Question
In: Statistics and Probability
A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples...
A botanist wishes to estimate the typical number of seeds for a
certain fruit. She samples 38 specimens and counts the number of
seeds in each. Use her sample results (mean = 56.1, standard
deviation = 21.1) to find the 98% 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.
98% C.I. =
Solutions
Expert Solution
Solution :
Given that,
Point estimate = sample mean = 
= 56.1
Population standard deviation = 
= 21.1
Sample size = n = 38
At 98% confidence level
= 1 - 98%
= 1 - 0.98 = 0.02
/2
= 0.01
Z/2
= Z0.01 = 2.326
Margin of error = E = Z/2
* ( 
/
= 2.326 * ( 21.1 /
= 7.962
At 98% confidence interval estimate of the population mean is,
- E < 
<
+ E
56.1 - 7.962 < < 56.1 + 7.962
48.138 <
< 64.062
( 48.138 , 64.062 )
98% C.I. = ( 48.138 , 64.062 )
orchestra answered 2 years agoRelated Solutions
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