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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. =

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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 * ( /n)

= 2.326 * ( 21.1 /  38 )

= 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 1 year ago
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