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Hi.I have an assignment that i would like it to be done.In the first file below you can find the instruction for the homework.In the second and third files (you can find them in the same folder) below you can find the files that you have to use to complete the work via Minitab 19.In the next tow files below you can find a chapters that you might need to done the work.The last file is the book which you have to use to answer some questions. ( you must check the book ASAP to make sure everything is good.)Note: the homework must be done in Minitab 19 then you can copy and paste to MS Word Document.Note: It is an ( IS business statistics ) course.Thanks. block_2_hw_instruction.docx untitled_folder.zip bbs13e_chapter07_ol.ppt bbs13e_chapter08_ol.ppt bbs13e_chapter09_ol.ppt basic_business_statistics__concepts_and_applications_13th_edition_b00hr7mssg_compressed__1_.pdf Unformatted Attachment Preview Homework Problems for Block 1-4 using Minitab version -19 This file includes the guidelines/instructions on the homework as well as Homework Problems Students can use IS 242 Business Statistics or Basic Business Statistics by Berenson, et al 13th ed. The homework file shows page numbers of both books, IS 242 Business Statistics and the original text - Basic Business Statistics by Berenson, Levine, Szabat 13th edition. Where BB means Basic Business Statistics 13th ed by Berenson et al. There are 4 sets of homework, one set for each block. Some problems in the homework require using the Minitab – version 19 program and data files. Minitab program is installed on all the PCs on the SCSU campus. Off-campus, users can access Minitab program through APPS ANYWHERE (virtual lab), All the data files for the Homework are posted on the D2L in Minitab Data for Homework or Excel Data for Homework module. Download the Minitab/Excel data files on your USB (or a PC with Minitab program installed). Log on a Campus PC and open Minitab 19 or access Minitab 19 through AppsAnywhere. After open Minitab-19, follow the steps explained in each problem. Refer “AppsAnywhere & Minitab 19 Using Tips” file posted on D2L The homework should be submitted in an MS Word format file. Use copy and paste to include Minitab outputs and graphs in the Word file to submit. Homework should be an independent work for each student. All the submitted contents are evaluated and graded, clear violations of homework guides, unnecessary or inconsistent contents will be penalized. Submit all questions in each block in one packet with clear identification information such as the name of the student, course number, homework number, problem number, sub-number, etc. For example, 1.4 Do problems 3 and 5 on page 262 (Problems for Section 2 in The Normal Distribution and Other Continuous Distributions Chapter) by showing all steps, 1.4 is Homework number; 3 and 5 are problem numbers, and (a) in 3 is sub-number. Block 2 Homework Problems (100 pts): 2.1 Describe the three properties (characteristics) of the sampling distribution of the sample mean which include properties on the shape, mean and standard error, limit the answer with 50 or fewer words, and number them (20 pts). Refer 7.2 or summary posted on D2L 2.2 The following problems are related to Section 2 in Sampling Distributions chapter and also similar with problem 1 or 2 in page 295 (Problem 7.1 page 261 in BB – Book by Berenson et al) (10 pts). Given a normal distribution of µ = 100 σ = 10 if you select sample of n=25, what is the probability that x̅ is a) Less than 95? b) between 95 and 97.5? c) above 90? d) there is a 65\% chance that x̅ is above what value? Refer that X has a normal distribution of mean = µ and standard deviation = σ, then z has the standard normal distribution, where z = ( ͞χ - µ)/ / n and χ ͞ is mean of X. 2.3 Do problems 1 and 2 in page 316 (Problem 8.1-2 page 278 in BB) (Problems for Section 1 in Confidence Interval Chapter), problem 11 on page 323 (Problem 8.11 page 285 in BB) (Problems for Section 2 in Confidence Interval Chapter) and problems 34 and 35 in page 332 (Problem 8.34-35 page 294 in BB) (Problems for Section 4 in Confidence Interval Chapter) (20 pts). Refer that the confidence interval is given as σ is known, use ͞ χ ± zα/2 / n For a 90\% confidence interval use zα/2 = z 0.05 = 1.645 For a 95\% confidence interval use zα/2 = z 0.25 = 1.96 For a 99\% confidence interval use zα/2 = z 0.005 = 2.58 σ is unknown, use ͞χ ± tα/2 , df s/ n Use tα/2 , df from t-table with df = n-1 (df is degrees of freedom) Sample size n = [ zα/2 σ/e]2 where zα/2 from a confidence level, e=size of error and σ population standard deviation; Specify the level of accuracy in terms of error (e) and confidence level and determine the required sample size. 2.4 Do problem 13 in page 363 (Problem 9.13 page 321 in BB) (Problems for Section 1 in Fundamentals of hypothesis testing Chapter) (10 