Suppose a landscaper wants to tell his clients the average length of the blades

Question

Suppose a landscaper wants to tell his clients the average length of the blades of grass in a freshly mowed lawn. He measures the length of 65 randomly selected blades of grass immediately after mowing a lavn. He does not know the population standard deviation of grass lengths in the lawn, but his data are normally distributed with no outliers. The table contains a summary of his data. Sample Sample Sample standard size mean (cm) deviation (cm) 65 3.57585 1.20657 Calculate the margin of error, m, for a 99% t-confidence interval for a population mean. Then, compute the upper and lower limits for the confidence interval. Give each of your answers with three decimal places of precision. m-U- lower limit = [ cm upper limit cm

Explanation / Answer

Degree of freedom, df = n-1 = 65-1 = 64

For 0.01 significance level, tcrit = 2.65

Margin of error, M = tcrit * s/sqrt(n) = 2.65 * 1.20657/sqrt(65) = 0.3966

lower limit = mean - M = 3.57585 - 0.3966 = 3.179

upper limit = mean + M = 3.57585 + 0.3966 = 3.972

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