Attributes are qualitative characteristic such as levels of literacy, employment status, etc., which are quantified in terms of levels/scores. The theory of attribute deals with qualitative types of characteristics that are calculated by using quantitative measurements. Therefore, the attribute needs slightly different kinds of statistical treatments, which the variables do not get. In science and research, an attribute is a characteristic of an object (person, thing, etc.). Attribute are closely related to variables. A variable is a logical set of attributes.

An experiment was conducted 144 times with tossing of four coins and the number of heads appeared at each throw are recorded. No. of heads 0 1 2 3 4 frequency 10 34 56 36 8 Fit binomial distribution to the above data.

Given that, an experiment was conducted 144 times with tossing of four coins. No. of heads 0 1 2 3 4 frequency 10 34 56 36 8 Null hypothesis H 0 : Fitting of binomial distribution is appropriate for the given data. Alternate hypothesis H 1 : Fitting of binomial distribution is not appropriate to the given data. Compute the expected frequencies: n = number of coins tossed at a time = 5 . Let X denote the number of heads (success) in n tosses. N = Number of times experiment is repeated = 640 . To find mean of the distribution: x f f x 0 10 0 1 34 34 2 56 112 3 36 108 4 8 32 Total 144 286 Mean : x ¯ = ∑ f x ∑ f = 286 144 = 1 . 986 . The probability mass function of binomial distribution is P ( x ) = C x n p x q n - x , x = 0 , 1 , … , n Mean of the binomial distribution is x ¯ = n p . ⇒ p = x ¯ n = 1 . 986 4 ≃ 0 . 5 , q = 1 - p = 0 . 5 . The expected frequency at x = N p ( x ) . The expected frequency at x = 0 : N × p ( 0 ) = 144 × ( 0 . 5 ) 4 = 9 . We use recurrence formulae to find the other expected frequencies. The expected frequency at x + 1 is n - x 1 + x p q × Expected frequency at x . x n - x 1 + x p q n - x 1 + x p q Expected frequency at x 0 4 1 4 9 1 1 . 5 1 1 . 5 36 2 0 . 6 1 0 . 6 54 3 0 . 25 1 0 . 25 36 . 2 4 0 1 0 9 . 1 Level of significance α = 0 . 5 . Test statistic χ 2 = ∑ i = 1 k O i - E i 2 E i . Calculation of the test statistic is : Observed frequency O i Expected frequency ( E i ) O i - E i O i - E i 2 ( O i - E i ) 2 E i 10 9 1 1 0 . 11 34 36 - 2 4 0 . 11 56 54 2 4 0 . 074 36 36 0 0 0 8 9 - 1 1 0 . 11 Total 0 . 407 χ 0 2 = ∑ i = 1 k O i − E i 2 E i = 0 . 407 . Critical value: Degrees of freedom = k – 1 – s = 5 – 1 – 1 = 3 , critical value for degree of freedom 3 at 5 % level of significance is 7 . 81 i.e., χ n - 1 , α 2 = χ 3 , 0 . 05 2 = 7 . 81 . As the calculated x 3 , 0 . 05 2 = 0 . 407 , is less than the critical value x 3 , 0 . 05 2 = 7 . 81 . We do not reject the null hypothesis. Hence, the fitting of binomial distribution is appropriate.

Give the test statistic for 2 × 2 contingency tables.

Test statistic for 2 × 2 , contingency table: Step 1 : Null hypothesis H 0 : The attributes are independent. Alternative hypothesis H 1 : The attributes are not independent. Step 2 :The observed frequencies are: A not A Total B a b a + b not B c d c + d Total a + c b + d N = a + b + c + d Step 3 : Level of significance α = 5 % . Step 4 : Calculate the value of the test statistic as χ 0 2 = N ( a d - b c ) 2 ( a + b ) ( c + d ) ( a + c ) ( b + d ) . Step 5 :From chi-square table the critical value at 5 % level of significance is χ 2 ( r - 1 ) ( c - 1 ) , α . Step 6 :The calculated value χ 2 0 is lesser than the critical value χ 2 ( r - 1 ) ( c - 1 ) , α , the null hypothesis H 0 is not rejected and vice-versa.

