Parametric tests S.No Type of Test Objectives Formulae 1 Z test a To find the significance of difference between mean of population and mean of sample when the sample size is large [> 30 units ] x Z / n b To find the significance of difference between the means of two large samples 1 2 2 2 1 2 1 2 x x Z x x n n c To find the significance of difference between population proportion and sample proportion ^ p p Z p.q n d To find the significance of difference between two sample proportions [In case of percentages, they are converted into proportion. eg. 60% as 0.6] ^ ^ 1 2 ^ ^ ^ ^ 1 1 2 2 1 2 p p Z p .q p .q n n 2 ‘t’ – test a To find the significance of difference x t s/ n
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Parametric tests
S.No Type of Test Objectives Formulae
1 Z test a To find the
significance of
difference
between mean of
population and
mean of sample
when the sample
size is large [>
30 units ]
xZ
/ n
b To find the
significance of
difference
between the
means of two
large samples
1 2
2 2
1 2
1 2
x xZ
x x
n n
c To find the
significance of
difference
between
population
proportion and
sample
proportion
^
p pZ
p.q
n
d To find the
significance of
difference
between two
sample
proportions [In
case of
percentages, they
are converted
into proportion.
eg. 60% as 0.6]
^ ^
1 2
^ ^ ^ ^
1 1 2 2
1 2
p pZ
p .q p .q
n n
2 ‘t’ – test a To find the
significance
of difference
xt
s / n
between mean of
population and
mean of sample
when the same
size is small [<
30 units] and
population
variance is not
known
b To find the
significance
of difference
between means
of two small
samples[or
atleast one is a
small sample]
1 2
x yt
1 1s
n n
Where
2 2
1 x 2 y
1 2
n 1 s n 1 ss
n n 2
3 Chi-Square
test as a
parametric
test
To find out the
significance of
difference
between
population
variance and
sample variance
2
2
2
s n 1
Types of commonly employed parametric tests, their objectives and the formulae
Notes :
N = Size of population [No. of units in the population]
n = Size of sample [No. of units in the sample]
1n = Size of sample 1
2n = Size of sample 2
= Mean of population
x = Mean of sample
1x = Mean of sample 1
2x = Mean of sample 2
2 = Variance of population
= Standard Deviation of population
s2 = Variance of sample
s= Standard Deviation of sample or combined S.D s of two samples or S.D of differences
2
1s = Variance of sample 1
2
2s = Variance of sample 2
1s = Standard Deviation of sample 1
2s = Standard Deviation of sample 2
d = Difference of value in the same sample unit [before and after treatment ]
d = Mean of differences
p = Proportion in population
q = 1 – p
q1 = 1 – p1 ( p1 = proportion in population 1)
q2 = 1 – p2 (p2 = proportion in population 2)
^
p = Proportion in sample [ ^
p is read as p hat ]
^
q = 1 – ^
p [^
q is read as q hat ]
^
1p = Proportion in sample 1
^
2p = Proportion in sample 2
^ ^
1 1q 1 p
^ ^
2 2q 1 p
r = Correlation coefficient
Z – test
Z- test, a parametric test, is employed to find out the significance of difference between
i. sample mean and population mean
ii. two sample means
iii. proportion in sample and proportion in population and
iv. two sample proportions.
Z test is used when the sample size is large [≥30 subjects) and population variable is known. If
population variance is not given, the sample variance can be taken as population variance as the
sample is large. In case the sample size is small (≤ 30) but normal population variance is known,
Z test can be used. To find out the critical Z value (table value) Z-distribution is used. As finding
out the table Z value is a little time consuming the table Z-values for two-tailed test and one-
tailed tests for commonly used levels of significance are presented in the table 9.2.
Type of
Test
Significance level
10% 5% 2.5% 2% 1%
Two – tailed 1.65 1.96 2.24 2.33 2.58
One-tailed
[left]
-1.28 -1.65 -1.96 -2.05 -2.33
One –tailed
[right]
+ 1.28 + 1.65 +1.96 +2.05 +2.33
Critical (table) Z values
To test the significance of difference between population mean and
sample mean [when sample size is large and population variance is known]
Example 1: A particular variety of wheat plants has shown a mean height of 82.63cm and
standard deviation of 3.89cm. From this population of plants, 50 plants are selected at random
and each plant is inoculated with a chemical which claims to increase the height of plants. After
inoculation the mean of the height of 50 plants is found to be 83.66cm. On the basis of this
evidence can it be now concluded that the chemical has a beneficial effect on the growth of
plants at 5% level of significance?
There is a numerical difference of 1.03cm (83.66 – 82.63cm) in the means. Now the
question is, “is the difference significant?” x
Solution
Null hypothesis (Ho): The chemical has no effect on the growth of wheat plants. That is,
the mean height of the sample 83.66cm does not differ significantly from the population mean of
82.63cm. The increase in height of 1.03cm is only due to chance factor and not due to the
chemical.
