test of hypothysis

1. SUBJECT: PROBABLITY AND INTRODUCTION TO STATISTICS (CODE-2142505)
BE Second Level Second Semester (Self Finance)
TOPIC: Test of Hypotheses
Prepared By:
GUJARAT TECHNOLOGY UNIVERSITY
BIRLA VISHVAKARMA MAHAVIDYALAYA
(ENGINEERING COLLEGE)
VALLABH VIDYANAGAR
Name Enrollment No
Krushal Kakadiya 130080125006
Dhruvit Kardani 130080125007
Tirth Lad 130080125008
Milan Lakhani 130080125009
Vishal Lapsiwala 130080125010
Guided By: Prof. A.H.Jariya

2. Significance Testing
 Also called “hypothesis testing”
 Objective: to test a claim about parameter μ
 Procedure:
A. State hypotheses H0 and Ha
B. Calculate test statistic
C. Convert test statistic to P-value and interpret
D. Consider significance level (optional)

3. Hypotheses
 H0 (null hypothesis) claims “no difference”
 Ha (alternative hypothesis) contradicts the null
 Example: We test whether a population gained weight on average…
H0: no average weight gain in population
Ha: H0 is wrong (i.e., “weight gain”)
 Next  collect data  quantify the extent to which the data provides
evidence against H0

4. One-Sample Test of Mean
 To test a single mean, the null hypothesis is
H0: μ = μ0, where μ0 represents the “null value” (null value comes from the
research question, not from data!)
 The alternative hypothesis can take these forms:
Ha: μ > μ0 (one-sided to right) or
Ha: μ < μ0 (one-side to left) or
Ha: μ ≠ μ0 (two-sided)
 For the weight gain illustrative example:
H0: μ = 0
Ha: μ > 0 (one-sided) or Ha: μ ≠ μ0 (two-sided)
Note: μ0 = 0 in this example

5. Illustrative Example: Weight Gain
 Let X ≡ weight gain
 X ~N(μ, σ = 1), the value of μ
unknown
 Under H0, μ = 0
 Take SRS of n = 10
 σx-bar = 1 / √(10) = 0.316
 Thus, under H0
x-bar~N(0, 0.316) .
Figure: Two possible
xbars when H0 true

6. T test for two samples
 What is the probability that two samples have the same mean?
Sample A Sample B
1 1
3 2
5 5
5 4
7 8
9 9
10 10
Sample
Mean 5.714286 5.571429

7. The T test Analysis
 Go to the Data
tab
 Click on data
analysis
 Select t-Test for
Two-Sample(s)
with Equal
Variance

8. With Our Data and .05 Confidence
Level
t stat = 0.08
t critical for two-
tail (H1 = not
equal) = 2.18.
T stat < t Critical,
so do not reject
the null
hypothesis of
equal means.
Also, α is 0.94,
which is far larger
than .05

9. T Test:
Two-Sample, Equal Variance
 If the variances of the two samples are believed to be the same, use this
option.
 It is the strongest t test—most likely to reject the null hypothesis of equality
if the means really are different.

10. T Test:
Two-Sample, Unequal Variance
 Does not require equal variances
 Use if you know they are unequal
 Use is you do not feel that you should assume equality
 You lose some discriminatory power
 Slightly less likely to reject the null hypothesis of equality if it is true

11. T Test:
Two-Sample, Paired
 In the sampling, the each value in one distribution is paired with a value in
the other distribution on some basis.
 For example, equal ability on some skill.

12. z Test for Two Sample Means
 Population
standard
deviation is
unknown.
 Must compute
the sample
variances.

13. z test
 Data tab
 Data
analysis
 z test
sample for
two means.
13
Z value is greater than z Critical for two tails (not equal),
so reject the null hypothesis of the means being equal.
Also, α = 2.31109E-08 < .05, so reject.

14. Chi-Square Test
 A fundamental problem is genetics is
determining whether the experimentally
determined data fits the results expected from
theory.

15. Goodness of Fit
 Mendel has no way of solving this problem. Shortly after the rediscovery of his
work in 1900, Karl Pearson and R.A. Fisher developed the “chi-square” test for
this purpose.
 The chi-square test is a “goodness of fit” test: it answers the question of how well
do experimental data fit expectations.
 We start with a theory for how the offspring will be distributed: the “null
hypothesis”. We will discuss the offspring of a self-pollination of a heterozygote.
The null hypothesis is that the offspring will appear in a ratio of 3/4 dominant to
1/4 recessive.

16. Formula
 The “Χ” is the Greek letter chi;
the “∑” is a sigma; it means to
sum the following terms for all
phenotypes. “obs” is the
number of individuals of the
given phenotype observed;
“exp” is the number of that
phenotype expected from the
null hypothesis.
 Note that you must use the
number of individuals, the
counts, and NOT proportions,
ratios, or frequencies.



exp
exp)( 2
2 obs