Dr.

Ajit Patil

Parametric Tests

Difference between Parametric and Non-Parametric Tests

Basis Parametric Non-Parametric

Certain Assumptions Yes No

Value of central Mean value Median value

tendency

Correlation Karl Pearson Spearman’s rank

correlation

Probability distribution Normal distribution Arbitrary

Population knowledge Required Does not required

Data Type Interval data Nominal data

Applicability Variables Attributes & Variables

Examples z-test, t-test, Anova etc. Chi-square, Kruskal-

Wallis etc.

Parametric Tests:

 The word Parametric is made from “Parameter”. An aspect of population

is known as Parameter whereas an aspect of sample is known as statistic.
 The Parametric tests includes making certain assumptions about the
population parameters and the distribution from where data is collected.
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 These parameters include population mean (μ), Population standard

deviation (σ), Population Variance (σ2) and Median, mode and proportion
(ρ).

Conditions for Parametric test:

I. The data should be normally distributed (Bell-shaped curve).

 If the data will be perfectly normally distributed, both the sides of curve
will be exact mirror of each other.

Properties of Normal Distribution Curve

1. It is a bell-shaped curve which ranges from Minus infinity (-α) to plus

infinity (+α).
2. Mean=Median=Mode and the Quartile 1 (Q1) is equidistant from Quartile
3 (Q3).
3. The standard normal distribution has Mean = 0 and S.D.=1 and the curve
drawn from this is known as Normal Standard Curve.
4. The normal curve must not touch the baseline.
5. Approximately 68% of the observation (68.27% to be exact) may lie in the
range of Mean ± 1SD which means Mean -1SD and Mean +1SD.
6. Approximately 95% of the observation (95.45% to be exact) may lie in the
range of Mean ± 2SD which means Mean - 2SD and Mean + 2SD.
7. Approximately 99% of the observation (99.37% to be exact) may lie in the
range of Mean ± 3SD which means Mean - 3SD and Mean + 3SD.
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8. There are 3 other aspects for the consideration of Normal distribution.

a) Modality- The distribution must have a single peak (Unimodal).
b) Skewness- The distribution must be symmetric.
(Mean=Median=Mode)
c) Kurtosis- The kurtosis should be 0.
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 This distribution is also known as Gaussian Distribution.

II. The measurement scale of the variables should be in interval.
III. The variances of the data collected from various groups should be
homogeneous.
IV. The data should be independent.

All these conditions enable the usage of parametric tests.

Types of Parametric Tests

• Z Test
To test Difference • 't' Test
• ANOVA (F-Test)

• Karl Pearson Correlation

To test Correlation

 How to make an appropriate choice in test statistic for testing the

hypotheses about means?
Sample Size Knowledge about population standard deviation (σ)
Known Not Known
Large (N>30) Z Z
Small (N<30) Z T

Test Concerning Means

1. Case of Single Population

A) Hypothesis testing of mean when the Population standard deviation is

known.
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 Z-test will be appropriate in case of sample size is large or small but

the population standard deviation is known.
 Given the null hypothesis H0: μ = μ0, the test statistic will be:

Where, x̄= Sample mean

σ= Population standard deviation
μ= the value under which the assumption of null hypothesis will be true.
n = Sample size

Steps for calculation:

1. Calculate the Test statistic using the above formula.
2. Obtain the critical values by using the below table with given level of
significance and region.
Rejection Level of Significance, α per cent
Region α = 0.10 α = 0.05 α=0.01 α=0.005
One-tailed ±1.28 ±1.645 ±2.33 ±2.58
Two-tailed ±1.645 ±1.96 ±2.58 ±2.81

3. Compare both the values and reject the null hypothesis if,
a. If calculate value > critical value, in case of right-tailed test.
b. If calculate value < critical value, in case of left -test.
c. If calculate value < lower critical value and calculate value > upper
critical value, in case of two-tailed test.

Example: 1 (two-tailed test)

The mean weight of the paper clips is 60 grams having standard deviation of 48.
Furthermore, A researcher thinks that material Z would have a significant impact
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on the weight of a paper clips. Through random sample of 64 clips collected and
the mean weight is 68 grams. At 95% confidence level test the material Z had an
effect.

