Question

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 = 67.4, standard deviation = 16.8) 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

Answer:

Confidence Interval:

where,

= mean of the sample , here sample is 60 specimens and counts number of seeds

z = z value for 95% of confidence interval

- standard deviation

n = number of intervals

Here, it is given that it is open interval hence we calculate value on both sides i.e. upper and lower side

therefore,

Confidence Interval:

= 67.4 1.96 x (16.8 / )

= 67.4 4.251

for Lower end C.I.  = 67.4 - 4.251 = 63.149

for Upper end C.I. = 67.4 + 4.251 = 71.651

^{Hence probability for number of seeds for certain fruits at 95% confidence Interval for lower end be 63.149}

and probability for number of seeds for certain fruits at 95% confidence Interval for upper end be 71.651