The distribution of the number of defective blades produced in a single shift in a factory over 100 shifts is given below. Number of defective blades 0 1 2 3 4 Number of shifts 12 14 23 18 33 Test whether the number of defective blades follows a Poisson distribution with mean = 0 . 44 Use α = 0 . 05 .

Given that, the distribution of the number of defective blades produced in a single shift in a factory over 100 shifts. Number of defective blades 0 1 2 3 4 Number of shifts 12 14 23 18 33 Null hypothesis H 0 : Fitting of Poisson distribution is appropriate for the given data Alternative hypothesis H 1 : Fitting of Poisson distribution is not appropriate for the given data. Data: The expected frequencies are computed as under: To find mean of the distribution: x f f x 0 12 0 1 14 14 2 23 46 3 18 54 4 33 132 Total 100 246 Mean : x ¯ = ∑ f x ∑ f = 246 100 = 2 . 46 . We know that, probability mass function of Poisson distribution is p ( x ) = e − m m x x ! ; x = 0 , 1 , … (i) In the case of Poisson distribution mean ( m ) = x ¯ = 2 . 46 . At x = 0 , equation (i) becomes p ( 0 ) = e − m m 0 0 ! = e m = e - 2 . 46 = 0 . 085 . The expected frequency at x = N p ( x ) . The expected frequency at x = 0 is N × p ( 0 ) = 100 × ( 0 . 085 ) 4 = 8 . 5 . We use recurrence formulae to find the other expected frequencies. The expected frequency at x + 1 is m 1 + x × Expected frequency at x . x m 1 + x Expected frequency at x 0 2 . 46 8 . 5 1 1 . 23 20 . 91 ≃ 21 2 0 . 82 25 . 72 ≃ 26 3 0 . 62 21 . 09 ≃ 21 4 0 . 5 13 . 07 ≃ 13 Level of significance α = 0 . 5 . Test statistic χ 2 = ∑ i = 1 k O i - E i 2 E i . Calculation of the test statistic is computed as under: Observed frequency O i Expected frequency ( E i ) O i - E i O i - E i 2 ( O i - E i ) 2 E i 12 9 3 9 1 14 21 - 7 49 2 . 33 23 26 - 3 9 0 . 4 18 21 - 3 9 0 . 43 33 13 20 400 30 . 88 Total 35 . 1 ∴ χ 0 2 = ∑ i = 1 k O i − E i 2 E i = 35 . 1 . Critical value: Degrees of freedom = k – 1 – s = 5 – 1 – 1 = 3 , critical value for degree of freedom 3 at 5 % level of significance is 7 . 81 i.e., x 3 , 0 . 05 2 = 7 . 81 . As the calculated x 3 , 0 . 05 2 = 35 . 1 is greater than the critical value x 3 , 0 . 05 2 = 7 . 81 .We reject the null hypothesis. Hence, we reject H 0 i.e., fitting of Poisson distribution is not appropriate for the given data

The eyesight was tested among 2000 randomly selected patients from a city and the following details are obtained. Gender Eyesight Total Poor Good Male 620 380 1000 Female 550 450 1000 Total 1170 830 2000 Can we conclude that there is an association between gender and quality of eye-sight at 5 % level of significance?

