Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size N , and the sample size is more than 5% of the population size ( n > 0.05 N ), better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by ( N − n ) / ( N − 1 ) . For the sample of 100 weights of M&M candies in Data Set 27 “M&M Weights” in Appendix B, we get x ¯ = 0.8565 g and s = 0.0518 g. First construct a 95% confidence interval estimate of μ , assuming that the population is large; then construct a 95% confidence interval estimate of the mean weight of M&Ms in the full bag from which the sample was taken. The full bag has 465 M&Ms. Compare the results.
Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size N , and the sample size is more than 5% of the population size ( n > 0.05 N ), better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by ( N − n ) / ( N − 1 ) . For the sample of 100 weights of M&M candies in Data Set 27 “M&M Weights” in Appendix B, we get x ¯ = 0.8565 g and s = 0.0518 g. First construct a 95% confidence interval estimate of μ , assuming that the population is large; then construct a 95% confidence interval estimate of the mean weight of M&Ms in the full bag from which the sample was taken. The full bag has 465 M&Ms. Compare the results.
Solution Summary: The author explains the 95% confidence interval estimate of mu , assuming that the population is large and the mean weight of M&Ms in the full bag.
Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size N, and the sample size is more than 5% of the population size (n > 0.05N), better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by
(
N
−
n
)
/
(
N
−
1
)
. For the sample of 100 weights of M&M candies in Data Set 27 “M&M Weights” in Appendix B, we get
x
¯
= 0.8565 g and s = 0.0518 g. First construct a 95% confidence interval estimate of μ, assuming that the population is large; then construct a 95% confidence interval estimate of the mean weight of M&Ms in the full bag from which the sample was taken. The full bag has 465 M&Ms. Compare the results.
Definition Definition Number of subjects or observations included in a study. A large sample size typically provides more reliable results and better representation of the population. As sample size and width of confidence interval are inversely related, if the sample size is increased, the width of the confidence interval decreases.
1.2.17. (!) Let G,, be the graph whose vertices are the permutations of (1,..., n}, with
two permutations a₁, ..., a,, and b₁, ..., b, adjacent if they differ by interchanging a pair
of adjacent entries (G3 shown below). Prove that G,, is connected.
132
123
213
312
321
231
You are planning an experiment to determine the effect of the brand of gasoline and the weight of a car on gas mileage measured in miles per gallon. You will use a single test car, adding weights so that its total weight is 3000, 3500, or 4000 pounds. The car will drive on a test track at each weight using each of Amoco, Marathon, and Speedway gasoline. Which is the best way to organize the study?
Start with 3000 pounds and Amoco and run the car on the test track. Then do 3500 and 4000 pounds. Change to Marathon and go through the three weights in order. Then change to Speedway and do the three weights in order once more.
Start with 3000 pounds and Amoco and run the car on the test track. Then change to Marathon and then to Speedway without changing the weight. Then add weights to get 3500 pounds and go through the three gasolines in the same order.Then change to 4000 pounds and do the three gasolines in order again.
Choose a gasoline at random, and run the car with this gasoline at…
AP1.2 A child is 40 inches tall, which places her at the 90th percentile of all children of similar age. The heights for children of this age form an approximately Normal distribution with a mean of 38 inches. Based on this information, what is the standard deviation of the heights of all children of this age?
0.20 inches (c) 0.65 inches (e) 1.56 inches
0.31 inches (d) 1.21 inches
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Hypothesis Testing - Solving Problems With Proportions; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=76VruarGn2Q;License: Standard YouTube License, CC-BY
Hypothesis Testing and Confidence Intervals (FRM Part 1 – Book 2 – Chapter 5); Author: Analystprep;https://www.youtube.com/watch?v=vth3yZIUlGQ;License: Standard YouTube License, CC-BY