Amanda Mueller MBA6018 – Data Analysis Unit 3 Activity 1 January 23‚ 2013 Practical Application Scenario 1 In 2010‚ Playbill Magazine contracted Boos Allen to conduct a survey aimed at determining the average annual household income of Playbill readers. 300 readers were randomly pulled and sampled from the list of customers provided by Playbill Magazine. From that sampling effort‚ Boos Allen was confident that the population average household income is $119‚155 and that the population
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(RSD) | 2.4658 | 8.3728 | Range (R) | 0.0376 0.0003 | 0.166 0.0003 | Relative Range (RR) | 6.9892 4.1935 | 30.817 18.490 | Confidence Limits (CL) | 5.3797 0.01392 | 5.3866 0.04222 | The measures of central
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ch9 Student: ___________________________________________________________________________ 1. The risk of incorrect acceptance and the risk of assessing control risk too low relate to the A. Preliminary estimates of materiality levels. B. Allowable risk of tolerable error. C. Efficiency of the audit. D. Effectiveness of the audit. 2. While performing a substantive test of details during an audit‚ the auditor determined that the sample results supported the conclusion that the recorded
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EXHIBIT TN.1 Coin Catapult Worksheet Exercise A EXERCISE A Data Table Sample Observation Sample Sample Number 1 2 3 4 5 Mean Range R 1 13.4 29 15.0 17 12.2 17.32 16.8 2 24.4 34 18.9 10 12 19.86 24 3 25.5 23.5 31 33.2 29.8 28.6 9.7 4 19.7 31.4 34.4 33.9 26.8 29.32 15.1 23.78 16.4 EXHIBIT TN.1 (Cont.) Coin Catapult Worksheet Data Table (for additional observations) Sample Sample Sample Observation Mean Range Number 1 2 3 4 5 R 5 29 28 27 17 32 26.6 15 6 32 32 12.6 23 23 24.52
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the survey results could be used by the magazine’s editors to identify topics that would be of interest to readers. Your report should address the follow- ing issues‚ but do not limit your analysis to just these areas. a. Develop appropriate descriptive statistics to summarize the data. b. Develop 95% confidence intervals for the mean age and
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Central Limit Theorem and Confidence Intervals Problem Sets Tiffany Blount QNT 561 September 7‚ 2010 Michelle Barnet University of Phoenix Central Limit Theorem and Confidence Intervals Problem Sets Chapter 8 Exercises: 21. What is sampling error? Could the value of the sampling error be zero? If it were zero‚ what would this mean? * Sampling error is the difference between the statistic estimated from a sample and the true population statistic. It is not impossible for the sampling
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250‚ 593‚ 540‚ 225‚ 177‚ 425‚ 318‚ 182‚ 275‚ and 228. Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation. Place your LOWER limit‚ in dollars rounded to 1 decimal place‚ in the first blank. Do not use a dollar sign‚ a comma‚ or any other stray mark. For example‚ 123.4 would be a legitimate entry. Correct199.26 Place your UPPER limit‚ in dollars rounded to 1 decimal place‚ in the second blank. Do not use a
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Problem N°1 1.Formulate the null and alternative hypotheses. Null Hypothesis: The average (mean) annual income was greater than or equal to $50‚000 H_0: μ≥50000 Alternate Hypothesis: The average (mean) annual income was less than $50‚000. H_a: μ 30 we will use the z-test. As Ha:μ0.40 the‚ test is a right tailed z-test. The critical value for significance level‚ α=0.05 for a right tailed z-test is given in the table as: 1.645. Decision Rule: Reject H_0‚if z>1.645 3. Calculate the test
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scatter diagram and the regression line. f. Calculate r and r2 and explain what they mean. g. Predict the monthly auto insurance premium for a driver with 10 years of driving experience. h. Compute the standard deviation of errors. i. Construct a 90% confidence interval for B. j. Test at the 5% significance level whether B is negative. k. Using α = .05‚ test whether ρ is different from zero. Solution a. Based on theory and intuition‚ we expect the insurance premium to depend on driving experience. Consequently
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hours. Construct a 95 confidence interval based on this sample and be sure to interpret this interval. Answer Since population standard deviation is unknown‚ t distribution can be used construct the confidence interval. The 95% confidence interval is given by X t / 2‚n 1 S S ‚ X t /2‚n 1 n n Details Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level 90 1870 84 95% Intermediate
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