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PART –B

(10 * 8 = 40)

1. A company has two plants to manufacture scooters. Plant I manufactures 80 % of the

scooters and Plant II manufactures 20%. At plant I , 85 out of 100 scooters are rated standard quality are better .At plant II only 65 out of 100 scooters are rated standard quality are better. What is the probability that scooter selected at random came from plant I, if it is known that the scooter is of standard quality? What is the probability that it came from plant II, if it is known that the scooter is of standard quality? Solution: Let A represent that the Scooter is manufactured from Plant I and B represent that the Scooter is manufactured from Plant II. Let R represent that the randomly selected scooters are having better standard quality rate. Therefore, we have P (A) = 0.8 P (B) = 0.2 P (R/A) = 0.85 P (R/B) = 0.65 P (A and R) = 0.85 * 0.8 = 0.68 P (B and R) = 0.65 * 0.2 = 0.13 The table given below shows the Joint distribution function of Plants and Better quality scooter rates Plant I Plant II Total Output Rated Standard Quality Better 0.68 0.13 0.81 Rated Standard Quality Poor 0.12 0.07 0.19 Total Output 0.8 0.2 1

From the above table, we have P (Scooter selected from Plant I\Rated Better Standard Quality) = P (A\R)

P( A ∩ R) 0.68 P( A \ R) = = = 0.8395 P( R) 0.81

P (Scooter selected from Plant II\Rated Better Standard Quality) = P (B\R)

P( B ∩ R) 0.13 P( B \ R) = = = 0.1605 P( R) 0.81
Therefore, The probability that scooter selected at random came from plant I, if it is known that the scooter is of standard quality is 0.8395 The probability that scooter selected at random came from plant II, if it is known that the scooter is of standard quality is 0.1605
2. A random sample of 500 pineapples was taken from a large consignment and 65 were

found to be bad. Show that the S.E of the proportion of bad ones in a sample of size is 0.015 and deduce that the percentage of bad pineapples in the consignment almost certainly lies between 8.5 and 17.5 Solution: Total number of Pineapples taken = N = 500 No of Bad Pineapples found = X = 65 Proportion of Bad Pineapples = p = X/N = 65/500 = 0.13 q = 1 – p = 1 – 0.13 = 0.87
S tan dard Error = p*q = N 0.13 * 0.87 = 500 0.1131 = 0.000226 = 0.015 500

The table given below shows the limits of the bad pineapples in the consignment almost certainly lies between 8.5 and 17.5 Alpha Level Standard Error Margin of Error Lower Limit Upper Limit 0.01 0.015 0.039 0.091 0.169 0.001 0.015 0.049 0.081 0.179 0.002 0.015 0.046 0.084 0.176 0.003 0.015 0.045 0.085 0.175

Therefore, at alpha = 0.3, that is at 3 Standard Error limits, the percentage of bad pineapples in the consignment almost certainly lies between 8.5 and 17.5. The 3 Standard error limits are almost certain

3. Random samples of 250 bolts manufactured by machine A and 200 bolts manufactured by

machine B showed 24 and 10 defective bolts respectively. Test the hypothesis that the machines are showing difference qualities of performance. Use 5 per cent level of significance Solution: In order to determine whether the machines are showing difference qualities of performance, we perform two proportion z test. The null and alternate hypotheses are given below: Null Hypothesis: H0: PA = PB That is, machines are showing no difference in qualities of performance Alternate Hypothesis: Ha: PA ≠ PB That is, machines are showing difference in qualities of performance Level of Significance: Let the level of significance be α = 0.05 Test Statistic: The test statistic is:
Z= p A − pB  1 1  P *Q * n + n   B   A

Where
P= n A * p A + nB * p B n A + nB

Total bolts manufactured from machine A = nA = 250 Total bolts manufactured from machine B = nB = 200 P (defective bolts from machine A) = pA = 24/250 = 0.096 P (defective bolts from machine B) = pB = 10/200 = 0.005
P= n A * p A + n B * p B 250 * 0.096 + 200 * 0.05 24 + 10 = = = 0.0756 n A + nB 250 + 200 450

Q = 1 – 0.0756 = 0.9244 On substituting these values in the above test statistic, we have

Z=

p A − pB  1 1  P *Q * n + n   B   A

=

0.096 − 0.05 1   1 0.0756 * 0.9244 *  +   250 200 
0.046 0.000629

=

=

0.046 =1.834691 0.025072

Therefore, the required value of z test statistic is 1.834691 Critical Value of Z: The table value of Z at 5% level of significance is: Zα/2 = Z0.05/2 = Z0.025 = ± 1.96 (By referring normal distribution table) The decision Rule is:  Reject the null hypothesis if the z test statistic is either less than – 1.96 or greater than 1.96  Fail to reject the null hypothesis if the z test statistic falls between – 1.96 and 1.96 Conclusion: Since the value of Z test statistic does not fall in the rejection region, we fail to reject the null hypothesis at 5% level of significance. Therefore, we there is no statistical evidence to conclude that the machines are showing difference in qualities of performance
4. Explain the process of hypothesis testing.

