Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 48 specimens and counts the number of seeds in each. Use her sample results (mean = 36.9, standard deviation = 16.5) to find the 98% 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: The open interval would be (31.4,42.5).

Step-by-step explanation:

Since we have given that

mean = 36.9

Standard deviation = 16.5

n = 48

At 98% confidence interval, z = 2.33

So, Interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)$

Hence, the open interval would be (31.4,42.5).