1. Choosing Sample Size Section 10.1.3

2. 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?

3. 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

4. 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).

5. 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.

6. 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.”

7. 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):

8. 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.

9. 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

10. 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

11. 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

12. 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.

13. Answer

15. 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

16. Homework Read pages 520 - 525 Do problems 9, 11, 13, 17