 State the null and alternative hypothesis

o Null hypothesis H0: µ = 60
o Alternative hypothesis H1: µ≠60
 Calculate the test statistics

o x̅ =68
o μ =60
o σ =48
o n =64

 Calculate the critical value by using the Z table at 0.05 level of

significance for two-tailed test.
o Z critical = ±2.81
 Compare Z cal. to Z critical
.

Interpretation: In hypothesis testing, Since, comparing Z cal. to Z critical, the

Z cal. value does not fall in rejection region. Hence, we fail to reject the null
hypothesis. We can say the mean weight of the Clips is 60 grams.

B) Hypothesis testing of mean when the Population standard deviation is

unknown.
 In this case, normality of distribution is required.
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 In case, if the population is not normally distributed, a large sample is

required because in a large sample, both the population and sample
standard deviation are almost same. So, the same method that was used
before (testing of mean when the Population standard deviation is known)
can be used.
 Therefore, the current method is often considered for small sample taken
out from a normal population.
 This distribution is student’s t-statistic, also known as t-test for single
sample.
 Thus, the test statistic will be:

 Where sample standard deviation (s), with (n-1) degrees of freedom can
be calculated as:

Steps for calculation:

1. Calculate the Test statistic using the above formula.

2. Obtain the critical values of t(n-1) distribution from the t-table for the given
significance level and degrees of freedom (d.f.).
3. Compare both the values and reject the null hypothesis if,
a. If calculate value > critical value, in case of right-tailed test.
b. If calculate value < critical value, in case of left -test.
c. If calculate value < lower critical value and calculate value > upper
critical value, in case of two-tailed test.
 Dr. Ajit Patil

Example:2 There is a certain type of Plinth beam of an under-construction

apartment in Noida and New Delhi that has an average thickness of Plinth
beams is µ = 67 cm. The team of the builder is studying these Plinth beams. A
random sample of Plinth beams of certain buildings gave the following
thickness (in cm).

59 51 76 38 65 54 49 62 68 55 64 67 63 74
65 79

Assume the thickness has an approximately normal distribution. Use a 1%

significance level to test the mean thickness in these certain buildings which is
different from that in Noida.

 State the null and alternative hypothesis

o Null hypothesis H0: µ = 67
o Alternative hypothesis H1: µ≠67
 Calculate the test statistics

o x̅ =61.8
o μ =67
o S =10.6
o n =16
o d.f. = 16-1=15

.
. = -1.962
√
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 Calculate the test statistic with the help of t-table at 0.01 level of
significance and 15 degrees of freedom.
o Test statistic = 2.947

 Compare the tcalculated with tcritical.

Interpretation: In hypothesis testing, since comparing t cal. to t critical, the t

cal. value does not fall in rejection region. Hence, we fail to reject the null
hypothesis. We can say the mean thickness of Plinth beams 67 cm.

2. Case of Two sample drawn

 Includes comparing of two sample means either from related or unrelated
groups or population.
 Arises two cases that are Paired tests and Unpaired Tests.
 Applied in the case of N<30.
 Also known as Paired t-test and Unpaired t-test.

a. Unpaired t-Tests
 Applied to compare the means of two unrelated or independent groups to
analyze if there is a significant difference between the two or not.
 Also referred as independent t-Test. In unpaired t-test, there are two cases
which need to be analyzed with the help of different statistics which are as
follows.

Case-1 Both the samples are unrelated or independent of each other and
both population variances are known.

The Test statistic will be:

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Since, the value of parameter (μ1-μ2) under null

hypothesis is zero (0), therefore the test statistic will be:

Where, x̄1 = Mean of sample 1 s1 = standard deviation of population1

x̄2 = Mean of sample 2 s2= standard deviation of population1

N1= No. of items in sample 1 N2= No. of items in sample 2

Steps of Calculation:

1. Calculate the t-statistic using above formula.

2. Obtain the critical values of distribution from the t-table for the given
significance level and degrees of freedom (d.f.). Degrees of freedom will
be calculated as:
{(N1-1) + (N2-1)} = (N1+N2-2)
3. Compare both the values and reject the null hypothesis according to the
conditions.

