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**Statistics Estimates and Sample Sizes**

Chapter 7, Part 2 Example Problems

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**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

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**Finding Margin of Error**

Question: n = 600, x = 120, 95% confidence Formula: We first need to find = therefore = 1 – 0.2 = 0.8

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**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 – = and find the z value of 1.96.

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**Finding Margin of Error**

Plugging everything into the formula

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**Finding Margin of Error with TI83/84**

In your TI-83/84 calculator: STAT TESTS 1-PropZInt Enter n 600, x 120 and confidence 0.95 as decimal Calculate You should get the interval (.16799, ). Subtract these = Divide this by 2 to get /2 = or rounded to the same answer as previously 0.032

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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?

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Confidence Intervals Use the same technique of finding the margin of error and then the confidence interval is the proportion +/- the error Then 𝑞 =1− 𝑝 = 1 – = 0.501 Thus, the interval is Low: – = 0.490 High: = 0.508 𝑝 = 𝑥 𝑛 = ,000 =0.499 𝐸𝑟𝑟𝑜𝑟= ∗ ,000 =0.009

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**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.

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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

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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

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**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 We plug everything into the formula. and we always round up one so n = 27.

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Sections 6-1 and 6-2 Overview Estimating a Population Proportion.

Sections 6-1 and 6-2 Overview Estimating a Population Proportion.

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