The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared
ID: 3152944 • Letter: T
Question
The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "During the average workday, how many hours do you spend conducting online research?" Let x = the number of hours a legal professional spends on online research during a typical workday. Suppose that x is normally distributed with a known standard deviation of 5. A sample of 250 legal professionals was surveyed, and the sample's mean response was 2.7 hours. Use the sample information to estimate mu, the mean number of hours a legal professional spends on online research during a typical workday. The sampling distribution of x is normal with a mean and a standard deviation. Use the following Distributions tool to develop a 99% confidence interval estimate of the mean number of hours a legal professional spends on online research during a typical workday. You can be 99% confident that the interval estimate to includes the population mean mu.Explanation / Answer
By central limit theorem, it has the same mean,
u(X) = 2.7
Swith a reduced standard deviation,
sigma(X) = sigma/sqrt(n) = 5/sqrt(250) = 0.316227766
Hence,
The sampling distribution of X is normal with a mean [[2.7]] and a standard deviation [[0.316227766]]. [ANSWER]
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Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 2.7
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 5
n = sample size = 250
Thus,
Margin of Error E = 0.814548746
Lower bound = 1.885451254
Upper bound = 3.514548746
Thus, the confidence interval is
( 1.885451254 , 3.514548746 ) [ANSWER, CONFIDENCE INTERVAL]
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