grams were 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477 (a) Construct a 90 percent confidence interval for the true mean weight. E = 1.96(s/sqrt(n)) = 1.96[0.131989/sqrt(10)]=1.96*0.41739 =0.081808 C.I. = (x-bar-E,x-bar+E) = (3.3048-0.0818,3.3048+0.0818) (b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? n=[z'*s/E]^2 n=[1.645*0.131989/0.03]^2 = 52.38; rounding up, n=53 (c) Discuss the factors which
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Question 1. (a) R²= 4.33% of the variation in the dependent variable RET is explained by the variation of the independent variables: GRI, SAT, TEN, Age, and MBA. We cannot assume from the low R squared that the model is useless because there might be one or more independent variables that are not significant to the dependent variable. Also, we cannot question the fact that the OLS are not accurate just by looking at the low R squared. In the opposite, adding new independent variables do increase
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be more centered, as mean, mode and median have similar values. 2. Develop a 95% confidence interval estimate of the mean assets, mean 7 day yield, and mean 30-day yield for the population of the money market funds. Provide a managerial interpretation of each interval estimate. Since the sample size is n=45 (>30), I can use a standard normal distribution to compute the interval estimation; 95% confidence level corresponds from the table to z=1.96. Since the standard deviation for the population
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and its references have been used in creating it. RSM is only responsible for supervision of completion of the work but not for the contents.” Table of content Abstract 3 Introduction 4 Theory 5 Current study 7 Effect sizes and confidence intervals 7 Research strategy 8 Populations and measurements 8 Populations 8 Practical relevance of the effect sizes 11 Critical Synthesis 13 Selection of studies 13 Critical evaluation of studies 13 Results and Discussion 14 An attempt
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cheating tendencies of 90 students. This test was given during and exit interview to help elevate any potential repercussions to the results. All statistical information can be found in Tables A1-A4. In addition to the statistical information, the confidence testing results and hypothesis testing can be found in Tables A5-A7. The students were asked three questions: * Did you ever copy work off the Internet as your own? * Did you ever copy answers off another student’s exam? * Did you
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M Project Report Israel Sharipo September 13, 2013 Abstract This paper shows the true proportions of M&M candies through the random sampling of 3 different bags of plain M&M 1.69oz bag of candy bought from three different stores. I believed one will be able to determine data analysis and how it works as it relates to our class project. Introduction The rationale of writing this paper is to show how
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the interval estimation of μ when σ is known and the sample is large, the proper distribution to use is | | the normal distribution | | | the t distribution with n degrees of freedom | | | the t distribution with n + 1 degrees of freedom | | | the t distribution with n + 2 degrees of freedom | 5 points QUESTION 2 1. An estimate of a population parameter that provides an interval of values believed to contain the value of the parameter is known as the | | confidence level
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From this case, there are two types of errors, which the consortium can make. A Type I Error is referred to as a “false positive.” A Type I error would be made when the null hypothesis is rejected when it should be accepted. This error may occur if the consortium defends any lawsuit against them if they are using 6% (6/100) as their surveying result. The results of the sample size of 100 people indicate that the percentage range is from 1.35% to 10.65%. The test results can be higher than 10%, but
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bootstrap sample? Explain why it wouldn’t be. a) 48, 55, 43, 61, 39 b) 43, 39, 56, 43, 61 c) 55, 48, 55, 48, 61 d) 39, 39, 39, 39, 39 2. The following bootstrap output is for mileage of a random sample of 25 mustang cars. Based on a 90% confidence interval, which of the following would not be a plausible value of the population mean, µ? Explain why it wouldn’t be. a) 60.01 b) 80.01 c) 55 d) 52 3. Which of the following p-values would provide more evidence in support of the alternate hypothesis
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verify my standard process is operating in a way that allows me to complete all of the tasks and affords me additional time. This paper will also outline the control limits of my morning process, the effects of any seasonal factors, and the confidence intervals involved. Statistical Process Control Data was recorded for a period of two weeks on how long it took me to complete my morning routine and get my kids off to school. I tracked the minutes it took me to complete each step of my morning
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