• Montgomery College

• MATH 117

Section 6.2-CI: Confidence Interval for a Mean

Example 1: Dark Chocolate for Good Health

Eleven people were given 46 grams (1.6 ounces) of dark chocolate every day for two weeks, and their vascular

health was measured before and after the two weeks. Larger numbers indicate greater vascular health, and the

mean increase for the participants was 1.3 with a standard deviation of 2.32. Assume a dotplot shows the data

are reasonably symmetric with no extreme values. Find and interpret a 90% confidence interval for the mean

increase in this measure of vascular health after two weeks of eating dark chocolate. Can we be 90% confident

that the mean change for everyone would be positive?

90% confidence interval with 10 degrees freedom

t* = 1.812 S tatistic± t¿∗SE

1.3± 1.812∗232

√11

1.3±1.268 = 0.032 to 2.568

90% sure that the mean change in this measure of vascular health for people who eat dark chocolate for two weeks is between 0.032 to

2.568.

We are 90% confident that the mean change is positive.

Example 2: Sample Size and Margin of Error for Dark Chocolate

(a) What is the margin of error for the confidence interval found in Example 1?

The margin of error is 1.268

(b) What sample size is needed if we want a margin of error within ±0.5, with 90% confidence? (Use the

standard deviation from the original sample to estimate σ.)

n=( z

¿

σ

ME

) = ( 1.645∗2.32

0.5

¿2=58.26

Sample size that needs to be used: at 59 (to achieve a level of accuracy)

Quick Self-Quiz: Cell Phone Calls

A survey of 1,917 cell phone users in May 2010 asked “On an average day, about how many cell phone calls do

you make and receive on your cell phone?” The mean number of calls was 13.10, with a standard deviation of

about 10.2. Find and interpret a 99% confidence interval for the mean number of cell phone calls for all cell

phone users.

99% sure that the mean number of calls per day for all cell phone users is between 12.50 and

13.70.