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# MGT 218 : Chapter 7 and chapter 8 and solutions for practice exercises MGT 218

Marketplace > Marshall University > Business > MGT 218 > MGT 218 Chapter 7 and chapter 8 and solutions for practice exercises
Winn
Marshall
GPA 3.72

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sampling distribution , estimation, point estimate ,and interval estimate
COURSE
Introductory Statistics
PROF.
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Statistics
KARMA
25 ?

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This 6 page Class Notes was uploaded by Winn on Tuesday March 8, 2016. The Class Notes belongs to MGT 218 at Marshall University taught by in Spring 2016. Since its upload, it has received 10 views. For similar materials see Introductory Statistics in Business at Marshall University.

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Date Created: 03/08/16
MGT 218 : Statistics questions for chapter 7 : Sample Distributions Chapter 8 : Estimation , Point Estimate and Internal Estimate Explaining Important Most Important 1) Theory : a) Estimation : The assignment of value(s) to a population parameter based on a value of the corresponding sample statistics is called estimation. b) Estimate and Estimator : The value(s) assigned to a population parameter based on the value of a sample statistics is called an estimate. The sample statistic used to estimate a population parameter is called an estimator. c) Point estimate : the value of a sample statistic that is used to estimate a population parameter is called a point estimate. d) Interval estimation : in interval estimation, an interval is constructed around the point estimate, and it is stated that this interval is likely to contain the corresponding population parameter. e) Confidence level and confidence interval :  Each interval is constructed with regard to a given confidence level and is called a confidence interval. The confidence interval is given as  Point estimate ( + - ) Margin of error  The confidence level associated with a confidence interval states how much confidence we have that this interval contains the true population parameter. The confidence level is denoted by ( 1 –a)100% f) Estimation of a population mean : standard deviation known  Confidence Interval for μ : the (1-a)100% confidence interval for μ under Cases I and II is The value of z used here is obtained from the standard normal distribution table ( Table IV of Appendix C) for the given confidence level.  Margin of Error : for the estimate for μ, denoted by E, is the quantity that is subtracted from and added to the value of x bar to obtain a confidence interval for μ , thus With Case 1 . If the following three conditions are fulfilled : 1- The population standard deviation σ is known. 2- The sample size is small ( n < 30 ) 3- The population from which the sample is selected is normally distributed. Case 2. If the following two conditions are fulfilled : 1- The population standard deviation σ is known 2- The sample size is large ( n >=30 ) n/N <= 0.05 Case 3. If the following three conditions are fulfilled 1. The population standard deviation σ is known. 2. The sample size is small ( n < 30 ) 3. The population from which the sample is selected is not normally distributed ( or its distribution is unknown ). 2) Exercises from the textbook : 7.4 ) a ) find μ = 15 + 13 +8+17+9+12/6 = 12.33 b) xx̄ = 13+8+9+12 / 4 = 10.2 Sampling error = 10.2 – 12.33 = -2.13 c) xx̄ = 13+8+6+12 / 4= 9.75 Nonsampling error = 9.75 – 10.2 = -0.45 d) Numbers : 17,9,12,8 xx̄ = (17+9+12+8)/4 = 11.5 Sampling error = 11.5 – 12.33 = -.83 7.8 ) The following data give the years of teaching experience for all ive faculty members of a department at a university. 7 8 14 7 20 a) Let x denote the years of teaching experience for a faculty member of this department. Write the population distribution of x b) List all the possible samples of size three ( without replacement ) that can be selected from this population. Calculate the mean for each of these samples. Write the sampling distribution of xx̄ c) Calculate the mean for the population data. Select one random sample of size three and calculate the sample mean xx̄. Compute the sampling error. a) Population probability distribution : X P(x) 7 .40 8 .20 14 .20 20 .20 = 1.00 b) Sample xx̄ F P(x) Sampling error 7,8,14 9.67 2 .2 -1.53 7,8,7 7.33 1 .1 -3.87 7,8,20 11.67 2 .2 0.47 8,14,20 14 1 .1 2.8 14,7,20 13.67 2 .2 2.27 7,14,7 9.33 1 .1 -1.87 7,7,20 11.33 1 .1 0.13 = 1 With μ = 11.2 7.14 ) Consider a large population with μ= 90 and σ=18. Assuming n/N <= .05, find the mean and standard deviation of sample mean for a sample size of : a) n= 18 => σ =M2.36 μ (x)= 60 b) n = 90 => σ M = 21.21 μ (x)= 60 7.16) A population of N = 100000 has σ = 40. In each of the following cases, which formula will you use to calculate σx for each of these cases. a) n = 2500 => n/N = 0.025 < 0.05 => σx = 0.8 b) n = 7000 => n/N = 0.357 < 0.05 => σx = 0.46 7.18 ) For a population , μ = 46 and σ = 10. a) For a sample selected from this population, μx = 46 and σx =2.0. Find the sample size. Assume n/N <= .05 => n = 25 b) n = 39.0625 7.20 ) The living spaces of all homes in a city have a mean of 2300 square feet and a standard deviation of 500 square feet. Let xx̄ be the mean living space for a random sample of 25 homes selected from this city. Find the mean and standard deviation of the sampling distribution of xx̄. Mean of sample = mean of population σx = 100 7.24) The standard deviation of the 2011 gross sales of all corporations is known to be \$139.50 million. Let xx̄ be the mean of the 2011 gross sales of a sample of corporations. What sample size will produce the σx to \$15.50 million? Assume n/N <=0.05 => n^2 = 139.50^2 / 15.50^2 = 81 8.24 ) A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 40 households from the city, which gave the mean water usage to be 3415.70 gallons over a one-month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is 389.60 gallons. Make a 95 % confidence interval for the average monthly residential water usage for all households in this city. 95% confidence interval : => z = 1.96  3415.70 ( +- ) 1.96 x ( 389.60 / 6.32 ) : the range is from 3294.96 to 3536.44 8.26) Lazurus Steel Corporation products iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to .10 inch. The quality control department takes a sample of 20 such rods every week, calculates the mean length of these rods, and makes a 99% confidence interval for the population mean. If either the upper limit of this confidence interval is greater than 36.05 inches or the lower limit of this confidence interval is less than 35.96 inches, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of 36.02 inches. Based on this sample, will you conclude that the machine needs an adjustments? Assume that the lengths of all such rods have a normal distribution. Solution : We have xx̄ = 36.02 and n = 20 with 99% confidence interval => z = 2.58 The range is from 36.02 (+-) 2.58 x (0.1 / 4.472) = 35.96 to 36.08.  The machine needs adjustment. 8.44) Find the value of t from the t distribution table for each of the following : a) Confidence level = 99% and df = 13 => t = 3.012 b) Confidence level = 95% and n=36 => df = 35 => t= 2.030 c) Confidence level = 90% and df = 16 => t= 1.746

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