Lecture 7 Hypothesis Formulation and testing - PowerPoint PPT Presentation

1 / 33
About This Presentation
Title:

Lecture 7 Hypothesis Formulation and testing

Description:

Survey results: 0.36. Var(p) 0.6. Stdev(p) 4.2 ... Survey research on Aquaculture Productivity: 0.86. 0.95. SD. 5.24. 3.05. Mean. District B ... – PowerPoint PPT presentation

Number of Views:1791
Avg rating:3.0/5.0
Slides: 34
Provided by: solo56
Category:

less

Transcript and Presenter's Notes

Title: Lecture 7 Hypothesis Formulation and testing


1
Lecture - 7Hypothesis Formulation and testing
Continued
2
Hypothesis testing tools
  • Testing for goodness of fit
  • Mostly nominal data e.g. counts or frequencies of
    items or events of groups but continuous data can
    also be tested
  • 1. Chi-square or ?2 test
  • - Karl Pearson (around 1900)
  • - Binomial normal distributions
  • - Discrete or continuous series
  • 2. Kolomogorov-Smirnov (K-S) test
  • - Normal distribution (Discrete or continuous)

3
Hypothesis testing tools
  • H0 there is no difference between observed (O)
    and expected (E) frequencies.
  • H1 there is a difference
  • Chi-square (?2) test
  • Example 1 In a plant breeding trial, if you get
    78 plants with yellow flowers and 22 green
    flowers (ratio 3.51). It was supposed to be 31
    ratio, test it whether they are statistically
    different or not.

4
Hypothesis testing tools
?2 test ?2 ? (O-E)2/E (78-75)2 /75
(25-22)2 / 25 32/7532/25 0.120.36
0.48 d.f. No. of classes 1 2-11 P0.49
which is 0.5 gt p gt 0.25
Therefore, Accept H0 which means the ratio of
yellow and green flowers is statistically similar
to the ratio found by Mendel i.e. 31
5
(No Transcript)
6
Hypothesis testing tools
Example 2
  • This comes from
  • 9331 ratio.
  • Y- Yellow, W-wrinkled
  • S smooth G-green
  • Here,
  • d.f. 4-1 3
  • 2 0.05,3 7.815 and ? 2 0.01,3 11.34 ,
    Plt0.05,
  • Therefore, Reject Ho, means the ratios are
    different

7
Hypothesis testing tools
Example 3 Application
Test whether a coaching class (intervention
program) has impacts on Students performance or
not?
,
,
,
,
8
Hypothesis testing tools
Example 3 contd..

22.01.
d.f. (r-1) (c-1) (2-1) ( 2-1) 1 ?2 0.01, 1
6.63 (from table) Plt0.05, therefore, reject Ho
which means the coaching (intervention program)
has positive impacts or has helped graduate more
students and should be recommended.
9
Hypothesis testing tools
Example 4 Continuous series
Fit the normal curve from the observed frequencies
use the following formula to find the
standardized frequencies
No. of fish (freq)
Fish size (g)
10
Hypothesis testing tools
Example 4 Continuous series e.g. test whether
size distribution of fish at harvest is normal
Fish size (g)
11
Hypothesis testing tools
Example 4 Continuous series e.g. test whether
size distribution of fish at harvest is normal
Fish size (g)
12
2. Kolmogorov-Smirnov (K-S) test - for discrete
  • Considers the cumulative frequency
  • More expanded form compared to Chi-square test
    (frequency classes or category)
  • D max maximum difference in cumulative freq.
    i.e. / d /
  • Example
  • A total of 35 dogs were tested individually which
    feed they prefer one by one (not in a group in
    which action of one may affect others)
  • H0 Moisture content of diet do not affect
    preference by dog
  • HA Dog prefer diets with high moisture content

13
2. Kolmogorov-Smirnov (K-S) test - for discrete
  • D max max. difference in cumulative freq. i.e.
    /d/ 9
  • No. of classes (k) 5 and total no. of data (n)
    35
  • Dmax, 0.05, 5, 35 7, reject H0