Example 3: A study was conducted to see the effect of two treatments on

certain group, the following results were obtained. Test statistically to see
whether the difference noted in the recovery time is significant statistically and
interpret from both the sample shown below.
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Group Methods of Sample size Mean no. of S. D

treatment recovery
time
I Placebo 13 8 3

II Vitamin 13 4 2

Given,

𝑛 = 13 & 𝑛 = 13 (equal sample size)

𝑥̅ = 8

𝑥̅ = 4

𝑠 =3

𝑠 =2

 State the null and alternative hypothesis

o Null hypothesis H0: µ1= µ2
o Alternative hypothesis H1: µ1≠ µ2
 Calculate the t statistic.

̅ ̅
𝑡 =

= = =4

tcal = 4
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 df = (N1+N2-2) = 13+13 – 2 = 24
 Obtain the critical values of distribution from the t-table for the given
significance level (10%) and degrees of freedom (24).
o t critical = 1.711
 Compare the tcalculated with tcritical.

Interpretation: In hypothesis testing, since comparing t cal. to t critical, the t

cal. fall in rejection region. Hence, we reject the null hypothesis and alternate
hypothesis is accepted. So, we can say that there is a significant difference in
the mean time of recovery by two methods of treatment.

Case-2 Both the samples are unrelated or independent of each other and
both population variances are unknown.

 Assume that both the populations have normal distribution and both the
population variances are equal. So, in this case unknown equal variances
are calculated through Pooled sample variance. Thus, the test statistic will
be:

Where, sp = Pooled sample variance , which can be calculated by using formula

below:

steps for calculation will remain same as the previous method.

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Example 4: In a study to assess the Symptoms of Covid-19 and its

complications, the following data were obtained. Test statistically whether the
difference in the mean symptoms of the two groups in significant?

Groups Sample size Mean S. D

Major 6 10 3
complications
16 12 1
Minor
complications

𝑛 =6 𝑛 = 16

𝑥̅ = 10 𝑥̅ = 12

𝑠 =3 𝑠 =1

Degree of freedom = 𝑛 + 𝑛 − 2

= 6 + 16 – 2

= 20

 Calculate the t statistic:

and

|𝑥̅ − 𝑥̅ |
𝑡 =
𝑛 + 𝑛 (𝑛 − 1)𝑆 + (𝑛 − 1)𝑆
[ ]
𝑛 ×𝑛 𝑛 +𝑛 −2

|10 − 12|
𝑡 =
6 + 16 5 × 9 + 15 × 1
6 × 16 20
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𝑡 =
96

2
𝑡 = = 2.4
√0.7
 Obtain the critical values of distribution from the t-table for the given
significance level (5%) and degrees of freedom (20).
o t critical = 2.09
 Compare the tcalculated with tcritical.

Interpretation: In hypothesis testing, since comparing t cal. to t critical, the t

cal. fall in rejection region. Hence, we reject the null hypothesis and alternate
hypothesis is accepted.

b. Paired t-Test (When two samples are related)

 Used when the two samples are not independent else.
 Applied to compare the means of two related groups to analyze if there is
a significant difference between the two or not.
 Takes in account before and after trial on related groups.
 Also known as correlated or dependent t-test.
 For this, the test statistic will be:

Where, d bar= Mean of difference

sd= Standard deviation of differences
n= No. of pair of observation
Steps for calculation:
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1. Calculate the difference in before and after observations is each set. (X1-
X2=d)
2. Calculate the mean of these differences. (d bar)
3. Calculate the standard deviation of those differences. (s d)
4. Find out the t-statistic using the above formula.
5. Obtain the critical values of distribution from the t-table for the given
significance level and degrees of freedom (n-1).
6. Compare both the values and reject the null hypothesis according to the
conditions.

Example 5: Sugar level of 9 normal individuals, who had been recumbent for 5
minutes was taken. Then 2ml of 0.5% solution of Insulin was given and the
sugar level recorded again. Did the Insulin shots lower the sugar level?

