Answer:
The right answer is "[tex]0.2<\mu d<7.2[/tex]".
Step-by-step explanation:
The given values are:
[tex]\bar{x_0} = 3.7[/tex]
[tex]s_o=4.95[/tex]
[tex]t_{\frac{0.05}{2} }=2.2621[/tex]
As we know,
95% confidence for [tex]\mu_0[/tex] will be:
= [tex]\bar{x_0} \pm t_{\frac{0.05}{2} },n-1\times \frac{s_o}{\sqrt{n} }[/tex]
The lower bound will be:
= [tex]3.7-2.2621\times \frac{4.95}{\sqrt{10} }[/tex]
= [tex]0.16\simeq 0.2[/tex]
The upper bound will be:
= [tex]3.7+2.2621\times \frac{4.95}{\sqrt{10} }[/tex]
= [tex]7.23\simeq7.2[/tex]
Thus the right answer is "[tex]0.2<\mu d<7.2[/tex]"