Margin Of Error Equation For Proportions
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How To Find Margin Of Error On Ti 84
for a sample proportion (if certain conditions are met) is where is the sample proportion, n is the sample size, and z* is the appropriate z*-value for your desired level of confidence (from the following table). z*-Values for Selected (Percentage) Confidence Levels Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 Note that these values are taken from the standard normal (Z-) distribution. The area between each z* value and the negative of that z* value is the confidence percentage (approximately). For example, the area between z*=1.28 and z=-1.28 is approximately 0.80. Hence this chart can be expanded to other confidence percentages as well. The chart shows only the confidence percentages most commonly used. Here are the steps for calculating the margin of error for a sample proportion: Find the sample size, n, and the sample proportion. The sample proportion is the number in the sample with the characteristic of interest, divided b
zc s x We can make a similar construction for a confidence interval for a population proportion. Instead of x, we can use p and instead of s, we use , hence, we can write the confidence interval for a large sample
Margin Of Error Sample Size
proportion as Confidence Interval Margin of Error for a Population Proportion Example 1000 margin of error calculator without population size randomly selected Americans were asked if they believed the minimum wage should be raised. 600 said yes. Construct a 95% confidence interval for the proportion margin of error formula algebra 2 of Americans who believe that the minimum wage should be raised. Solution: We have p = 600/1000 = .6 zc = 1.96 and n = 1000 We calculate: Hence we can conclude that between 57 and 63 http://www.dummies.com/education/math/statistics/how-to-calculate-the-margin-of-error-for-a-sample-proportion/ percent of all Americans agree with the proposal. In other words, with a margin of error of .03 , 60% agree. Calculating n for Estimating a Mean Example Suppose that you were interested in the average number of units that students take at a two year college to get an AA degree. Suppose you wanted to find a 95% confidence interval with a margin of error of .5 for m knowing s = 10. How many people should we ask? https://www.ltcconline.net/greenl/courses/201/estimation/ciprop.htm Solution Solving for n in Margin of Error = E = zc s/ we have E = zcs zc s = E Squaring both sides, we get We use the formula: Example A Subaru dealer wants to find out the age of their customers (for advertising purposes). They want the margin of error to be 3 years old. If they want a 90% confidence interval, how many people do they need to know about? Solution: We have E = 3, zc = 1.65 but there is no way of finding sigma exactly. They use the following reasoning: most car customers are between 16 and 68 years old hence the range is Range = 68 - 16 = 52 The range covers about four standard deviations hence one standard deviation is about s @ 52/4 = 13 We can now calculate n: Hence the dealer should survey at least 52 people. Finding n to Estimate a Proportion Example Suppose that you are in charge to see if dropping a computer will damage it. You want to find the proportion of computers that break. If you want a 90% confidence interval for this proportion, with a margin of error of 4%, How many computers should you drop? Solution The formula states that Squaring both sides, we get that zc2 p(1 - p) E2 = n Multi
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