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 32 blades of grass was measured, and the amount of growth in one week was recorded. The sample mean weekly growth was found to be 0.52 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.

Explanation / Answer

mean is 0.52 and s is 0.15

for sample size of 32, the standard error SE is s/sqrt(N)=0.15/sqrt(32)=0.0265

z for 99% confidence is 2.576

thus lower bound is mean-z*SE=0.52-2.576*0.0265=0.451736

upper bound is mean+z*SE=0.52+2.576*0.0265=0.588264

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