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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.

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JeanaShupp

Answer with explanation:

Let [tex]\mu[/tex] be the population mean .

Given : Sample size : n=22

Population standard deviation: [tex]\sigma=2[/tex]

Sample mean : [tex]\overline{x}=46[/tex]

z-value for 91% confidence level : [tex]z_c=1.645[/tex]

The formula to find the error bound (EBM) :

[tex]E=z_c\cdot\dfrac{\sigma}{\sqrt{n}}[/tex]

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

[tex]E=(1.645)\cdot\dfrac{2}{\sqrt{22}}=0.70143035681\approx0.7014[/tex]

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

Furthermore , the 90% confidence interval will be :-

[tex](\overline{x}-E ,\ \overline{x}+E)\\\\=(46-0.7014,\ 46+0.7014)\\\\=(45.2986,\ 46.7014)[/tex]

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