Answer:
19
Explanation:
The minimum sample size is given by
[tex]p(|x\bar-\mu|<E) \geq 1-\alpha P(\frac {x\bar-\mu}{\frac {\sigma}{\sqrt n}}<-\frac {E}{\frac {\sigma}{\sqrt n}} \leq \frac {\alpha}{2})[/tex]
[tex]-\frac {E}{\frac {\sigma}{\sqrt n}} \leq -z_{\alpha/2}[/tex] hence making n the subject
[tex]n \geq (\frac {z_{\alpha/2}\times \sigma}{E})^{2}[/tex]
Standard deviation, [tex]\sigma=\sqrt variance[/tex] hence [tex]\sigma=\sqrt 800[/tex]
Significance level, [tex]\alpha[/tex]=1-Confidence=1-0.95=0.05
Critical value=[tex]z_{\apha/2}=z_{0.025}[/tex] and from z table the critical value is 1.96
[tex]n \geq (\frac {z_{\alpha/2}\times \sigma}{E})^{2}=(\frac {1.96\times \sqrt 800}{13})^{2}= 18.18509\approx 19[/tex]
The minimum n has to be an integer hence we round it off to the nearest whole number