Given that, the eyesight was tested among 2000 randomly selected patients from a city and the following details are obtained. Gender Eye-sight Total Poor Good Male 620 380 1000 Female 550 450 1000 Total 1170 830 2000 Null hypothesis H 0 : The attributes Gender and Eyesight are independent. Alternative hypothesis H 1 : The attributes Gender and Eyesight are not independent. Compute expected frequencies E i j corresponding to each cell of the contingency table using the formula E i j = R i × C j N , i = 1 , 2 ; j = 1 , 2 , 3 , 4 . Here, N = Total sample size, R i = Row sum corresponding to i t h row, C j = Column sum corresponding to j t h column. Gender Eye -sight Total Poor eyesight observed Frequency ( O i j ) Poor eyesight expected Frequency ( E i j ) Good Eyesight observed Frequency ( O i j ) Good eyesight expected Frequency ( E i j ) Male O 11 = 620 E 11 = 1000 × 1170 2000 = 585 O 12 = 380 E 12 = 1000 × 830 2000 = 415 1000 Female O 21 = 550 E 21 = 1000 × 1170 2000 = 585 O 22 = 450 E 22 = 1000 × 830 2000 = 415 1000 Total 1170 830 2000 Level of significance α = 5 % . We know that, test statistic χ 0 2 = ∑ i = 1 2 ∑ j = 1 2 O i j − E i j 2 E i j This chi-square test statistic is calculated as follows: χ 0 2 = 620 - 585 2 585 + 550 - 585 2 585 + 380 - 415 2 415 + 450 - 415 2 415 ⇒ 2 . 094 + 2 . 094 + 2 . 952 + 2 . 952 = 10 . 09 , with degrees of freedom ( r - 1 ) ( c - 1 ) = ( 2 - 1 ) ( 2 - 1 ) = 1 . From the chi-square table the critical value at 5 % level of significance is χ 2 ( 2 - 1 ) ( 2 - 1 ) , 0 . 05 = χ 2 = 1 , 0 . 05 3 . 84 . Decision: As the calculated value χ 0 2 = 10 . 09 is greater than the critical value χ 2 = 1 , 0 . 05 3 . 84 , the null hypothesis H 0 is rejected. Hence, the attributes Gender and Eyesight are not independent.

The quality grade of electric components produced in two factories is given in the table given below. Factory Quality of grade Total Poor Medium Good Excellent A 136 165 151 148 600 B 31 58 55 36 180 Total 167 223 206 184 780 Test whether there is any association between factories and quality of grades.

Given that, the quality grade of electric components produced in two factories is given in the table given below. Factory Quality of grade Total Poor Medium Good Excellent A 136 165 151 148 600 B 31 58 55 36 180 Total 167 223 206 184 Null hypothesis H 0 : The attributes Factory and quality of grade are independent. Alternate hypothesis H 1 : The attributes Factory and quality of grade are not independent. Compute expected frequencies E i j corresponding to each cell of the contingency table using the formula E i j = R i × C j N , i = 1 , 2 ; j = 1 , 2 , 3 , 4 . Here, N = Total sample size, R i = Row sum corresponding to i t h row, C j = Column sum corresponding to j t h column. Factory Quality of grade Total Quality of grade observed Frequency ( O i j ) Quality of grade expected Frequency ( E i j ) Quality of grade observed Frequency ( O i j ) Quality of grade expected Frequency ( E i j ) Quality of grade observed Frequency ( O i j ) Quality of grade expected Frequency ( E i j ) Quality of grade observed Frequency ( O i j ) Quality of grade expected Frequency ( E i j ) A O 11 = 136 E 11 = 600 × 167 780 = 128 . 46 O 12 = 165 E 12 = 600 × 223 780 = 171 . 5 O 13 = 151 E 13 = 600 × 206 780 = 158 . 5 O 14 = 148 E 14 = 600 × 184 780 = 141 . 5 600 B O 21 = 31 E 21 = 180 × 167 780 = 38 . 54 O 22 = 58 E 22 = 223 × 180 780 = 51 . 5 O 23 = 55 E 23 = 180 × 206 780 = 47 . 5 O 24 = 36 E 24 = 180 × 184 780 = 42 . 5 180 Total 167 830 206 780 Level of significance α = 5 % . Calculate the value of the test statistic as χ 0 2 = ∑ i = 1 2 ∑ j = 1 4 O i j − E i j 2 E i j This chi-square test statistic is calculated as follows: χ 0 2 = 136 - 128 . 5 2 128 . 5 + 31 - 38 . 5 2 38 . 5 + 165 - 171 . 5 2 171 . 5 + 58 - 51 . 5 2 51 . 5 + 151 - 158 . 5 2 158 . 5 + 55 - 47 . 5 2 47 . 5 + 148 - 141 . 5 2 141 . 5 + 36 - 42 . 5 2 42 . 5 = 0 . 44 + 1 . 46 + 0 . 25 + 0 . 82 + 0 . 36 + 1 . 18 + 0 . 30 + 1 = 5 . 81 , with degrees of freedom ( r - 1 ) ( c - 1 ) = ( 2 - 1 ) ( 4 - 1 ) = 3 . Critical value: From the chi-square table the critical value at 5 % level of significance is χ 2 ( 2 - 1 ) ( 4 - 1 ) , 0 . 05 = χ 2 = 3 , 0 . 05 7 . 81 . Decision: As the calculated value χ 0 2 = 5 . 81 is lesser than the critical value χ 2 = 3 , 0 . 05 7 . 81 the null hypothesis H 0 is not rejected. Hence, the attributes Gender and Eyesight are independent.

