The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "How many hours do you work at your job on a typical workday?" Let X = the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6. A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours. Use the sample information to estimate mu , the mean number of hours a legal professional works on a typical workday. The sampling distribution of X bar is normal with a mean equal to the unknown population mean and a standard deviation of 0.7969. The 95% interval estimate of the population mean mu is LCL = to UCL = .

Therefore, the sampling distribution of is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is  

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

The sample mean is  

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

The sample standard deviation would be  

Where   is the population standard deviation and n is the sample size.

Therefore, the sampling distribution of is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

Where the margin of error is given by

Where n is the sampling size, s is the sample standard deviation and  is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

So the required 95% confidence interval is

The 95% interval estimate of the population mean is

LCL = 7.431 hours to UCL = 10.569 hours

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