Question

1010000003fb1 km 15250 O radea Assignment Due Sunday 29 18 at 13 Attempts: 1. Estimating a population mean (population standard devlation Iknown) The 2008 workplace Productivity Survey, commissioned by LexisNexis and prepared by wodne Reserch, included the question, "How many work related e-mails do you receive during the typical workday Let x-the number of work-related e-mails a legal professional receives during a typical workday. Suppose that x is normally distributed with a known standard deviation of 22.7. A sample of 250 tegal professionals was surveyed, and the sample information to estimate p, the mean number of work-related e-mails a legal typical workday sample's mean response was 36.7 e-mails. Use the and a standard The sampling distribution of & is normal with a mean use the following Distributions tool to develope 90% conhdence interval for the mean number of e-mails a legal professional receives during a typical warkday Distributions ncu des th4 pop.iston maon μ.

Please answer all parts

Answer 1

1. The sampling distribution of sample mean $\bar{x}$ is normal with mean 36.7 and a standard deviation

  $\sigma/\sqrt{n}=22.7/\sqrt{250}=1.4356$

2. Here we use Normal distribution to develop 90% confidance interval for the mean number of work-related emails during a typical workday.

The critical z-value for ?=0.1 is zc?=1.645.

The 90% confidence for the population mean ? is computed using the following expression

$CI = \left(\bar X - \frac{z_c \times \sigma}{\sqrt{n}} , \bar X + \frac{z_c \times \sigma}{\sqrt{n}} \right)$

Therefore, based on the information provided, the 90 % confidence for the population mean ? is

  $CI = \left(36.7 - \frac{ 1.645 \times 22.7}{\sqrt{ 250}} , 36.7 + \frac{ 1.645 \times 22.7}{\sqrt{ 250}} \right)$

  =(36.7?2.361,36.7+2.361)

CI= (34.339, 39.061).

You can 90% confidant that the interval 34.339 to 39.061 indicates the population mean ?.