Presentation on theme: "Statistics Estimates and Sample Sizes"— Presentation transcript:
1Statistics Estimates and Sample Sizes Chapter 7, Part 2Example Problems
2Finding 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% confidenceAnswer: The margin of error is found with the formula
3Finding Margin of Error Question: n = 600, x = 120, 95% confidenceFormula:We first need to find =therefore = 1 – 0.2 = 0.8
4Finding Margin of Error The z is found by 1 – 0.95 (confidence) = 0.05 and divided in half 0.05/2 = 0.025Therefore in the z table you look up 1 – = and find the z value of 1.96.
5Finding Margin of Error Plugging everything into the formula
6Finding Margin of Error with TI83/84 In your TI-83/84 calculator:STATTESTS1-PropZIntEnter n 600, x 120 and confidence 0.95 as decimalCalculateYou should get the interval (.16799, ).Subtract these =Divide this by 2 to get /2 = or rounded to the same answer as previously 0.032
7Confidence IntervalsQuestion: 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?
8Confidence IntervalsUse the same technique of finding the margin of error and then the confidence interval is the proportion +/- the errorThen 𝑞 =1− 𝑝 = 1 – = 0.501Thus, the interval isLow: – = 0.490High: = 0.508𝑝 = 𝑥 𝑛 = ,000 =0.499𝐸𝑟𝑟𝑜𝑟= ∗ ,000 =0.009
9Confidence Intervals TI83/84 In your TI-83 calculator:STATTESTS1-PropZIntEnter x=5988, n=12000 and confidence as decimal 0.95CalculateYou should get (.490, .508) for the interval.
10Confidence IntervalsBased 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 holidayIf 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
11Finding Sample SizeQuestion: 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
12Finding 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 isWe plug everything into the formula.and we always round up one so n = 27.