Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 54 specimens and counts the number of seeds in each. Use her sample results (mean = 81.1, standard deviation = 8.4) to find the 90% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

Answer 1

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

$(x^{-} -t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} -t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} })$

Step(ii):-

Degrees of freedom

ν = n-1 = 54-1 =53

$t_{\frac{0.10}{2} } = t_{0.05} = 2.0057$

$(81.1 -2.0057\frac{8.4}{\sqrt{54} } , 81.1 +2.0057\frac{8.4}{\sqrt{54} })$

(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)