A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 60 specimens and counts the number of seeds in each. Use her sample results (mean = 17.1, standard deviation = 16.4) to find the 95% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
Solution :
Given that,
Point estimate = sample mean = = 17.1
sample standard deviation = s = 16.4
sample size = n = 60
Degrees of freedom = df = n - 1 = 60 - 1 = 59
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
t/2,df
= t 0.025, 59 = 2.001
Margin of error = E = t/2,df * (s /n)
= 2.001 * ( 16.4/ 60)
Margin of error = E = 4.2
The 99% confidence interval estimate of the population mean is,
± E
= 17.1 ± 4.2
=( 12.9, 21.3 )
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