Section_3.3_and_5.1[1]

# Section_3.3_and_5.1[1] - 29 29 σ ≈ ≈ = 61 39 25 098 9...

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Unformatted text preview: ( 29( 29 σ ≈ ≈ = . . . . 61 39 25 098 9 8% Homework for Section 3.3 and 5.1 - The Expected Value and Standard Deviation of Sample Percents Summary Suppose that in a population p is the percent that has some characteristic that we are interested in. And suppose we will get a simple random sample of individuals from the population. The percent in the sample will vary from one sample to the next - it is a random variable. We call it . \$ p The mean of = p \$ p The standard deviation of is \$ p σ ≈- p p n ( ) 1 (In this formula, p must be expressed as a decimal or fraction.) Example : Suppose 61% of the students at SF State are women. Suppose I pick a simple random sample of 25 students at SF State. My is the percent of the sample \$ p who are women. In this example p = .61 and n = 25. The expected mean value of is p = .61 = 61% \$ p The standard deviation is So we expect the sample percent to be around 61%, give or take 9.8%. The margin of error of the sample percent is two standard deviations: 2(.098) = .196, or 19.6%. This number is the farthest distance I expect my sample percent to be from the actual population percent. (There is a 5% chance that the sample percent could be farther away than this. Where does that 5% come from?) So in this example, the margin of error is 2(.098) = .196, or about 20%. We do not expect that the sample percent will be more than 20% away from the population value...
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Section_3.3_and_5.1[1] - 29 29 σ ≈ ≈ = 61 39 25 098 9...

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