91024 
0

Thanks Nauseated.     

Did you enjoy the icecream?

Melody  Apr 19, 2015
 #4
avatar+1035 
0

I didn’t have any ice cream, because I was stuffed with your birthday celebration coconut cake. That was delicious!

Nauseated  Apr 21, 2015
 #5
avatar+91024 
0

Hi Nauseated,

Yes it was!!  Want another peice?

Bertie is on now, maybe you would like the icecream Bertie ? 

It is of a no melt variety.  Its makers are very ingenious :)

Melody  Apr 21, 2015

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