Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 41 specimens and counts the number of seeds in each. Use her sample results (mean = 54.7, standard deviation = 5.3) to find the 90% 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).

Answer 1

(53.3; 56.1)

Explanation:

Given that:

Sample size, n = 41

Mean, xbar = 54.7

Standard deviation, s = 5.3

Confidence level, Zcritical at 90% = 1.645

Confidence interval :

Xbar ± Margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 1.645 * 5.3/sqrt(41)

Margin of Error = 1.362

Lower boundary = 54.7 - 1.362 = 53.338

Upper boundary = 54.7 + 1.362 = 56.062

(53.3 ; 56.1)