EASY

Earn 100

Define contingency table?

Important Questions on Tests Based on Sampling Distributions-I

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

σ^{2}

is the variance of normal population, then the degrees of freedom of the sampling distribution of the test statistic for testing

H_{0} : σ^{2} = {σ_{0}}^{2}

is:

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Write the test procedure for testing the independence of attributes

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Define an attribute.

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

An experiment was conducted $144$ times with tossing of four coins and the number of heads appeared at each throw are recorded.

No. of heads	$0$	$1$	$2$	$3$	$4$
frequency	$10$	$34$	$56$	$36$	$8$

Fit binomial distribution to the above data.

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

In a sample of

200

households in a colony, two brands of hair oils A and B are applied by

90

, females. Further,

60

females and

70

males are using brand A. To test whether there is any association between the gender and brand of hair oil used, by constructing a contingency table.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

State the properties of χ2 distribution

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

If chi-square is performed for testing goodness of fit to a data with k classes on estimating ‘s’ parameters then degrees of freedom of test statistic is.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Define chi-square statistic

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Give the test statistic for

2 \times 2

contingency tables.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

What are the minimum requirements of chi-square test?

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

The distribution of the number of defective blades produced in a single shift in a factory over $100$ shifts is given below.

Number of defective blades	$0$	$1$	$2$	$3$	$4$
Number of shifts	$12$	$14$	$23$	$18$	$33$

Test whether the number of defective blades follows a Poisson distribution with mean = $0.44$ Use $α = 0.05 .$

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

The eyesight was tested among $2000$ randomly selected patients from a city and the following details are obtained.

Gender	Eyesight		Total
Gender	Poor	Good	Total
Male	$620$	$380$	$1000$
Female	$550$	$450$	$1000$
Total	$1170$	$830$	$2000$

Can we conclude that there is an association between gender and quality of eye-sight at $5 %$ level of significance?

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Write the procedure to test the population variance.

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

The quality grade of electric components produced in two factories is given in the table given below.

Factory	Quality of grade				Total
Factory	Poor	Medium	Good	Excellent	Total
A	$136$	$165$	$151$	$148$	$600$
B	$31$	$58$	$55$	$36$	$180$
Total	$167$	$223$	$206$	$184$	$780$

Test whether there is any association between factories and quality of grades.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Give the recurrence formula for binomial distribution

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Write down the computational steps for testing the significance of goodness of fit.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

n

is the degree of freedom of chi-square distribution then its variance is

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

Write the applications of chi-square distribution.

EASY

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

What is a contingency table?

MEDIUM

Mathematics>Statistics>Tests Based on Sampling Distributions-I>Chi-Square Distribution and Its Applications

The weights (in kg) of

10

students from a school are

38, 40, 45, 53, 47, 43, 55, 48, 52, 49

. Can we say that variance of the distribution of weights of all students from the above school is equal to

20 kg

Define contingency table?

Important Questions on Tests Based on Sampling Distributions-I

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