T distribution with 20 degrees of freedom t distribution with 10 degrees of freedom 0 t distribution with degrees of freedom t Distribution The t-distribution

t distribution with 20 degrees of freedom

t distribution with 10 degrees of freedom

0

t distribution with ∞ degrees of freedom

t Distribution

The t-distribution is used when n is small and s is unknown.

Standard normaldistribution

0 t.0250 = ?

t Distribution

If n is large there is no difference between the standard normal and t distributions.


.0250


Degrees Area in Upper Tail

of Freedom .20 .10 .05 .025 .01 .005

. . . . . . .

97 0.845 1.290 1.661 1.985 2.365 2.627

98 0.845 1.290 1.661 1.984 2.365 2.627

99 0.845 1.290 1.660 1.984 2.3646 2.626

100 .845 1.290 1.660 1.984 2.364 2.626

.842 1.282 1.645 1.960 2.326 2.576

t Distribution


P( > 1.96) = .0250t

0 1.96

t Distribution


.0250


P( > 1.96) = .0250t


Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

-2.2 .0139.0136

.0132 .0129 .0125

.0122 .0119 .0116 .0113 .0110

-2.1 .0179.0174

.0170 .0166 .0162

.0158

.0154 .0150 .0146

.0143

-2.0 .0228.0222

.0217 .0212 .0207

.0202

.0197 .0192 .0188

.0183

-1.9 .0287.0281

.0274 .0268 .0262

.0256

.0250 .0244 .0239

.0233

-1.8 .0359.0351

.0344 .0336 .0329

.0322

.0314 .0307 .0301

.0294

-1.7 .0446.0436

.0427 .0418 .0409

.0401

.0392 .0384 .0375

.0367

t Distribution


P( > 1.96) = .0250tz

t Distribution


of Freedom .20 .10 .05 .025 .01 .005

. . . . . . .

97 0.845 1.290 1.661 1.985 2.365 2.627

98 0.845 1.290 1.661 1.984 2.365 2.627

99 0.845 1.290 1.660 1.984 2.3646 2.626

100 .845 1.290 1.660 1.984 2.364 2.626

.842 1.282 1.645 1.960 2.326 2.576


0 1.96

t Distribution



.0250


0

t Distribution


t.0250 = ?

.0250

with df = 20

t Distribution


of Freedom .20 .10 .05 .025 .01 .005

. . . . . . .

18 0.862 1.330 1.734 2.101 2.552 2.878

19 0.861 1.328 1.729 2.093 2.539 2.861

20 0.860 1.325 1.725 2.086 2.528 2.845

21 0.859 1.323 1.721 2.080 2.5176 2.831

. . . . . . .

P( > 2.086) = .0250t


with df = 20

0

t Distribution


2.086

.0250

with df = 20

0

t Distribution

with df = 10


.0250

t.0250 = ?

t Distribution


of Freedom .20 .10 .05 .025 .01 .005

. . . . . . .

8 0.889 1.397 1.860 2.306 2.896 3.355

9 0.883 1.383 1.833 2.262 2.821 3.250

10 0.879 1.372 1.812 2.228 2.7638 3.169

11 0.876 1.363 1.796 2.201 2.7181 3.106

. . . . . . .


with df = 10

0

t Distribution

with df = 10


.0250

2.228

Interval Estimate of a Population Mean

s is the known population standard deviation

/ 2x zn

z/2 is the z value providing an area of /2 in the upper tail of the standard normal distribution

OROR

n > 2 if the data is normally distributedn > 30 if the data is roughly symmetricn > 50 if the data is heavily skewed

and

The 1 - confidence interval for unknown m

/ 2p zn

Interval Estimate of a Population Proportion

(1 )p p

z/2 is the z value providing an area of /2 in the upper tail of the standard normal distribution

andn > 5/pn > 5/(1 – p)

and

The 1 - confidence interval for unknown p


The 1 - confidence interval for unknown m

s estimates s because it is unknown

/ 2xn

st

t/2 is the t value providing an area of /2 in the upper tail of the t-distribution

OROR

n > 2 if the data is normally distributedn > 15 if the data is roughly symmetricn > 30 if the data is slightly skewedn > 50 if the data is heavily skewed

and

OR

Example: Discount Sounds

Discount Sounds has 260 retail outlets throughout the United States. The firm is evaluating a potential location for a new outlet, based in part, on the mean annual income of the individuals in the marketing area of the new location.

