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!
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