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Statistics Estimates and Sample Sizes. Chapter 7, Part 2 Example Problems. Finding Margin of Error. Question: Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.
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StatisticsEstimates and Sample Sizes Chapter 7, Part 2 Example Problems
Finding Margin of Error • Question: Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. n = 600, x = 120, 95% confidence • Answer: The margin of error is found with the formula
Finding Margin of Error • Question:n = 600, x = 120, 95% confidence • Formula: • We first need to find = • therefore = 1 – 0.2 = 0.8
Finding Margin of Error • The z is found by 1 – 0.95 (confidence) = 0.05 and divided in half 0.05/2 = 0.025 • Therefore in the z table you look up 1 – 0.025 = 0.975 and find the z value of 1.96.
Finding Margin of Error • Plugging everything into the formula
Finding Margin of Error with TI83/84 • In your TI-83/84 calculator: • STAT • TESTS • 1-PropZInt • Enter n 600, x 120and confidence 0.95as decimal • Calculate • You should get the interval (.16799, .23201). • Subtract these .23201-.16799 = .06402. • Divide this by 2 to get .06402/2 = .03201 or rounded to the same answer as previously 0.032
Confidence Intervals • Question: In the week before and the week after a holiday, there were 12,000 total deaths, and 5988 of them occurred in the week before the holiday. • Construct a 99% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday. • Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?
Confidence Intervals • Use the same technique of finding the margin of error and then the confidence interval is the proportion +/- the error • Then = 1 – 0.499 = 0.501 • Thus, the interval is • Low: 0.499 – 0.009 = 0.490 • High: 0.499 + 0.009 = 0.508
Confidence Intervals TI83/84 • In your TI-83 calculator: • STAT • TESTS • 1-PropZInt • Enter x=5988, n=12000 and confidence as decimal 0.95 • Calculate • You should get (.490, .508) for the interval.
Confidence Intervals • Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday? • This would be no because the interval is not less than 0.5 (might close to it) and thus there is no proof that the number of deaths is different the week before and the week after a holiday • If the confidence interval only contains proportions below 0.5 then there is evidence that the number of deaths the week before the holiday is different than the week after the holiday
Finding Sample Size • Question: Use the margin of error, confidence level, and population standard deviation to find the minimum sample size required to estimate an unknown population mean. Margin of error = 0.9 inches Confidence Level = 90% σ = 2.8 inches
Finding Sample Size • We need the sample size formula • We know σ = 2.8 inches and E = 0.9 inches, so we just need zα/2 which can be found in the lower right corner of a Positive Z values for 0.90 confidence level is 1.645. • We plug everything into the formula. and we always round up one so n = 27.