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
+1038 ....
\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}
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+118703 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=√p∗qnp=6270,q=870,andn=70SE=√p∗qnSE=√6270∗87070SE=0.0380$4decplaces)
√((6270)×(870))70=0.0380271500370681
6270=3135=0.8857142857142857
0.8857 is the mean of this distribution.
A 95% confidence interval is within 2 standard errors. Mmm
So the 95% conficence limits will be
0.8857±2×0.0380=0.8857±0.076$Sothelowerlimitis$0.8097$andtheupperlimitis$0.9617
Now I am guessing EVEN more. :)
NO I am not going to bother because I really have no idea what I am doing :((
Maybe Bertie can help :/
+1038 ....
\Displaystyle \Text$ {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}
