Spring 2016 - WUSTL Math
... with two degrees of freedom. We have P(A > 2) = P(χ22 > 2). With software, such as the command pchisq(2, df = 2, lower.tail = FALSE) in R, or something similar using a statistics calculator, we obtain 0.3678794 for the requested probability. If the given table is used instead, then we interpolate be ...
... with two degrees of freedom. We have P(A > 2) = P(χ22 > 2). With software, such as the command pchisq(2, df = 2, lower.tail = FALSE) in R, or something similar using a statistics calculator, we obtain 0.3678794 for the requested probability. If the given table is used instead, then we interpolate be ...
Process and Measurement System Capability Analysis
... • Natural Tolerances= µ+3σ and µ-3σ = UNTL (Upper Natural Tolerance Limit) and LNTL (Lower Natural Tolerance Limit) • Normal Distribution includes 99.73% of the variable • Or 0.27% of the process output will fall outside natural tolerance limits • Process Capability Analysis a study to estimate PC ( ...
... • Natural Tolerances= µ+3σ and µ-3σ = UNTL (Upper Natural Tolerance Limit) and LNTL (Lower Natural Tolerance Limit) • Normal Distribution includes 99.73% of the variable • Or 0.27% of the process output will fall outside natural tolerance limits • Process Capability Analysis a study to estimate PC ( ...
learning objectives for Introductory Statistics
... F1. Given a study, describe how the law of averages (law or large numbers) applies. F2. Describe the sampling distribution of a statistic. F3. Given a study, describe the sampling distribution of x-bar as specifically as possible. This involves stating whether this distribution is at least approxima ...
... F1. Given a study, describe how the law of averages (law or large numbers) applies. F2. Describe the sampling distribution of a statistic. F3. Given a study, describe the sampling distribution of x-bar as specifically as possible. This involves stating whether this distribution is at least approxima ...
Inference for Regression
... show a linear pattern. 2. Independence Assumption: The residuals must be mutually independent. A check of a residuals plotted against the x values show no patters, trends, or clumping. 3. Equal variance assumption: The variability of y should be be about the same for all values of x. A check of the ...
... show a linear pattern. 2. Independence Assumption: The residuals must be mutually independent. A check of a residuals plotted against the x values show no patters, trends, or clumping. 3. Equal variance assumption: The variability of y should be be about the same for all values of x. A check of the ...
Interpreting Confidence Intervals
... where the statistic we use is the point estimator for the parameter. Properties of Confidence Intervals: •The “margin of error” is the (critical value) • (standard deviation of statistic) •The user chooses the confidence level, and the margin of error follows from this choice. •The critical value de ...
... where the statistic we use is the point estimator for the parameter. Properties of Confidence Intervals: •The “margin of error” is the (critical value) • (standard deviation of statistic) •The user chooses the confidence level, and the margin of error follows from this choice. •The critical value de ...
Document
... preliminary statement based upon a confidence interval for a proportion but an actual inference should only be made and stated as a conclusion to a hypothesis test based on the traditional or p-value method introduced in Chapter 9. The reason for this is that the distributions are not the same for ...
... preliminary statement based upon a confidence interval for a proportion but an actual inference should only be made and stated as a conclusion to a hypothesis test based on the traditional or p-value method introduced in Chapter 9. The reason for this is that the distributions are not the same for ...
Confidence Intervals
... Another Example ((The Level of CI)) • Question: There is a sample of 50 micro-drills with an average lifetime (expressed as the number of holes drilled before failure) was 12.68 with a standard deviation of 6.83. Suppose an engineer reported a confidence interval of (11.09, 14.27) but neglected to ...
... Another Example ((The Level of CI)) • Question: There is a sample of 50 micro-drills with an average lifetime (expressed as the number of holes drilled before failure) was 12.68 with a standard deviation of 6.83. Suppose an engineer reported a confidence interval of (11.09, 14.27) but neglected to ...
1 ISC 207 Practice questions for Quiz-3 (Solutions)
... a. What is the value of the test statistic? (103-100)/(18/sqrt(18)) = 3/2 = 1.5 b. What is the rejection region at alpha of 0.05? >1.96 or <-1.96 c. What will be your decision at alpha of 0.05? Fail to reject H0 d. What is the p-value for this study? 2*P(z>1.5) = 2*(1-0.93319) = 2*0.06681=0.13362 e. ...
... a. What is the value of the test statistic? (103-100)/(18/sqrt(18)) = 3/2 = 1.5 b. What is the rejection region at alpha of 0.05? >1.96 or <-1.96 c. What will be your decision at alpha of 0.05? Fail to reject H0 d. What is the p-value for this study? 2*P(z>1.5) = 2*(1-0.93319) = 2*0.06681=0.13362 e. ...
German tank problem
In the statistical theory of estimation, the problem of estimating the maximum of a discrete uniform distribution from sampling without replacement is known in English as the German tank problem, due to its application in World War II to the estimation of the number of German tanks.The analyses illustrate the difference between frequentist inference and Bayesian inference.Estimating the population maximum based on a single sample yields divergent results, while the estimation based on multiple samples is an instructive practical estimation question whose answer is simple but not obvious.