Testing of hypothesis

Type I
 To test the significance difference
between sample proportion p’ and
population proportion p, the test
statistic 𝒛 =
𝒑′−𝒑
𝒑𝒒
𝒏
, 𝒒 = 𝟏 − 𝒑 ,
n – sample size

Type II
between two sample proportions
𝒑 𝟏
′
𝒂𝒏𝒅 𝒑 𝟐
′
 The test statistic 𝒛 =
𝒑 𝟏
′
−𝒑 𝟐
′
𝒑𝒒
𝟏
𝒏 𝟏
+
𝟏
𝒏 𝟐
, 𝒒 =
𝟏 − 𝒑, 𝒑 =
𝒏 𝟏 𝒑 𝟏
′
+𝒏 𝟐 𝒑 𝟐
′
𝒏 𝟏+𝒏 𝟐
,
𝒏 𝟏& 𝒏 𝟐 sample sizes

Type III
between sample mean 𝒙 and
population mean 𝝁
𝒙−𝝁
𝝈/ 𝒏
, if 𝝈 is
known
𝒙−𝝁
𝒔/ 𝒏
, 𝐢𝐟 𝐬𝐚𝐦𝐩𝐥𝐞 𝐒. 𝐃 𝐬 𝐢𝐬 𝐤𝐧𝐨𝐰𝐧

Type IV
 To test the significance difference between two
sample mean 𝒙 𝟏 𝒂𝒏𝒅 𝒙 𝟐
𝒙 𝟏−𝒙 𝟐
𝝈 𝟏
𝟐
𝒏 𝟏
+
𝝈 𝟐
𝟐
𝒏 𝟐
, if 𝝈 𝟏 ≠ 𝝈 𝟐 and 𝝈 𝟏&𝝈 𝟐
are known
𝒙 𝟏−𝒙 𝟐
𝝈
𝟏
𝒏 𝟏
+
𝟏
𝒏 𝟐
, if 𝝈 𝟏 = 𝝈 𝟐 = 𝝈
𝒙 𝟏−𝒙 𝟐
𝒔 𝟏
𝟐
𝒏 𝟏
+
𝒔 𝟐
𝟐
𝒏 𝟐
, if 𝝈 𝟏&𝝈 𝟐 are not
known, 𝒔 𝟏
𝟐
& 𝒔 𝟐
𝟐
are known

Type V
 To test significance difference
between sample S.D s and population
S.D 𝝈
𝒔−𝝈
𝝈 𝟐
𝟐𝒏

Type VI
between sample S.D’s 𝒔 𝟏 𝒂𝒏𝒅 𝒔 𝟐
𝒔 𝟏−𝒔 𝟐
𝒔 𝟏
𝟐
𝟐𝒏 𝟏
+
𝒔 𝟐
𝟐
𝟐𝒏 𝟐

STUDENT’S t- TEST
 If 𝑥1, 𝑥2 … . 𝑥 𝑛 are the random samples of size n from a Normal
population with mean 𝜇 and variance 𝜎2
(denoted by 𝒮2
)
 Sample mean 𝑥 =
1
𝑛 𝑖=1
𝑛
𝑥𝑖 and sample variance denoted by 𝓈2
 Population variance 𝒮2
=
1
𝑛−1 𝑖=1
𝑛
𝑥𝑖 − 𝑥 2
 The Student’s t- statistic is defined by =
𝑥−𝜇
𝑆/ 𝑛
, where n is the sample
size
 And the degrees of freedom is (𝑛 − 1)
 The relation between sample variance and population variance is 𝑛𝑠2
=
𝑛 − 1 𝑆2

t –Test - Type I
 To test the significance difference between
sample mean 𝒙 and population mean
𝝁 (single mean)
 n- sample size, 𝑥- sample mean , 𝜇 –
population mean
 The test statistic t =
𝒙−𝝁
𝑺/ 𝒏
, if S – population
S.D ( 𝑺 𝟐
population variance) is known,
 Degrees of freedom = (𝒏 − 𝟏)
 Test statistic t =
𝒙−𝝁
𝒔/ 𝒏−𝟏
, if s – sample S.D
(𝒔 𝟐
sample variance) is known

