the logic that allows you to be 95% confident that the confidence interval contains the population parameter? We know from the CLT that sample means are normally distributed around the real population mean (). Any time you have a sample mean within E (margin of error) of then the confidence interval will contain . Since 95% of the sample means are within E of then 95% of the confidence interval constructed in this way will contain. • Why do we use confidence intervals verses point estimates
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JET Copies Case Problem Assignment 1 Professor Dr. Elena Klimova MAT 540 – Quantitative Methods Janeiro 28‚ 2013 5. Model number of days to repair In regard to the first part of the Case Problem (The average number of days needed to repair the copier)‚ I worked on the Excel to find the number of the days required to repair the copier (Repair Time (days). In Excel‚ I wrote down the table information given from the case study to make it easier to find it and copy‚ if necessary. I used
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Unit 6 Assignment 1 Case Analysis and Report MBA 6018; Data Analysis for Business Decisions Dr. Hannon Lynette Capella University December 26th‚ 2012 Table of contents introduction: Destination marketing ………………………………………………. Pg. 3 Honeymoon Destinations Market ……………………………………………………. Pg 3 Data Analysis ……………………………………………………………………… PGS 3-4 First case scenario ……………………………………………………………….. pgs 4-7 recommendation…………………………………………………………….... pgs 7-9 second case scenario………………………………………………………………
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Chapter 2. Instead of estimating the two forms of average values in the population‚ they would be measuring directly. Of course‚ when measuring everyone in a population‚ the true value is known; thus there is no need for confidence intervals. After all the purpose of the confidence interval is to tell how certain the author is that a presented interval brackets the true value in the population. With everyone measured‚ the true value would be known‚ unless of course there were measurement or calculation
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the airline has to be profitable and efficient at the same time. The airline can chose to satisfy 95% of the people. The company will still be able to satisfy a large portion of the population and can be profitable. We find the minimum/ maximum sitting distance for men/ women that will satisfy 95% of the people: Because of the normal distribution the mean= 0‚ standard deviation=1. 95 percent means z= +/- 1.96 Max sitting distance for men = 23.5 +(1.96*1.1)= 25.656 inches Min sitting
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Pearson’s Correlation 4 Spearman’s Rho 4 Probability 4 Binomial Distribution 4 Assumptions: 5 Subjective Probability 5 Normal Distribution 5 Standard Normal Distribution 5 Sampling Distribution 5 Standard Error of Statistic 5 Central Limit Theorem 5 Area under the Sampling Distribution of the Mean 6 Sampling Distribution‚ Difference between Independent means 6 Sampling Distribution of a Linear Combination of Means 6 Sampling Distribution of Pearson’s R 7 Sampling Distribution
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1. As the degrees of freedom increase‚ the t distribution approaches the b. normal distribution 2. If the margin of error in an interval estimate of μ is 4.6‚ the interval estimate equals b. [pic] 3. The t distribution is a family of similar probability distributions‚ with each individual distribution depending on a parameter known as the c. degrees of freedom 4. The probability that the interval estimation procedure will generate an interval that does
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EC/315 Midterm Examination Directions: This test is open-book and open notes and covers the content from weeks 1 through week 4 of EC/315. The test will be typed and submitted in the Dropbox marked Midterm Exam. The midterm is due the last day of Week 4. PROBLEM 1 (Weight 40 points). NBC TV news‚ in a segment on the price of gasoline‚ reported last evening that the mean price nationwide is $1.50 per gallon for self-serve regular unleaded. A random sample of 35 stations in the Milwaukee
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ABOUT THE SLOPE? (D) INTERPRET THE 95 PERCENT CONFIDENCE LIMITS FOR THE SLOPE. (E) VERIFY THAT _F_ = _T_2 FOR THE SLOPE. (F) IN YOUR OWN WORDS‚ DESCRIBE THE FIT OF THIS REGRESSION. R2 0.202 STD. ERROR 6.816 N 35 ANOVA TABLE _SOURCE SS DF MS F P-VALUE_ REGRESSION 387.6959 1 387.6959 8.35 .0068 RESIDUAL 1‚533.0614 33 46.4564 TOTAL 1‚920.7573 34 REGRESSION OUTPUT _CONFIDENCE INTERVAL_ _VARIABLES COEFFICIENTS STD. ERROR T (DF_ = _33) P-VALUE 95% LOWER 95% UPPER_ INTERCEPT 30.7963 6.4078 4
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Using a 95% confidence level plot the overall percentage of late flights p‚ and the upper and lower control limits. p = Total Number of errors Total Number of Sample p= 120 = 0.04 (100)*(30) sigma= SQRT(((.04)*(1-.04))/100) sigma= 0.02 UCL= p + 2*.02 = 0.08 LCL= p - 2*.02 = 0 Using 95% Confidence level (z=2)‚ we plot a control chart for the fraction of late flights for ezch week. We also superimpose the industry upper and lower control limits of .1 and
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