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

Using the table

• For the 99% CI: find z.005

• For the 95% CI: find z.025

• For the 90% CI: find z.05

Review of Theory for 95% Confidence Intervals

• Recall that P(-1.96<z<1.96)=.95• That is, there’s a 95% chance that

-1.96<z<1.96• Since the CLT says that z= then

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

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?

Example #2- bowling scoresSuppose 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?

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.

45 55 52 62 72 88 46 62 81 4273 48 45 52 81 66 68 59 5245 48 36 42 48 49

Sample Size for estimating µ

Let the margin of error be E=z*σ/nSolve 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.

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

Example #1: Bowling scores– using t

From the following sample, find the 99% confidence interval for the true mean µ bowling score:

• 170 175 166 108 180 155 95 20062 88 52 107 191

• Note: The CI for the is:

Example #2: Placement scores– using t

From the following sample, find the 90% confidence interval for the true mean µ placement score:

• 65 42 88 21 90 65 72 8771

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.

Theory

• For large samples, the distribution of is approximately normal with

• mean µ = p and • standard error = (pq/n)

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 *

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.

Sample Size for p

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

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?