calculated in order to obtain a point estimate of the mean overpayment. The estimated mean is then extrapolated to the overpayment amount to the population of all 1,000 claims. http://pages.wustl.edu/montgomery/lecture-7 Point Estimate vs. Interval Estimate To estimate population parameters, statisticians use sample statistics. For example, we use sample means to estimate population means and we use sample proportions to estimate population proportions. An estimate of a population parameter
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to Construct a Confidence Interval Instructions on the left Instructions on the right pertain to means pertain to proportions 1. POPULATION a. Identify the parameter of interest: π : proportion µ : Mean Numerical (Measurement) Categorical (success-failure) b. Describe the variable in context with the problem: µ = mean of the amount of drying time of a particular paint. π = proportion of people in the community who prefer smoking 2. STATISTICAL METHOD a. Determine the confidence level (1 - α) and
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MAT 300 M&Ms® Project Part 3 (21 pts) We will be constructing confidence intervals for the proportion of each color as well as the mean number of candies per bag. You will use the methods of 6.3 for the proportions and 6.1 for the mean. For the Bonus, you will use the sample size formula on page 338. You can use StatCrunch to assist with the calculations. A link for StatCrunch can be found under Tools for Success in Course Home. Here is also a link: http://statcrunch.pearsoncmg.com/statcrunch/larson_les4e/dataset/index
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Question: In a poll of 600 voters in a campaign to eliminate non-returnable beverage containers, 210 of the voters were opposed. Develop a 92% confidence interval estimate for the proportion of all the voters who opposed the container control bill. Answer: 92% of confidence interval (CI) for p is given by p±Z0.08 √(pq/n) Here, p= 210/600= 0.35 q= (1-p) = 0.65 Z0.08 = 1.75609 n= 600 Hence, CI is 0.35±1.75609 √((0.35*0.65)/600) = 0.35±0.03409 (0.31591,0.38409) Question: A random sample
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Introduction In-Process Technology was set up in 1985 to produce and sell a thermal oxidation process that could be used to reduce industrial pollution. For years, however, the company performed dismally. In 1992, the current CEO was hired to turn things around, and the company was reorganized and renamed Thermatrix. Eventually, Thermatrix was able to attract good employees using stock options and other competitive compensation and currently has 60 employees and annual sales of approximately $15
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“A confidence interval for the mean is an interval estimate around a sample mean that provides us with a range of where the true population mean lies” (Donnelly, 2013, p. 337). In this scenario, the marketing team from the Mi Casa Front Porch restaurant is trying to determine the average bill for a party of two for lunch and dinner. The marketing team sampled a range of different bills from lunch and dinner and is trying to determine what price ranges they should use for advertising. Since they have
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process a random sample of 36 bolts manufactured has a mean of 3 inches w/ a standard deviation of .3 inches. What is the 99% confidence interval for the true mean length of the bolt? Confidence interval = [pic] = (2.87, 3.13) 2. When the population is normally distributed, population standard deviation σ is unknown, and the sample size is n = 15; the confidence interval for the population mean σ is based on the ________. Sample standard deviation, s. 3. Suppose that the percentage returns
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Statistics 4/29/15 Homework Question 1 – What is the essence of the confidence interval? Analyze the relationship between the confidence interval and central limit theorem. Question 2 – Explain the essence of Hypothesis testing. How related are null hypothesis and Alternative Hypothesis. How do you apply confidence interval in hypothesis testing? Question 3 – Explain the difference between T distribution and Z distribution. When and how do we use T distribution? What is the meaning of the number
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477 A) Construct a 90% confidence interval for the true mean weight Assuming the population has a normal distribution, we should use the Student t distribution E= z(sigma)/ sqrt(n) E= 1.645(0.13199/ sqrt(10) E= 0.06866 x-bar= 3.3048 90% CI = 3.3048-.06866= 3.23614, 3.3048 + 0.06866 = 3.37346 Confidence interval: 3.23614 to 3.37346 B) What sample size would be necessary to estimate the true weight with an error of +,- 0.03 grams with 90% confidence ? E = z*sigma/sqrt(n)
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Math 221 **** Example Format **** Week 6 Lab Submitted by: (Insert Name) Part 1. Normal Distributions and Birth Weights in America 1(a) 37 to 39 weeks as mean is around 7.33 lb. 1(b) 40 weeks as mean is around 7.72 lb. 1(c) 28 to 31 weeks as mean is 4.07 lb. 2(a) 99.88%, Excel command used was NORMDIST(5.5,1.88,1.19,TRUE). 2(b) 43.83% 2(c) 4.66% 2(d) 2.75% 3(a) Above 8.7269, Excel command used was NORMINV(0.9,7
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