Statistic Seminar, Chapter.7 TESTING THE SIGNIFICANCE

OF A SINGLE MEAN-The Single-Sample z and t tests-

Jun.13 2008

Yoshioka

Chapter 7

In this chapter, we deal with case I research. (cf. pp.126)

Case I Case II

z score

= the mean of the sample

= the hypothesized mean

= the standard deviation of the population

X

X

z

z score (reminder)

0

e.g. For alpha levels .05

z +3.0+1.0 +2.0-3.0 -2.0 -1.0

68%

95%Normal distribution

z statistics

xobt

Xz

x

X

= the mean of the sample

= the hypothesized mean

= the standard error of the mean x n

The sampling distribution

100104105

Hypothesized μ True μ

Sample X

Due to sampling error??

The acceptance and rejection regions

0

e.g. For alpha levels .05

+1.96-1.96z

95% 2.5%2.5%

Accept H0 Reject H0Reject H0

(Critical region)

t statistics (Formula 7.2 pp.154)

xobt s

Xt

x

ss

n

The mean of the sample

The hypothesized mean

The sample standard deviation

Sample size

Type I and Type II Errors

Correct

Correct

Type IIError

Type IError

True state of affairs

YourDecision

H0 is true H0 is false

Retain H0

Reject H0

Degree of Freedom

e.g.

You can 4 numbers whatever you like.

note the mean oh them should be 10.

a, b, c, 10-(a+b+c)

Standard Deviation

2( )

1

X Xs

n

2( )X

N

Sample SD Population SD

Example

Each Okadaken member takes 20 minutes to eat lunch on average. According 2weeks of investigation, Ikeuchi takes 22 minutes to eat lunch on average. Standard deviation of him was 2 minutes.

Compared to other members, is Ikeuchi a slow eater or not?

note: this is an imaginary story.

Solution

Identify the null and alternative hypotheses

H0 : μ ＝ 20

H1 : μ≠ 20

Set alpha as .05

Solution

Compute tobt using formulax

obt s

Xt

22 20

2 / 103.16

xobt s

Xt

Solution

Find the critical value for and The critical value is 2.262.

Compare the tobt of 3.16 to the critical value of 2.262. tobt is larger than the critical value, so H0 is rejected.

So Ikeuchi is ….

9df .05

Thank you for your attention!