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

Answer:

99% Confidence interval:   (50.9,61.3)        

Step-by-step 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:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

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

$56.1 \pm 2.64(\dfrac{16.7}{\sqrt{72}} ) \\\\= 56.1 \pm 5.19\\\\ = (50.91 ,61.29)\approx (50.9,61.3)$