A sample of size of 36 was taken; the sample mean income is $31,100. The population is not believed to be highly skewed. The population standard deviation is estimated to be $4,500, and the confidence coefficient to be used in the interval estimate is .95.

n

s

1 - a

x


The margin of error is:

/ 2z

n

Thus, at 95% confidence, the margin of error is $1,470.

4,500

1.9636

1,470

Interval estimate of is:

We are 95% confident that the interval contains m.

$31,100 + $1,470

$29,630 to $32,570


Example: Discount Sounds

Political Science, Inc. (PSI) specializes in voter polls andsurveys designed to keep political office seekers informedof their position in a race. Using telephone surveys, PSI interviewers ask registered voters who they would vote for if the election were held that day.

Example: Political Science, Inc.

In a current election campaign, PSI has just found that 220 registered voters, out of 500 contacted, favor McSame. PSI wants to develop a 95% confidence interval estimate for the proportion of the population of registered voters that favor McSame.

1 - a

n


where: n = 500, p

PSI is 95% confident that the proportion of all voters

that favor McSame is between .3965 and .4835.

.44(1 .44).44 1.96

500= .44 + .0435

= 220/500 = .44, z.025 = 1.96

/ 2

(1 )p pp z

n



A reporter for a student newspaper is writing an article on the cost of off-campus housing. A sample of 16 efficiency apartments within a half-mile of campus resulted in a sample mean of $650 per month and a sample standard deviation of $55.

Example: Apartment Rents

Compute a 95% confidence interval estimate of the mean rent per month for the population of efficiency apartments within a half-mile of campus. We will assume this population to be normally distributed.

s1 - ax

n


At 95% confidence,


of Freedom .20 .100 .050 .025 .010 .005

. . . . . . .

13 0.870 1.350 1.771 2.160 2.650 3.012

14 0.868 1.345 1.761 2.145 2.624 2.977

15 0.866 1.341 1.753 2.131 2.602 2.947

16 0.865 1.337 1.746 2.120 2.583 2.921

. . . . . . .

t.025 is based on df = n - 1 = = .05, /2 = .025.

16 - 1 = 15


x tsn

.025

We are 95% confident that the mean rent per monthfor the population of efficiency apartments within ahalf-mile of campus is between $620.70 and $679.30.

Interval Estimate

55650 2.131

16

Est. Marginof Error

650 29.30 (620.70, 679.30)

Point Estimate of the population mean

Interval Estimate of the population meanInterval Estimate of the population mean


Recall that Discount Sounds is evaluating a potential location for a new retail outlet, based in part, on the mean annual income of the individuals in the marketing area of the new location.

Suppose that Discount Sounds’ management teamwants an estimate of the population mean such thatthere is a .95 probability that the margin of error is $500or less. How large should the sample be? E

1-a

n?

Recall that = 4,500.

Sufficient Sample Size for the Interval Estimateof a Population Mean

At 95% confidence, z.025 = 1.96.

311.172

22025. )(

E

zn

2 2

2

(1.96) (4,500)(500)

312

Suppose that PSI would like a .99 probability that the sample proportion is within .03 of the population proportion. How large should the sample be to meet the required precision? Recall that in the previous example, a sample of 500 similar units yielded a sample proportion of .44.

1-a

E

p*


Sufficient Sample Size for the Interval Estimateof a Population Proportion

At 99% confidence,

2 * *.005

2

( ) (1 )z p pn

E

a = .01, a/2 = .005, z.005 = 2.575

2

2

(2.575) (.44)(.56)(.03)

1816

.025.025

.950

1.25x

s = 10n = 64 a = 0.05

.025 xz 2.45

z.025 = 1.96

1.25x

1 31.25x

33.728.8 31.25

Repeated Sampling

2 28.15x

.025.025

.950

1.25x

30.625.7 28.15

s = 10n = 64 a = 0.05

.025 xz 2.45

z.025 = 1.96

1.25x

Repeated Sampling

3 33.25x

.025.025

.950

1.25x

30.8 35.733.25

s = 10n = 64 a = 0.05

.025 xz 2.45

z.025 = 1.96

1.25x

Repeated Sampling


http://www.halsnarr.com/ppt/data_election_polling.xlsx

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T distribution with 20 degrees of freedom t distribution with 10 degrees of freedom 0 t distribution with degrees of freedom t Distribution The t-distribution