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busa3321_CIMeanProp

# busa3321_CIMeanProp - Confidence Intervals for Means and...

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Confidence Intervals for Means and Proportions Point estimates Margins of Error Estimation for the average of a population ( σ known) Estimation for the average of a population ( σ unknown) Estimation for the proportion of successes in a population Determining sample size 1. Point Estimates For population average use sample average For population proportion use sample proportion 2. Margins of Error Knowledge – inversely related Variability – directly related Confidence – directly related 3. Estimation for the average of a population ( σ known) 3.1 Requirements Simple random sample The sample size is large or sampling from a normal σ known 3.2 Margin of error Knowledge –sample size Variance – population variance Confidence – related to empirical rule Margin of error Standard normal table value times the population standard deviation divided by square root of n 3.3 Estimate sample mean plus or minus the margin of error n z x σ ±

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3.4 Example: Suppose from a random sample of size 49, we find a sample mean of 30. It is known that the typical distance a value is from the population (standard deviation) is 35. What is the population mean with 95% confidence? Solution: First calculate the typical error in a sample mean. This is value is 35 divided by the square root of 49 = 5. Therefore when using this sample mean the typical error you would expect is five. Next determine how far you have to go either side of the sample mean for the specified confidence. With 95% confidence you have to go 1.96 standard errors (1.96*5=9.8) either side of the sample mean to have 95% confidence that the population mean is within the interval. 8 . 9 30 5 96 . 1 30 49 35 96 . 1 30 ± = ± = ± = ± n z x σ With 95% confidence we can say that the population mean is 30 with a maximum possible error of ± 9.8 For other examples, double click the embedded Excel file below. Scroll down to see solution. Press F9 to see another example. Estimate the mean age of all buyers with 98% confidence. After collecting a random sample of 6 buyers you find a sample mean of 35. Assume distribution of age for all buyers is normal with a standard deviation of 3.
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