Answer: 48
Step-by-step explanation:
As per given , we have
Confidence level : [tex]1-\alpha=0.95[/tex]
Significance level : [tex]\alpha=0.05[/tex]
Using z-value table , the critical z-value =[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]
Margin of error : E =0.57 gallon
Prior standard deviation : [tex]\sigma=2\text{ gallons}[/tex]
Required minimum sample would be :
[tex]n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2\\\\ =(\dfrac{1.96\times2}{0.57})^2\\\\=47.2957833179\approx48[/tex]
Hence, the required minimum sample size = 48
