1. Statistics Estimates and Sample Sizes
Chapter 7, Part 2
Example Problems

2. 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

3. Finding Margin of Error
Question: n = 600, x = 120, 95% confidence
Formula:
We first need to find =
therefore = 1 – 0.2 = 0.8

4. 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.

5. Finding Margin of Error
Plugging everything into the formula

6. 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

7. 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?

8. 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

9. 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.

10. 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

11. 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

12. 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.