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Ch. 8 – Practical Examples of Confidence Intervals for z, t, p

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Using the table For the 99% CI: find z.005 For the 95% CI: find z.025 For the 90% CI: find z.05

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Review of Theory for 95% Confidence Intervals Recall that P(-1.96

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Confidence Intervals -1.96 < <1.96 A few steps of algebra show That< < So the 95% CI for µ is In general, the CI for µ is

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Example #1- administrator salaries Suppose you’d like to estimate the population mean µ for the salary of administrators at the local college. A. If a sample of 35 administrators showed that the sample mean x-bar = $56,000. If the population standard deviation is known to be $17,000, find 95% confidence interval for µ. B. For A, identify the point estimate and the margin of error. C. Without changing any other variables in “A”, find the margin of error and the 99% confidence interval for µ. D. When the confidence level is increased, what happens to the margin of error?

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Example #2- bowling scores Suppose you’d like to estimate the population mean µ bowling score for a league. A sample showed that x-bar = 180, and we’re assuming that the population standard deviation is = 10. A.If the sample size that produced x-bar = 180 was 30, find the 95% confidence interval for the true mean bowling score µ. B. If the sample size was instead 100, find the 95% confidence interval for the true mean bowling score µ. C. What happens to the margin of error when the sample size is increased? D. Keep n=100, = 180, and let = 20. Calculate the 95% confidence interval for the true mean bowling score µ. E. What happens to the margin of error when the standard deviation increases?

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Example #3-placement scores Incoming math students were all given a placement exam. Assume that the test results were normally distributed and that the population standard deviation on the exam is 14. From the following sample, calculate the 95% confidence interval for the true mean placement score µ. 455552627288466281 4273484552816668 5952454836424849

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Sample Size for estimating µ Let the margin of error be E=z*σ/ n Solve for n: n=( z* /E) 2 Example: In the placement score example, we wish to calculate a 95% confidence interval, and we are content to be within 2 points of the actual answer. A. If a previous study stated that = 20, find the sample size that we should use in this study. B. If = 40 instead, find the new sample size. C. If = 10, find the new sample size. D. If = 10, and we wish to increase our accuracy to be within 1 point, find the new n.

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8.2: Confidence Intervals in practice Most of the time, µ is not known. In these circumstances, we use s (sample standard deviation) as an estimate for µ (population standard deviation), and we use the t distribution instead of the z. Practice finding t values: When n=20, find t.01, t.05, t.025 When n=12, find t.01, t.05, t.025 If n=8 and you wish to find a 95% confidence interval, find the appropriate t If n=22 and you wish to find a 90% confidence interval, find the appropriate t

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Example #1: Bowling scores– using t From the following sample, find the 99% confidence interval for the true mean µ bowling score: 17017516610818015595200 628852107191 Note: The CI for the is:

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Example #2: Placement scores– using t From the following sample, find the 90% confidence interval for the true mean µ placement score: 654288219065728771

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8.3: Estimating p in the Binomial Distribution The point estimates for p and q are: p-hat =r/n and q-hat= 1- p-hat Ex: A sample of 200 people at a large corporation were asked if they enjoyed their jobs. 142 said yes. Find a point estimate for the true proportion of people who enjoy their jobs. Find a point estimate for the proportion of those who do not.

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Theory For large samples, the distribution of is approximately normal with mean µ = p and standard error = (pq/n)

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Example #1: Job Satisfaction -- p example For the previous example, find a 95% confidence interval for the true proportion of people who enjoy their jobs. Note: CI for p is ± z *

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Example #2: Voting A random sample of 800 adults reveals that 200 planned to vote in the mid-term elections. Find the 99% confidence interval of the true proportion of adults that plan to vote in the mid-term elections. Find the 90% confidence interval of the true proportion of adults that plan to vote in the mid-term elections.

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Sample Size for p Recall that E=z* (pq/n). Solve for n n=p*q*(z/E) 2

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Sample size ex for p Example: A previous study showed that 25% of adults vote in the midterm elections. If you wish to be within 1 percentage point of the true proportion, what sample size would you need in order to create a 95% confidence interval of the true proportion of those who will vote. If you don’t have a previous study, what would be a reasonable assumption for p-hat ? Recalculate, and see how this changes the required sample size?

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