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Assignment 13 - Lesson 17 &amp; 18 - Confidence Intervals - Fall 2009

Assignment 13 - Lesson 17 & 18 - Confidence Intervals - Fall 2009

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Problem 1. WEE-426-25: Women's Weights (Lesson 17) Summary of Weights No Selector Count: 40 Mean: 136.875 Median: 136.5 MidRange: 141 StdDev: 12.7685 Range: 52 IntQRange: 19 The U.S. National Center for Health Statistics estimates mean weights of Americans by age, height, and sex and publishes the results in Vital and Health Statistics. 40 U.S. women, 5 ft 4 in. tall and age 18-24, are randomly selected. Their weights, in pounds, are recorded in Data Desk. Assume that the population standard deviation of all such weights is 10.0 lb . d. What is the sample size? Answer: n = 40 e. To construct a confidence interval, which procedure is more appropriate, z-interval or t-interval? Explain briefly. Answer: z-Interval, coz it’s more appropriate for larger samle size. z-Interval for Individual µ’s No Selector Sigma = 10,00 Individual Confidence 95,00% Bounds: Lower Bound < µ < Upper Bound With 95,00% Confidence, 133,77602 < µ(Weights) < 139,97398 f. Construct 95% and 99% confidence intervals for the mean weight of all U.S. women 5 ft 4 in. tall and in the age group 18-24 years. Answer : z-Interval for Individual µ’s No Selector Sigma = 10,00 Individual Confidence 99,00% Bounds: Lower Bound < µ < Upper Bound With 99,00% Confidence, 132,80226 < µ(Weights) < 140,94774 g. Interpret your answers in Part (c). Answer: We are 99 % confident that estimated weight for Forty U.S. women, 5 ft 4 in. tall and age 18-24 h. Which confidence interval is wider, 95% or 99%? Answer: 99% - is wider. i. What is the margin of error for each confidence interval?

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Answer: For 95 % confidence interval, the margin of error is 3.09898; For 99 % confidence interval, the margin of error is 4.07274 Problem 2. Assume that the population standard deviation is unknown , repeat Problem 1. a.
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Assignment 13 - Lesson 17 & 18 - Confidence Intervals - Fall 2009

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