Answer:Yes
Step-by-step explanation:
Given
[tex]mean \mu =2.1 lb[/tex]
[tex]standard deviation \sigma =4.8lb[/tex]
n=40
[tex]Standard error=\frac{\sigma }{\sqrt{n}}[/tex]
[tex]S.E.=\frac{4.8}{\sqrt{40}}=0.76[/tex]
[tex]\alpha =0.1[/tex]
[tex]Z_{\frac{\alpha }{2}}=1.645[/tex]
Margin of error=[tex]\pm Z_{\frac{\alpha }{2}}\cdot S.E.[/tex]
Margin of error=[tex]\pm 1.645\times 0.76[/tex]
Margin of error=[tex]\pm 1.2502[/tex]
Confidence interval [tex]\mu \pm Z_{\frac{\alpha }{2}}\cdot S.E.[/tex]
[tex]\left ( 0.8498,3.3502\right )[/tex]
it is effective since it is greater than zero