t-Test - Type II
between two sample means 𝒙 𝟏 and 𝒙 𝟐
with sample size respectively 𝒏 𝟏 and 𝒏 𝟐
 The test statistic t =
𝒙 𝟏−𝒙 𝟐
𝑺
𝟏
𝒏 𝟏
+
𝟏
𝒏 𝟐
, if 𝒔 𝟏 & 𝒔 𝟐
sample S.D (𝒔 𝟏
𝟐
& 𝒔 𝟐
𝟐
− sample variances)
of respective samples are known and
S=
𝒏 𝟏 𝒔 𝟏
𝟐
+𝒏 𝟐 𝒔 𝟐
𝟐
𝒏 𝟏+𝒏 𝟐−𝟐
or S =
( 𝒙 𝟏−𝒙 𝟏
𝟐+ 𝒙 𝟐−𝒙 𝟐
𝟐)
𝒏 𝟏+𝒏 𝟐−𝟐
 Degrees of freedom is 𝒏 𝟏 + 𝒏 𝟐 − 𝟐

t- Test -Type III
 To test the significance difference in
means – Paired Data
 Given two samples are paired and same
size n ,
 The test statistic 𝒕 =
𝒅
𝑺 𝒏
, where d is
the difference between each
corresponding samples
 And 𝑺 =
𝒅− 𝒅
𝟐
𝒏−𝟏
 Degrees of freedom = (𝒏 − 𝟏)

NULL HYPOTHESIS 𝑯 𝟎 OF t-
TEST
 𝑯 𝟎 ∶ 𝝁 𝟏 = 𝝁 𝟐 , 𝑯 𝟏 ∶ 𝝁 𝟏 ≠ 𝝁 𝟐 𝒐𝒓 𝝁 𝟏 <
𝝁 𝟐 𝒐𝒓 𝝁 𝟏 > 𝝁 𝟐
 If calculated t < tabulated t with given
significance level, then 𝑯 𝟎 is
accepted
 If calculated t > tabulated t with given
significance level, then 𝑯 𝟎 is rejected

APPLICATION OF t- TEST
 TESTING THE SIGNIFICANCE OF THE
DIFFERENCE BETWEEN
 (1) THE MEAN OF A SAMPLE AND
THE MEAN OF THE POPULATION.
 (2) THE MEANS OF TWO SAMPLES

 VARIANCE RATION TEST (OR)
F- TEST FOR EQUALITY OF VARIANCE

 If 𝑥1, 𝑥2, … . 𝑥 𝑛1
𝑎𝑛𝑑 𝑦1, 𝑦2 … . 𝑦 𝑛2
are two independent
random sample from normal population.
 The test statistic 𝑭 =
𝑺 𝟏
𝟐
𝑺 𝟐
𝟐 𝒊𝒇 𝑺 𝟏
𝟐
> 𝑺 𝟐
𝟐
𝒐𝒓 =
𝑺 𝟐
𝟐
𝑺 𝟏
𝟐 (𝒊𝒇 𝑺 𝟐
𝟐
>
𝑺 𝟏
𝟐
)
 Where 𝑺 𝟏
𝟐
=
𝟏
𝒏 𝟏−𝟏 𝒊=𝟏
𝒏 𝟏
𝒙𝒊 − 𝒙𝒊
𝟐
& 𝑺 𝟐
𝟐
=
𝟏
𝒏 𝟐−𝟏 𝒊=𝟏
𝒏 𝟐
( 𝒚𝒊 −

NULL HYPOTHESIS 𝑯 𝟎 OF F-
TEST
 𝑯 𝟎 ; 𝝈 𝟏
𝟐
= 𝝈 𝟐
𝟐
𝑯 𝟏 ∶ 𝝈 𝟏
𝟐
≠ 𝝈 𝟐
𝟐
 If calculated F < tabulated F with
given significance level, then 𝑯 𝟎 is
accepted
 If calculated F > tabulated F with
given significance level, then 𝑯 𝟎 is
rejected

APPLICATION OF F- TEST
 (1) TESTING THE SIGNIFICANCE OF
THE DIFFERENCE BETWEEN THE
VARIANCES OF TWO POPULATIONS
FROM WHICH TWO SAMPLES ARE
DRAWN.
 (2) ANALYSIS OF VARIANCE.

