Question

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. =

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 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 ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,41 = 2.020

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.020* (10.3 / $\sqrt$ 42)

= 3.210

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

79.1 - 3.210 < $\mu$ < 79.1 + 3.210

75.890 < $\mu$ < 82.310

95% C.I. = ( 75.890 , 82.310)

Answer 2

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 67 specimens and counts the number of seeds in each. Use her sample results (mean =49.8, standard. deviation = 9.7) to find the 95% confidence interval for the number of seeds for the species. Enter your answer as an open- interval.