pts) 2.5 Confidence intervals; use MINITAB for the FORCE data file which is in the Minitab Data for Homework, construct confidence intervals of the population mean force (90\%, 95\%, and 99\%) using t-distribution (10 pts). To get the Minitab output: To open Force file on a PC with Minitab installed: Select.File → Open, select the data file (Force) → click open and OK( if Minitab format file does not open, open Excel format file) Go to top-line menu, and click Stat → Basic statistics → 1t – 1 Sample t In the dialog box, choose One or more Samples, each in a column from the top-right pulldown menu box, and double-click Force in the left column, to copy in the second-right box. Click Options in the dialogue box (1-Sample t – Options), enter 90.0 in the Confidence level Box. Click Ok and Ok. Go to top-line menu and repeat for confidence level 95.0 and 99.0 The session window shows output (results), Select (mark) results, copy and paste on Word file to submit. 2. 6. Hypothesis tests --Use Minitab, answer problem 30, part a) on page 368 (Problem 9.30 page 326 in BB) (applying the concepts in Fundamentals of hypothesis testing Chapter), use both t and p-value for DRINK file use α = 0.05. Submit your own Minitab output with Hypothesis test using t and showing all 6 steps as in pages 364-5 (page 322-324 in BB) (critical value approach section in Fundamentals of hypothesis testing Chapter) (to simplify combine steps 3 and 4 together, then the sixstep approach becomes a five-step approach which is more common) and p-value (5 steps as in page 366 in P-value approach). Test the hypothesis using the t following the 5 steps, and repeat the test hypothesis using p-Value following 5 steps. (30 pts) To get the Minitab output: To open Drink file in a PC with Minitab installed: Select.File → Open, select the data file (Drink) → click open and OK Go to topline menu – click Stat → Basic statistics –> 1t 1 Sample t In the dialog box, choose One or more Samples, each in a column from the top-right pulldown menu box, and double-click Amount in the left column to enter the name of the column Amount to One or more Samples, each in a column box (or simply type in Amount) Check to Perform hypothesis test, type 2 in the Hypothesized mean box Click Ok You will get the following output (T-Value and P-value are deleted intentionally) Descriptive Statistics N Mean StDev SE Mean 50 2.00072 0.04456 90\% CI for μ 0.00630 (1.99015, 2.01129) μ: mean of Amount Test Null hypothesis H₀: μ = 2 Alternative hypothesis H₁: μ ≠ 2 T-Value P-Value ----------- --------- Use the T-value and p-value for the hypothesis test by following 5-step below 5 steps using t-value are: 1. State H0 and H1 2. Determine the level of significance of test: α such as 0.01. 0.05 or 0.1 3. Set-up critical value(s) – this means determining what test statistic to use and rejection area(s) – this step is a combination of steps 3 and 4 in the 6 steps Critical value approach in page 364 of text in Fundamentals of Hypothesis Testing: OneSample Tests Chapter. {For p-value, reject H0 if p-value < α} 4. Calculate the test statistic (use t = 0.11 from above Minitab output) 5. Make inference – conclusion 5 steps using p-Value (as a test statistic) are: 1. State H0 and H1 2. Determine the level of significance of test: α such as 0.01. 0.05 or 0.1 3. Set-up critical value(s): reject H0 if p-value < α 4. Calculate the test statistic (use p-value = 0.910 from above Minitab output) 5. Make inference – conclusion Both methods, using t or p-value for the same set of data., should lead the same conclusion. Chapter 7 Sampling Distributions Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 1 Learning Objectives In this chapter, you learn: ◼ The concept of the sampling distribution ◼ Three properties of sampling distribution of sample mean – slide 18 ◼ To compute probabilities related to the sample mean ◼ The importance of the Central Limit Theorem Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 2 7.1 Sampling Distributions DCOVA ◼ A sampling distribution is a distribution of all of the possible values of a sample statistic such as sample mean for a given sample size selected from a population. ◼ For example, suppose you sample 50 students from your college regarding their mean GPA. If you obtained many different samples of size 50, you will compute a different mean for each sample. We are interested in the distribution of all potential mean GPAs we might calculate for any sample of 50 students. Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 3 Developing a Sampling Distribution DCOVA ◼ Assume there is a population … ◼ Population size N=4 ◼ Random variable, X, is age of individuals ◼ Values of X: 18, 20, 22, 24 (years) Copyright © 2015, 2012, 2009 Pearson Education, Inc. A B C D Chapter 07, Slide 4 7.2 Sampling Distribution of Sample Mean Learn: How to construct sampling distribution ◼ Slides 6-9 3 properties of sampling distribution (property on shape, mean, and variation) Normal populations - summarized in slide 12 Non Normal population – summarized in slide 18, 20 Application examples Slides 16,17, 24 Notations: Refer slide 10 to note the differences between Population and Sampling distribution in mean and standard deviation. Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 5 Developing a Sampling Distribution (continued) Summary Measures for the Population Distribution: X  μ= P(x) i N .3 18 + 20 + 22 + 24 = = 21 4 σ= DCOVA  (X − μ) i N .2 .1 0 2 = 2.236 18 20 22 24 A B C D x Uniform Distribution Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 6 Developing a Sampling Distribution (continued) Now consider all possible samples of size n=2 1st Obs DCOVA 16 Sample Means 2nd Observation 18 20 22 24 18 18,18 18,20 18,22 18,24 20 20,18 20,20 20,22 20,24 1st 2nd Observation Obs 18 20 22 24 22 22,18 22,20 22,22 22,24 18 18 19 20 21 24 24,18 24,20 24,22 24,24 20 19 20 21 22 16 possible samples (sampling with replacement) Copyright © 2015, 2012, 2009 Pearson Education, Inc. 22 20 21 22 23 24 21 22 23 24 Chapter 07, Slide 7 Developing a Sampling Distribution (continued) DCOVA Sampling Distribution of All Sample Means Sample Means Distribution 16 Sample Means 1st 2nd Observation Obs 18 20 22 24 18 18 19 20 21 20 19 20 21 22 22 20 21 22 23 24 21 22 23 24 _ P(X) .3 .2 .1 0 18 19 Copyright © 2015, 2012, 2009 Pearson Education, Inc. 20 21 22 23 24 _ X (no longer uniform) Chapter 07, Slide 8 Developing A Sampling Distribution (continued) DCOVA Summary Measures of this Sampling Distribution: 18 + 19 + 19 +  + 24 μX = = 21 16 σX = (18 - 21) 2 + (19 - 21) 2 +  + (24 - 21) 2 = 1.58 16 Note: Here we divide by 16 because there are 16 different samples of size 2. Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 9 Comparing the Population Distribution to the Sample Means Distribution DCOVA Population N=4 μ = 21 σ = 2.236 Sample Means Distribution n=2 μX = 21 σ X = 1.58 _ P(X) .3 P(X) .3 .2 .2 .1 .1 0 18 20 22 24 A B C D X 0 18 19 Copyright © 2015, 2012, 2009 Pearson Education, Inc. 20 21 22 23 24 _ X Chapter 07, Slide 10 Sample Mean Sampling Distribution: Standard Error of the Mean ◼ ◼ DCOVA Different samples of the same size from the same population will yield different sample means A measure of the variability in the mean from sample to sample is given by the Standard Error of the Mean: (This assumes that sampling is with replacement or sampling is without replacement from an infinite population) σ σX = n ◼ Note that the standard error of the mean decreases as the sample size increases Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 11 Sample Mean Sampling Distribution: If the Population is Normal DCOVA ◼ If a population is normal with mean μ and standard deviation σ, the sampling distribution of X is also normally distributed with μX = μ and σ σX = n Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 12 Z-value for Sampling Distribution of the Mean DCOVA ◼ Z-value for the sampling distribution of X : Z= where: ( X − μX ) σX ( X − μ) = σ n X = sample mean μ = population mean σ = population standard deviation n = sample size Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 13 Sampling Distribution Properties DCOVA μx = μ (i.e. x is unbiased ) Normal Population Distribution μ x μx x Normal Sampling Distribution (has the same mean) Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 14 Sampling Distribution Properties (continued) DCOVA As n increases, σx Larger sample size decreases Smaller sample size μ Copyright © 2015, 2012, 2009 Pearson Education, Inc. x Chapter 07, Slide 15 Determining An Interval Including A Fixed Proportion of the Sample Means DCOVA Find a symmetrically distributed interval around µ that will include 95\% of the sample means when µ = 368, σ = 15, and n = 25. ◼ ◼ ◼ Since the interval contains 95\% of the sample means 5\% of the sample means will be outside the interval Since the interval is symmetric 2.5\% will be above the upper limit and 2.5\% will be below the lower limit. From the standardized normal table, the Z score with 2.5\% (0.0250) below it is -1.96 and the Z score with 2.5\% (0.0250) above it is 1.96. Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 16 Determining An Interval Including A Fixed Proportion of the Sample Means (continued) ◼ Calculating the lower limit of the interval DCOVA σ 15 XL = μ+Z = 368 + (−1.96) = 362.12 n 25 ◼ ◼ Calculating the upper limit of the interval σ 15 XU = μ + Z = 368 + (1.96) = 373.88 n 25 95\% of all sample means of sample size 25 are between 362.12 and 373.88 Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 17 Main idea: 3 Properties of Sample Mean Sampling Distribution: If the Population is not Normal DCOVA Property on shape. We can apply the Central Limit Theorem: ◼ ◼ Even if the population is not normal, …sample means from the population will be approximately normal as long as the sample size is large enough. Property on mean. μx = μ Property on variation. σ σx = n Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 18 Central Limit Theorem As the sample size gets large enough… n↑ DCOVA the sampling distribution of the sample mean becomes almost normal regardless of shape of population x Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 19 Sample Mean Sampling Distribution: If the Population is not Normal (continued) DCOVA Population Distribution Sampling distribution properties: Central Tendency μx = μ σ σx = n Variation μ x Sampling Distribution (becomes normal as n increases) Larger sample size Smaller sample size Copyright © 2015, 2012, 2009 Pearson Education, Inc. μx x Chapter 07, Slide 20 How Large is Large Enough? DCOVA ◼ For most distributions, n > 30 will give a sampling distribution that is nearly normal ◼ For fairly symmetric distributions, n > 15 ◼ For a normal population distribution, the sampling distribution of the mean is always normally distributed Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 21 Example DCOVA ◼ ◼ Suppose a population has mean μ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected. What is the probability that the sample mean is between 7.8 and 8.2? Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 22 Example (continued) DCOVA Solution: ◼ ◼ Even if the population is not normally distributed, the central limit theorem can be used (n > 30) … so the sampling distribution of approximately normal x is ◼ … with mean μx = 8 ◼ σ 3 …and standard deviation σ x = n = 36 = 0.5 Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 23 Example (continued) Solution (continued): DCOVA    7.8 - 8 X -μ 8.2 - 8  P(7.8  X  8.2) = P    3 σ 3   36 n 36   = P(-0.4  Z  0.4) = 0.6554 - 0.3446 = 0.3108 Population Distribution ??? ? ?? ? ? ? ? ? μ=8 Sampling Distribution Standard Normal Distribution Sample Standardize ? X 7.8 μX = 8 8.2 Copyright © 2015, 2012, 2009 Pearson Education, Inc. x -0.4 μz = 0 0.4 Z Chapter 07, Slide 24 Chapter Summary In this chapter we discussed ◼ ◼ ◼ Sampling distributions The sampling distribution of the mean ◼ For normal populations ◼ Using the Central Limit Theorem Calculating probabilities using sampling distributions Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 25 Chapter 8 Confidence Interval Estimation Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 1 Learning Objectives ◼ Confidence Intervals for the Population Mean, μ : To construct and interpret confidence interval estimates for the population mean ◼ ◼ ◼ when Population Standard Deviation σ is Known 8.1 when Population Standard Deviation σ is Unknown 8.2 Determining the Required Sample Size To determine the sample size necessary to develop a confidence interval for the population mean 8.4 Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 2 Outlines: Point and Interval Estimates DCOVA ◼ A point estimate is a single number, ◼ a confidence interval provides additional information about the variability of the estimate Lower Confidence Limit Point Estimate Upper Confidence Limit Width of confidence interval Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 3 Point Estimates DCOVA We can estimate a Population Parameter … with a Sample Statistic (a Point Estimate) Mean μ X Proportion π p Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 4 Confidence Intervals DCOVA ◼ How much uncertainty is associated with a point estimate of a population parameter? ◼ An interval estimate provides more information about a population characteristic than does a point estimate ◼ Such interval estimates are called confidence intervals Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 5 Confidence Interval Estimate DCOVA ◼ An interval gives a range of values: ◼ Takes into consideration variation in sample statistics from sample to sample ◼ Based on observations from 1 sample ◼ Gives information about closeness to unknown population parameters ◼ Stated in terms of level of confidence ◼ e.g. 95\% confident, 99\% confident ◼ Can never be 100\% confident Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 6 Confidence Interval Example DCOVA Cereal fill example ◼ Population has µ = 368 and σ = 15. ◼ If you take a sample of size n = 25 you know ◼ ◼ 368 ± 1.96 * 15 / 25 = (362.12, 373.88). 95\% of the intervals formed in this manner will contain µ. When you don’t know µ, you use X to estimate µ ◼ ◼ If X = 362.3 the interval is 362.3 ± 1.96 * 15 / 25 = (356.42, 368.18) Since 356.42 ≤ µ ≤ 368.18 the interval based on this sample makes a correct statement about µ. But what about the intervals from other possible samples of size 25? Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 7 Confidence Interval Example (continued) DCOVA Sample # X Lower Limit 1 362.30 356.42 368.18 Yes 2 369.50 363.62 375.38 Yes 3 360.00 354.12 365.88 No 4 362.12 356.24 368.00 Yes 5 ... Purchase answer to see full attachment