SUBJECT: PROBABLITY AND INTRODUCTION TO STATISTICS (CODE-2142505)

BE Second Level Second Semester (Self Finance)

TOPIC: Test of Hypotheses

Prepared By:

GUJARAT TECHNOLOGY UNIVERSITYBIRLA 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

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)

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 populationHa: H0 is wrong (i.e., “weight gain”)

Next collect data quantify the extent to which the data provides evidence against H0

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) orHa: 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

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

T test for two samples

What is the probability that two samples have the same mean?

Sample A Sample B1 13 25 55 47 89 9

10 10Sample Mean 5.714286 5.571429

The T test Analysis

Go to the Data tab

Click on data analysis

Select t-Test for Two-Sample(s) with Equal Variance

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

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.

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

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.

z Test for Two Sample Means

Population standard deviation is unknown.

Must compute the sample variances.

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.

Chi-Square Test

A fundamental problem is genetics is determining whether the experimentally determined data fits the results expected from theory.

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.

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)( 22 obs