14
(No Transcript)
15
(No Transcript)
16
Hypothesis testing tools
2. Kolmogorov-Smirnov (K-S) test (For
continuous data)
Chi-square test
K-S test
17
2. Kolmogorov-Smirnov (K-S) test continuous
series (grouped data) H0 Distribution of moth in
a tree trunk is not different (or the
distribution is homogenous). HA More moths are
found lower parts of the tree trunk
  • D max max. difference in cumulative freq. i.e.
    /d/ 4
  • No. of classes (k) 5 and total no. of data (n)
    15
  • Dmax, 0.05, 5, 15 5, Do not reject H0 ie.
    Distribution of moths on the tree trunk is
    similar (not heterogeneous)

18
Comparison of two samples
No sig. difference No sig.
difference Sig. different (overlap is lt5)
19
Comparison of two means
  • Population mean (?) and sample mean (x)
  • Hypothetical mean and sample mean
  • Two sample means (e.g. Sample A B)

20
Z - test
  • Comparison between two means
  • Population mean (?) and sample means (x )
  • Use population standard deviation
  • Sample size should be more than 30 so that the
    distribution need not be normal

Z ( x - µ ) / s vn
21
Students t - tests
Student" (real name W. S. Gossett
1876-1937) - developed the method.
  • Comparison between two means
  • Population mean (?) and sample means (x)
  • One treatment sample vs hypothetical or standard
    value
  • Two treatment means
  • Sample size can be lt30 but they have to have
    normal distribution
  • Use sample Standard Deviation

22
Students t - tests
  • Types
  • One-sample t-test
  • Paired/matched sample t-test
  • Independent sample t-test
  • One-sample or single sample t-test
  • Comparison of a sample mean against a known
    standard or known number (often 0).
  • Example
  • H0 Tilapia do not grow in cold season lt20C (0
    g/fish/day)
  • HA Tilapia grow even below 20C (gt0 g/fish/day)
  • A trial was conducted using 4 ponds stocked with
    1 fish/m2, we found 1.0 g DWG and 0.1 SD

23
  • One-sample or single sample t-test
  • H0 Test the hypothesis whether tilapia really
    grew or not or the 1.0 g is significantly higher
    than 0 or not?
  • Here, µ 0, x 1.0g, s 0.1g

24
(No Transcript)
25
Two-sample t-test
  • Calculate the pooled variance
  • As the sample size can be different, pooled
    variance is calculated taking weighted average of
    the variance.
  • Population pooled Variance s12 /n1 s22 /n2
  • Sample Pooled Variance S12/n1 S2 2 /n2
  • Degree of freedom n1 n2 -2
  • Calculate test statistics and compare with the
    tabulated value for P

26
Example Rats injected with a drug to increase
pain sensitivity from a heat source in 4.3
seconds, The mean difference of 2.8 seconds is
statistically significant (t(12) 2.33, p
.038). Thus, the drug indeed increased pain
sensitivity.
Control   2.6   10   9.5   7.4   6.9   8.5   5.2
Drug    2.2   3.8   7.1   2.7   6.2   5.3   
3.1
MS Excel Results
27
Survey research on Aquaculture Productivity
  • Suppose, a student gathered data from 35 farmers
    in two districts using survey method from to
  • - compare productivity between districts A B
  • compare District A and B separately against the
    national average.

District A District B
28
(No Transcript)
29
Results
30
Paired-sample t-test
  • To compare means on the same or related subject
    over time or in differing circumstances
  • t Mean difference between pairs / SE of mean
    difference
  • d / svn
  • N no. of pairs

Data set 1 Data set 2
Pond A Pond B
1 2 3 4 5 6
Sampling points (weeks - pairs)
31
Paired-sample t-test
  • Note Another aspect of these type of data is to
    see the correlation
  • - There can be both or only one
  • A B correlated (may or may not differ)
  • A C No correlation, sig. difference
  • B C No correlation, no difference

32
Paired-sample t-test
  • Example for paired t-test try in the lab
    session.

33
Lab session after 15 minutes
Statistical principles and t-tests http//www.ph
ysics.csbsju.edu/stats/t-test.html http//www.stat
soft.com/textbook/stathome.html
Thank you!
Write a Comment
User Comments (0)
About PowerShow.com