Alternative Hypothesis (Ha): The chemical has a positive effect on the growth of wheat
plants and the mean height of the sample 83.66cm significantly differs from the population mean
of 82.63cm. The increase in height of 1.03cm is due to the effect of chemical and not due to
chance factor.
Given Data: µ = 82.63cm, x = 83.66cm, n = 50, S.D = 3.89
Test criterion: Here, the sample mean is to be compared with population mean. As S.D of
population is known and the sample size is large (>30), Z test is used.
Calculated Z value:
xZ
/ n
= 83.66 82.63
1.873.89 / 50
Table value : Table Z value at 5% level of significance is 1.96 [As finding the table Z
value at different levels of significance is a little difficult it is better to remember the table Z
values for the commonly used levels of significance. Table Z value at 10% level is 1.65; at 5%
level 1.96 and at 1% level 2.58.
Interpretation: As the calculated Z value (1.87) is less than the table Z value of 1.96, the
null hypothesis is accepted.
Inference: Though there is an increase in the height of wheat plants consequent to the
application of the chemical, the increase is not statistically significant. That is, the observed
increase of 1.03 cm is mostly due to chance factor and not entirely due to the chemical.
Therefore, it is inferred that the chemical has no effect to increase the height of wheat plants.
Example 2: The overweight ladies taking brisk walks in the Race Course in Coimbatore,
had a mean weight of 87kg two years ago. Now, a sample of 40 ladies is taken at random and the
mean weight is found to be 72kg. Variance is found to be 25kg. Test at 1% level of significance
whether brisk walk is effective in reducing weight.
Solution
Null hypothesis (Ho): Brisk walk has no effect in reducing weight x where is the
mean weight of ladies 2 years ago and x is the mean weight of 40 ladies after 2 years of
walking.
Alternative hypothesis (Ha): Brisk walk reduces body weight:
Given data: = 87 kg. x : 72 kg, 2 = 25kg, S.D = 2 = 25 = 5; n = 40
Test criterion: Here, the sample mean is to be compared with the population mean. As S.D of
population is known and the sample size is large (>30), Z test is used
Calculated Z value:
.x
Z/ n
= 87 72 15
0.795/ 40
= 18.97
[In test values the sign is ignored]
Table value: The table Z value at 1% level of significance is 2.58
Interpretation: As the calculated Z value of 18.97 is greater than the table value of 2.58, the null
hypothesis is rejected.
Inference: The mean reduction in weight of 15kg consequent to brisk walk is not a chance factor.
It is inferred that continuous brisk walk helps in reducing body weight.
To test the significance of difference between means of two large samples.
Example 1: A social worker desires to find out whether there is any significant difference in
daily wages between the plantation workers in Valparai [plantation Zone in Coimbatore dt. of
Tamilnadu) and Idukki in Kerala. The data collected are given below:
Plantation Zone Sample size
Sample mean
(Rs)
Population variance
(Rs)
Valparai 80 82 125
Idukki 70 85 136
Differences in daily wages in plantations (Example)
Test the difference at 5% level of significance.
Solution
Null Hypothesis (Ho): There is no significant difference in daily wages between the workers in
Valparai and that of Idukki i.e the wages are equal (1 = 2 ). As the samples are large (>30), it is
assumed 1= 1x and 2= 2x .
Alternative hypothesis (Ha): There is difference in daily wages i.e. they are not equal (1 ≠ 2).
Given data : 1x = 82 , 2x = 85 , 1n = 80
2n = 70 , 2
1 = 125, 2
2 = 136
Test criterion : Here, two sample means are compared for significance of difference. As the
standard deviations of the populations are known and the samples are large, Z test is used.
1 2
2 2
1 2
1 2
x xZ
x x
n n
= 82 85 3 3
1.87125 136 1.56 1.94
80 70
= -1.60 = 1.60 (sign is ignored)
Table Z value at 5% level of significance is 1.96
Interpretation: As the calculated Z value of 1.60 is less than the table value of 1.96, the N.H. is
accepted.
Inference: Though the average daily wage of Idukki workers is Rs.85 and that of Valparai
workers is Rs.82, statistically there is no significant difference. The numerical difference is only
a chance factor.
Example 2: Intelligence test given to one group of girls and another group of boys showed the
following results :
Gender Number of Students
Tested Standard Deviation
Mean
Intelligence Score
Girls 50 10 75
Boys 100 12 70
Scores of intelligence tests of students (example)
Is the difference in the mean scores statistically different at 1% level of significance?
Solution
N.H : There is no difference in the intelligence test scores of girls and boys i.e. they are equal
(1 = 2)
A.H : There is difference i.e they are not equal (1 ≠2)