Answer:
[tex]CI=(15.2,25.1)[/tex]
Step-by-step explanation:
From the Question We are told that
Sample size [tex]n=11[/tex]
Mean [tex]\=x =20.2[/tex]
Standard Deviation [tex]\sigma=5.19[/tex]
Generally the equation for Critical Value is mathematically given by
[tex]Critical\ value = t_{\alpha/2,(n-1)}[/tex]
[tex]Critical\ Value=t_{0.01/2,10}[/tex]
[tex]Critical\ Value=3.1693[/tex]
Generally he 99% confidence interval for the mean yield is
[tex]CI=\bar{X}\pm t_{\alpha/2,(n-1)}S/{\sqrt{n}}[/tex]
[tex]CI=20.2\pm 3.1693*5.19/{\sqrt{11}}[/tex]
[tex]CI=20.2\pm4.9595[/tex]
[tex]CI=(15.2,25.1)[/tex]
