Choosing Sample Size

Section 10.1.3

Starter 10.1.3

• A coin is weighted so that it comes up heads 80% of the time. You bet $1 that you can make it come up tails within the first 3 tosses.– Is this problem binomial, geometric, or neither?– What is the probability that you will win the bet?

Today’s Objective

• Express confidence intervals as part of a three-phrase sentence in context

• Determine how large a sample size is needed to achieve a given margin of error

California Standard 17.0

Students determine confidence intervals for a simple random sample from a normal distribution of data and determine the sample size required for a desired margin of error

The phrasing of a C.I.

• Work the problem on the slide I am about to show you.

• Write a sentence that summarizes what you find in the following form:– “I am xx% confident that…– the “parameter being estimated” is…

• Write this part in context

– between (lower bound) and (upper bound).

Estimate Snickers Weights• Do snickers 1-oz “fun-size” candy bars

really weigh 1 oz?

• Suppose we know that the weights vary normally with σ = .005, and that the weights of 8 randomly chosen bars are

.95 1.02 .98 .97 1.05 1.01 .98 1.00

• Find a 90% confidence interval for the true mean weight of fun-sized Snickers.

• Write a sentence that summarizes your results using the form I just showed you.

Answer

• The sample mean is .995 oz

• Margin of error is 1.645x.005/√8=.003 oz

• The C.I. is (.992, .998)

• So write:

“I am 90% confident that the mean weight of Snickers bars is between 0.992 oz and 0.998 oz.”

Sample size for margin of error

• A confidence interval is formed by taking an estimate ± the margin of error

• The formula for margin of error is

• So if there is a certain maximum margin of error M desired, re-write as an inequality and solve for n (sample size):

*z mn

*z Mn

Example• Let’s continue the example we did

yesterday about the active ingredient in a painkiller.

• We had three specimens and formed a 99% confidence interval in which the margin of error turned out to be 0.0101.– Recall that σ was 0.0068

• Suppose the client wants a margin of error that is no more than 0.005 g.

• How many measurements do we need?– Use the formula and figure it out.

Answer• We still have σ

= .0068• z* = 2.576• M has now been

specified as .005 at most

• So solve the inequality and state n as an integer

*

*

.0068.005 2.576

z Mn

M zn

n

Answer

• We still have σ = .0068

• z* = 2.576• M has now been

specified as .005 at most

• So solve the inequality and state n as an integer

*

*

.0068.005 2.576

2.576 .00683.50336

.005

z Mn

M zn

nx

n

Answer

• We still have σ = .0068

• z* = 2.576• M has now been

specified as .005 at most

• So solve the inequality and state n as an integer

*

*

.0068.005 2.576

2.576 .00683.50336

.00512.27

So use: n = 13

z Mn

M zn

nx

n

n

What About Proportions?

• For national polling, how many people do we need to talk to in order to have a margin of error of no more than 3% in a 95% confidence interval?

• Now the margin of error term becomes:m = z* √(pq/n)

For purposes of this calculation, use p = 0.5

• Do the calculation now.

Answer

*

(.5)(.5).03 1.96

pqM z

n

n

*

(.5)(.5).03 1.96

(1.96)(.5).03

(1.96)(.5)32.66

.031067.1 so use n = 1068 or more

pqM z

n

n

n

n

n

Today’s Objective

• Express confidence intervals as part of a three-phrase sentence in context

• Determine how large a sample size is needed to achieve a given margin of error

California Standard 17.0

Students determine confidence intervals for a simple random sample from a normal distribution of data and determine the sample size required for a desired margin of error

Homework

• Read pages 520 - 525

• Do problems 9, 11, 13, 17