CHI-SQUARE ( 𝝌 𝟐 ) DISTRIBUTION
 𝑂𝑖 (i=1,2…n) are set of observed
(experimental) frequencies and 𝐸𝑖
(i=1,2…n) are the corresponding set of
expected (theoretical or hypothetical)
frequencies, then the test statistic CHI-
SQUARE (𝜒2
) is defined by 𝝌 𝟐
=
𝒊=𝟏
𝒏 𝑶 𝒊−𝑬 𝒊
𝟐
𝑬 𝒊
 Degrees of freedom is 𝝂 = 𝒏 − 𝟏

 To test of independence of attributes (or) for the
(m x n) contingency table
 Test statistic CHI-SQUARE (𝜒2) is
𝜒2
=
𝑖=1
𝑛
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
 Where 𝑬𝒊 =
𝒓𝒐𝒘 𝒕𝒐𝒕𝒂𝒍 × 𝒄𝒐𝒍𝒖𝒎𝒏 𝒕𝒐𝒕𝒂𝒍
𝑮𝒓𝒂𝒏𝒅 𝒕𝒐𝒕𝒂𝒍
,
 Degrees of freedom(𝒎 − 𝟏)(𝒏 − 𝟏), m- no. of
rows and n- no. of columns
 For fitting Binomial distribution – degrees of
freedom = (𝒏 − 𝟏)
 For fitting Poisson distribution – degrees of
freedom = (𝒏 − 𝟐)

NULL HYPOTHESIS (𝑯 𝟎) OF CHI-
SQUARE
 The null hypothesis (𝐻0) of the Chi-
Square test is that no relationship exists
on the categorical variables in the
population; they are independent.

APPLICATION OF CHI-SQUARE(𝝌 𝟐
)
TEST
 (1) IT IS USED TO TEST THE
GOODNESS OF FIT.
 (2) IT IS USED TO TEST THE
INDEPENDENCE OF ATTRIBUTES.
 (3) TO TEST THE HOMOGENEITY OF
A GIVEN DATA

CONDITIONS FOR THE APPLICATION OF
CHI-SQUARE (𝝌 𝟐)TEST
 (1) THE EXPERIMENTAL DATA (OR SAMPLE
DEVIATIONS) MUST BE INDEPENDENT OF EACH
OTHER.
 (2) THE SAMPLE SIZE SHOULD BE REASONABLY
LARGE, ≥ 50.
 (3) THE THEORETICAL CELL FREQUENCY
SHOULD BE ATLEAST 5. IF IT IS LESS THAN 5, IT
IS COMBINED WITH ADJACENT FREQUENCIES
SO THAT THE POOLED FREQUENCY IS > 5.
 (4) THE CONSTRAINTS ON THE CELL
FREQUENCIES SHOULD BE LINEAR.
 EG., 𝑂𝑖 = 𝐸𝑖 = 𝑁 ≥ 50

2 × 2 CONTIGENCY TABLE.
 LET A AND B TWO ATTRIBUTES. DIVIDING A INTO 𝐴1 AND 𝐴2 AND B
INTO 𝐵1, 𝐵2, WE GET THE FOLLOWING 2 × 2 TABLE, CALLED THE 2 × 2
TABLE.

FORMULA FOR THE CHI-SQUARE ( 𝝌 𝟐
)TEST OF INDEPENDENCE FOR
B A 𝑨 𝟏 𝑨 𝟐 Total
𝑩 𝟏 a b a + b
𝑩 𝟐 c d c + d
Total a + c b + d N=a+ b+ c+ d
THE VALUE OF 𝝌 𝟐
=
𝑵 𝒂𝒅−𝒃𝒄 𝟐
𝒂+𝒃 𝒄+𝒅 𝒂+𝒄 𝒃+𝒅

VARIOUS STEPS INVOLVED IN TESTING
OF HYPOTHESIS
 Step 1. State the null hypothesis 𝐻0.
 Step 2. Decide the alternate hypothesis 𝐻1.
 Step 3. Choose the level of significance
α(α = 5% or α = 1%)
 Step 4. Compute the test statistic 𝑍 =
𝑡 − 𝐸 𝑡
𝑆.𝐸 𝑜𝑓 (𝑡)
.
 Step 5. Compare the computed value of
|𝑍| with the table value of Z and decide
the acceptance or the rejection of 𝐻0.
 Step 6. Inference.

Testing of hypothesis

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