As part of a project to develop better lawn fertilizers, a research chemist want

Question

As part of a project to develop better lawn fertilizers, a research chemist wanted to determine the mean weekly growth rate of Kentucky bluegrass, a common type of grass. A random sample of 30 blades of grass was measured, and the amount of growth in one week was recorded. The sample mean weekly growth was found to be 0.53 inches and the sample standard deviation was found to be 0.15 inches. Assuming that the weekly growth is normally distributed, estimate with 99% confidence the mean weekly growth of Kentucky bluegrass.

Complete the following sentence to provide an interpretation of the confidence interval in the context of the problem:

We are 99% confident that the  (Click to select)samplepopulation   (Click to select)standard deviationmean  weekly growth of  (Click to select)30all blades of Kentucky bluegrass is between    and  inches.

Explanation / Answer

Mean is 0.53 and s is 0.15

for sample size of 30, the standard error of mean is SE=s/sqrt(N)=0.15/sqrt(30)=0.02739

z for 99% is 2.576

thus lower bound is mean-z*SE=0.53-2.576*0.02739=0.45944

upper bound is mean+z*SE=0.53+2.576*0.02739=0.6006

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