Chapter 7Confidence Intervals for a

Population Mean ; t distributions

• t distributions• t confidence intervals for a

population mean • Sample size required to

estimate

The Importance of the Central Limit Theorem

When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is

,Nn

Since the sampling model for x is the normal model, when we standardize x we get the

standard normal z

n

xz

nxSD

)( that Note

If is unknown, we probably don’t know either.

The sample standard deviation s provides an estimate of

the population standard deviation

For a sample of size n,

the sample standard deviation s is:

n − 1 is the “degrees of freedom.”

The value s/√n is called the standard error of x , denoted

SE(x).

nxSD

)(

2)(1

1 xxn

s i

nsxSE )(

Standardize using s for

Substitute s (sample standard deviation) for

n

xz ssss sss s

n

xz

Note quite correct

Not knowing means using z is no longer correct

t-distributionsSuppose that a Simple Random Sample of size n is drawn

from a population whose distribution can be approximated by

a N(µ, σ) model. When is known, the sampling model for

the mean x is N(, /√n).

When is estimated from the sample standard deviation s, the sampling model for the mean x follows a t distribution t(, s/√n) with degrees of freedom n − 1.

is the 1-sample t statistic

t x s n

Confidence Interval Estimates CONFIDENCE

INTERVAL for

where: t = Critical value from t-

distribution with n-1 degrees of freedom

= Sample mean s = Sample standard

deviation n = Sample size

For very small samples (n < 15), the data should follow a Normal model very closely.

For moderate sample sizes (n between 15 and 40), t methods will work well as long as the data are unimodal and reasonably symmetric.

For sample sizes larger than 40, t methods are safe to use unless the data are extremely skewed. If outliers are present, analyses can be performed twice, with the outliers and without.

nstx

x

t distributions Very similar to z~N(0, 1) Sometimes called Student’s t

distribution; Gossett, brewery employee Properties:i) symmetric around 0 (like z)ii)degrees of freedom

if > 1, E( ) = 0if > 2, = - 2, which is alwaysbigger than 1.

t

-3 -2 -1 0 1 2 3

Z

0 1 2 3-1-2-3

z = x - x

x

t =

x - s

, s = sn

x

x

x

Student’s t Distribution

-3 -2 -1 0 1 2 3

Z

t

0 1 2 3-1-2-3

nx-x = z

ns - x =t x

Student’s t Distribution

Figure 11.3, Page 372

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

ns

x - x =t

Student’s t Distribution

Figure 11.3, Page 372

Degrees of Freedoms = s2

s = (X X)

n -12

i2

i=1

n

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

t7

Student’s t Distribution

Figure 11.3, Page 372

ns

x - x =t Degrees of Freedoms = s2

s = (X X)

n -12

i2

i=1

n

Degrees of Freedom1 3.0777 6.314 12.706 31.821 63.6572 1.8856 2.9200 4.3027 6.9645 9.9250. . . . . .. . . . . .

10 1.3722 1.8125 2.2281 2.7638 3.1693. . . . . .. . . . . .

100 1.2901 1.6604 1.9840 2.3642 2.62591.282 1.6449 1.9600 2.3263 2.5758

0.80 0.90 0.95 0.98 0.99

t-Table: text- inside back cover

90% confidence interval; df = n-1 = 10

118125.1 :interval confidence%90 sx

0 1.8125

Student’s t Distribution

P(t > 1.8125) = .05

-1.8125.05.05

.90

t10

P(t < -1.8125) = .05

Comparing t and z Critical Values

Conf.level n = 30

z = 1.645 90% t = 1.6991

z = 1.96 95% t = 2.0452

z = 2.33 98% t = 2.4620

z = 2.58 99% t = 2.7564

Example – An investor is trying to estimate the return

on investment in companies that won quality awards last year.

– A random sample of 41 such companies is selected, and the return on investment is recorded for each company. The data for the 41 companies have

– Construct a 95% confidence interval for the

mean return.

18.875.14 sx

40141freedom of degrees18.875.14

sx

awards.quality that wincompaniesfor investmenton return

mean population thecontains 17.36) (12.14,interval that theconfident 95% are We

36.17,14.1261.275.144118.80211.275.14

2.0211 t table,- tfrom

nstx

1.. nfdnstx

Example Because cardiac deaths increase after heavy

snowfalls, a study was conducted to measure the cardiac demands of shoveling snow by hand

The maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation were

Find a 90% CI for the population mean max. heart rate for those who shovel snow.

15,175 sx

Solution 1.. nfdnstx

shovelers snowfor rateheart maximummean thecontains 183.70) (166.30,

interval that theconfident 90% are We)70.183,30.166(

70.817510

158331.1175

1.8331 ttable,- t theFrom1015,175

nsx

EXAMPLE: Consumer Protection Agency

Selected random sample of 16 packages of a product whose packages are marked as weighing 1 pound.

From the 16 packages: a. find a 95% CI for the mean weight

of the 1-pound packages b. should the company’s claim that the

mean weight is 1 pound be challenged ?

1.10 pounds, .36 poundx s

EXAMPLE

95% CI, n=16, df=15, x=1.10s=.36critical value of t is 2.1315

becomes

.361.10 (2.1315) 1.10 .19 .91, 1.2916

Since 1 pound is in the interval, the company'sclaim appears reasonable.

tsx tn

1.. nfdnstx