Serial No. Sugar level Sugar level Difference Square

without with
Insulin Insulin d =𝒙𝟏 − 𝒙𝟐 𝒅𝟐
𝒙𝟏 𝒙𝟐
1. 122 120 2 4
2. 121 118 3 9
3. 120 115 5 25
4. 115 110 5 25
5. 126 122 4 16
6. 130 130 0 0
7. 120 116 4 16
8. 125 124 1 1
9. 128 125 3 9
Total ∑𝑑 = 27 𝛴𝑑 = 105
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 Calculate the mean of differences

∑
Mean of difference 𝑑̅ =

= =3
 Calculate the standard deviation of those differences

(∑𝒅)𝟐
∑𝒅𝟐
𝒏
S. D =
𝒏 𝟏

𝟐𝟕×𝟐𝟕
𝟏𝟎𝟓 𝟏𝟎𝟓 𝟖𝟕
𝟗
=
𝟗 𝟏
= 𝟖

𝟐𝟒
S. D =
𝟖
= √𝟑 = 𝟏. 𝟕𝟑

 Calculate the standard error

𝐒⋅𝐃
S. E =
√𝐧
. .
S. E = = = 0.58
√

Degree of freedom = n -1 = 9 – 1= 8

 Find out the t-statistic

𝑑̅
𝑡 =
𝑆. 𝐸

=
.
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𝑡 = 5.17

 Obtain the critical values of distribution from the t-table for the given
significance level of 5% and degrees of freedom (8).

At 8 degree of freedom, 5% significant limit of ‘t’ is 2.306.

 Compare the tcalculated with tcritical.

Interpretation: In hypothesis testing, since comparing t cal. to t critical, the t

cal. fall in rejection region. Hence, we reject the null hypothesis and alternate
hypothesis is accepted. Hence, there is no doubt that the Insulin shots produced
positive effect, t = 5.17.

Test Concerning Population Proportion

 Describes whether respondents or a proportion of population possess a

specific characteristic or not.
 For example, one can have the knowledge regarding the proportion of
students skilled in a particular subject or proportion of consumers who
make use of a specific product.

Case of Single Population proportion

In this case, the test statistic will be,

Where,
p bar= sample proportion
p0- Probability of success
q0 = Probability of failure
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= (1-p)
n= Sample size

1.1 Case of Two Population proportions

In this case, the test statistic will be, -

Where,

P1 bar= sample proportion of population 1

P2 bar= sample proportion of population 2
N1= Sample size 1
N2= Sample size 2

ANOVA (ANALYSIS OF VARIANCE)

 Simple method of comparing means of more than two populations at single

time i.e. ANOVA.
 Helps to find out whether the different samples drawn from population
have the same means or not.
 The theory of ANOVA was developed by R A Fisher.
 The important principle in this technique is that the total amount of
variation in the dependent variable is broken into two parts, the one is
amount which can be attributed to some specified causes and the other is
amount which can be attributed to chance.
 Variation may be present between samples and within samples as well.
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 Variation between sample refers to amount which can be attributed to some

specified causes while variation within sample refers to amount which can
be attributed to chance.
 The independent variables are measured on Categorical (Nominal) scale.
 The dependent variables are measured on Interval or ratio scale.
 If one independent variable (one factor) is categorized into numerous
groups, then one can use one -way ANOVA or one-factor ANOVA.
 If the two-independent variable (two factor) is categorized into numerous
groups, then two-way ANOVA or two-factor ANOVA is utilized.

Assumptions in ANOVA

1. Sample is drawn from a normally distributed population.

2. Each of these population has an equal variance.
3. All the variables are controlled or constant except the one, which is tested.

 Thus, to measure the variances, the total variance in the total (joint) sample
is broken into two parts.
1. Between sample variances (Caused by the different treatments)
2. Within sample variances (Caused by the random unexplained disturbance)

 With the help of these two variances, the test statistic will be:

Fc= Between sample Variance

Within Sample Variance

H0: All the population means are equal. (μA= μB= μC).

H1: All the population means are not equal. (μA= μB ≠ μC).
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1. Annexures
1.1 Some critical values of “t” (t-table)
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1.2 Critical values for F-distribution (at 10 percent)

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1.3 Critical values for F-distribution (at 5 percent)

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6.4 Critical values for F-distribution (at 1 percent)