Answer:
90% confidence interval for the number of seeds for the species
(78.8073 ,83.3927)
Step-by-step explanation:
Step(i):-
Given sample size 'n' =54
Mean of the sample x⁻ = 81.1
standard deviation of the sample 'S' = 8.4
90% confidence interval for the number of seeds for the species
[tex](x^{-} -t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} -t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} })[/tex]
Step(ii):-
Degrees of freedom
ν = n-1 = 54-1 =53
[tex]t_{\frac{0.10}{2} } = t_{0.05} = 2.0057[/tex]
[tex](81.1 -2.0057\frac{8.4}{\sqrt{54} } , 81.1 +2.0057\frac{8.4}{\sqrt{54} })[/tex]
(81.1- 2.2927 ,81.1 + 2.2927 )
(78.8073 ,83.3927)
Final answer:-
90% confidence interval for the number of seeds for the species
(78.8073 ,83.3927)
