A botanist wishes to estimate the typical number of seeds for a
certain fruit. She samples 42 specimens and counts the number of
seeds in each. Use her sample results (mean = 79.1, standard
deviation = 10.3) 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 3 decimal places.
95% C.I. =
Solution :
Given that,
Point estimate = sample mean = = 79.1
sample standard deviation = s = 10.3
sample size = n = 42
Degrees of freedom = df = n - 1 = 41
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,41 = 2.020
Margin of error = E = t/2,df * (s /n)
= 2.020* (10.3 / 42)
= 3.210
The 95% confidence interval estimate of the population mean is,
- E < < + E
79.1 - 3.210 < < 79.1 + 3.210
75.890 < < 82.310
95% C.I. = ( 75.890 , 82.310)
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