Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 72 specimens and counts the number of seeds in each. Use her sample results (mean = 56.1, standard deviation = 16.7) to find the 99% 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

99% Confidence interval: (50.9,61.3)

Explanation:

We are given the following in the question:

Sample mean,
`\bar{x}` = 56.1

Sample size, n = 72

Alpha, α = 0.01

Sample standard deviation, σ = 16.7

Degree of freedom = n - 1 = 71

99% Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))`

Putting the values, we get,

`t_(critical)\text{ at degree of freedom 71 and}~\alpha_(0.01) = \pm 2.64`

`56.1 \pm 2.64((16.7)/(√(72)) ) \\\\= 56.1 \pm 5.19\\\\ = (50.91 ,61.29)\approx (50.9,61.3)`