Test of Significance

Tests of significance Dept of AGB Veterinary College, SHIMOGA

Common types of Comparisons to be made • Comparison of sample mean with population mean • Comparison of two sample means • Comparison of sample proportions with the population proportions • Comparison of two sample proportions

STAT 512 2018-19 VCS AGB RJ

Procedure for estimation • Finding out the type of problem, size of sample and the question to be answered.

• Stating or setting up of Null hypothesis.

• Determining the correct sampling distribution and calculating standard error.

• Calculating the critical difference • Comparison of estimated or calculated value with table value.

• Drawing conclusion STAT 512 2018-19 VCS AGB RJ

The procedure is as follows • • • •

Hypothesis Test statistics Comparison of calculated value Inference

The hypothesis(Ho) is formulated as there is no significant differences between the sample mean and the population mean or the sample can be regarded as the representative of the population.

STAT 512 2018-19 VCS AGB RJ

Large sample test when the population variance is known The test statistic for testing null hypothesis is ‘Z’

Z=

x SEx

or

Z

x S

2

n

X= Sample mean, µ= Population mean, Σ2 or S2 = standard deviation of sample, n= no. of units in sample STAT 512 2018-19 VCS AGB RJ

Test for small sample size The test statistics for testing the hypothesis is students ‘t’ test whenever the sample size is less than 30.

x t / n

t

x S

2

n

Whenever the sample size is more than 30 it can be considered as a large sample and approximated to normal population.

STAT 512 2018-19 VCS AGB RJ

Problem 1 • The yield of butter fat produced in a random sample of ten cows are as follows:65,39,28,36,50,34,46,36,54,52.

• Test the hypothesis that the population mean is equal to 40 lbs (level of significance is 0.05) •Comparison of sample mean with population mean STAT 512 2018-19 VCS AGB RJ

Comparing sample mean with population mean.

Calculate



x

 xx 2 S  n 1



t

x S

2

n

and S2

2 or

2 

Σ x  2  Σ x n S2  n 1

STAT 512 2018-19 VCS AGB RJ

  

For comparing the means of two independent samples ‘t’ test for this is given by

x x 1 2 t Sp2(n1  n1 ) 1 2 STAT 512 2018-19 VCS AGB RJ

s1 n1  1  s 2 n2  1 n1  n2  2 2

Sp2

=

2

Sp2 = Pooled variance S12 = Variance of 1st group S22

= Variance of 2nd group

The calculated ‘t’ value is compared with table value at n1+n2-2 degrees of freedom STAT 512 2018-19 VCS AGB RJ

A group of seven chickens were reared on a high protein diet weighed 13,12,15,17,14,17 and 15 ounces respectively. The weights of the second group consisting of 5 chicks similarly treated except they were on low protein diet was 9,15,14,11 and 11 ounces.

Test whether there is any significant difference between high and low protein diet.

STAT 512 2018-19 VCS AGB RJ

x x 1 2 t Sp2(n1  n1 ) 1 2

s1 n1  1  s 2 n2  1 n1  n2  2 2

Sp2 =

2

STAT 512 2018-19 VCS AGB RJ



 xx 2 S  n 1



2

or S 2 

 x1  

x

x1 S1  2

n1 n1  1

 x   n 1

2

2

2

2

n

 x2  

2

x2 S2  2

STAT 512 2018-19 VCS AGB RJ

2

n2 n2  1

Paired ‘t’ test for difference of mean • The test consists of testing the average of the differences in the pairs of values recorded on the same variable on same character at two different periods.

d t SE d

d= difference Xi & Yi

d

=

d n

STAT 512 2018-19 VCS AGB RJ

SE d 

S d  n

d= mean of the difference between the paired values

S d  

d

2

 d   n 1

2

n

The calculated ‘t’ value is compared with table value at n-1 degrees of freedom STAT 512 2018-19 VCS AGB RJ

Ten experimental animals were subjected to conditions simulating disease. The number of heart beat per minute before and after the experimentation was recorded as follows Before: 70,84,88,110,105,100,110,67,79,86

After: 115,148,176,191,158,178,179,140,161,157 Do these data provide sufficient evidence to indicate that the experimental condition increased the number of heart beats per minute?

STAT 512 2018-19 VCS AGB RJ

d t SE d

SE d 

S d  

d

2

S d  n

 d   n 1

n

STAT 512 2018-19 VCS AGB RJ

2

Comparison of sample proportion with population proportion • If ‘X’ is the number of individuals possessing the given character or attribute in a random sample of size ‘n’ from a large infinite population then • P=X/n, P is the population proportion.

Z

P p PQ n

P= proportion of individuals having this attribute in the population Q=1-P STAT 512 2018-19 VCS AGB RJ

In random sample of 500 cows, 65 cows were found to be unproductive. Test whether population proportion is equal to 0.15(P).

Z

P p PQ n

p=X/n p=65/500 =0.13 P=0.15 Q=1-0.15 =o.85

STAT 512 2018-19 VCS AGB RJ

Comparison of two sample proportions • If two independent large samples of n1 and n2 are taken from population A and population B respectively and let X1 and X2 be observed units of individuals possessing the given attribute in the above sample respectively • Then

x1 p1  n1

x2 p2  n2

STAT 512 2018-19 VCS AGB RJ

Test statistics for difference of proportions

Z

p1  p2 1 1 PQ    n1 n2 

n1 p1  n2 p2 P n1  n2 STAT 512 2018-19 VCS AGB RJ

• For the following data find out whether the mortality rate is higher in local cattle in comparison to exotic cattle.

Local cattle Exotic cattle Total animals 246 349 No. of deaths 36 61

STAT 512 2018-19 VCS AGB RJ

Z

p1  p2 1 1 PQ    n1 n2 

n1 p1  n2 p2 P n1  n2 STAT 512 2018-19 VCS AGB RJ

x1 p1  n1 x2 p2  n2