Question

Suppose germination periods, in days, for grass seed are normally distributed and have a known population standard deviation of 2 days and an unknown population mean. A random sample of 22 types of grass seed is taken and gives a sample mean of 46 days. Find the error bound (EBM) of the confidence interval with a 90% confidence level.

Answer 1

Answer with explanation:

Let
`\mu` be the population mean .

Given : Sample size : n=22

Population standard deviation:
`\sigma=2`

Sample mean :
`\overline{x}=46`

z-value for 91% confidence level :
`z_c=1.645`

The formula to find the error bound (EBM) :

`E=z_c\cdot(\sigma)/(√(n))`

Then , the error bound (EBM) of the confidence interval with a 90% confidence level will be :-

`E=(1.645)\cdot(2)/(√(22))=0.70143035681\approx0.7014`

Thus , the error bound (EBM) of the confidence interval with a 90% confidence level: E=0.7014

Furthermore , the 90% confidence interval will be :-

`(\overline{x}-E ,\ \overline{x}+E)\\\\=(46-0.7014,\ 46+0.7014)\\\\=(45.2986,\ 46.7014)`