5. The number of customers appears at the ticket counter of PVR theatre at a rate of

120 per hour. Find the probability that during a given minute

(i) (ii) Solution:

No customer appear. At least two customers appear.

Number of customer appears at the ticket counter of PVR theatre = λ = 120/60 = 2 Here, X follows Poisson distribution with λ = 2 In general, the probability mass function of Poisson distribution is:
P( X = x ) = e −λ * λx , x = 0,1,2,..... x!

(i) No customer appear P (X = 0) =
e −2 * 2 0 = 0.135335. 0!

Therefore, the probability that no customer appear during a given minute is 0.135335 (ii) At least two customers appear. P (X >=2) = 1 – P (X < 2) = 1 – P (X = 0) – P (X = 1)
=1 − e −2 * 2 0 e −2 * 21 − = 1 − 0.135335. − 0.270671 = 0.594 0! 1!

Therefore, the probability that at least 2 customers appear during a given minute is 0.594
6. On an average six people per hour use an electronic teller machine during the prime

shopping hours in a department store. What is the probability that (i) (ii) Exactly six people will use the machine during a randomly selected hour. Fewer than five people will use the machine during a randomly selected hour.

Solution: Number of customer use an electronic teller machine = λ = 6 Here, X follows Poisson distribution with λ = 6 In general, the probability mass function of Poisson distribution is:
P( X = x ) = e −λ * λx , x = 0,1,2,..... x!

(i)

Exactly six people will use the machine during a randomly selected hour.
e −6 * 6 6 = 0.160623 6!

P (X = 6) =

Therefore, the probability that exactly six people will use the machine during a randomly selected hour is 0.160623 (ii) Fewer than five people will use the machine during a randomly selected hour.
e −6 * 6 0 e −6 * 61 e −6 * 6 2 e −6 * 6 3 e −6 * 6 4 + + + + 0! 1! 2! 3! 4!

P (X <2) = P (X = 0) + P (X = 1)P (X =2) + P (X = 3) + P (X = 4)
=

= 0.002479 + 0.014873 + 0.044618 + 0.089235 + 0.133853 = 0.285057 Therefore, the probability that fewer than five people will use the machine during a randomly selected hour is 0.285057
7. The following contingency table presents the reactions of legislators to a tax plan according

to party affiliation. Test whether party affiliation influences the reaction to the tax plan. Reaction Party Party A Party B Party C Total Solution: In order to determine whether there is any association between Party affiliation and legislator’s reaction to a tax plan, we perform chi – square test for independence. The null and alternate hypotheses are given below: Null Hypothesis: H0: There is no association between Party affiliation and legislator’s reaction to a tax plan Alternate Hypothesis: H1: There is an association between Party affiliation and legislator’s reaction to a tax plan Level of Significance: Let the level of significance be α=1% or 0.01 Test Statistic: The test statistic is: In Favour 120 50 50 Neutral 20 30 10 Opposed 20 60 40 120 Total 160 140 100 400

220 60 (Critical x2 for df = 4 is 13.28 at α=1%)

χ = ∑∑
2 j =1 i =1

n

m

(O

ij

− Eij ) Eij

2

Where Oij  Observed Frequency of ith row and jth column Eij  Expected Frequency of ith row and jth column Under the assumption that the null hypothesis is true, the expected frequencies are calculated as follows:
E ij =

[ Row Total ] i * [Column Total ] j
Grand Total

For example, E11 = (160 * 220)/400 = 88 E12 = (160 * 60)/400 = 24 Similarly, the expected frequencies for the remaining cells are calculated and the output table is given below: In Favor Party A Party B Party C Total 88 77 55 220 Neutral 24 21 15 60 Opposed 48 42 30 120 Total 160 140 100 400

The table given below shows the workings of Chi – Square test statistic Observed Frequency 120 50 50 20 30 10 20 60 40 Expected Frequency 88 77 55 24 21 15 48 42 30 O-E 32 -27 -5 -4 9 -5 -28 18 10 (O - E)^2 1024 729 25 16 81 25 784 324 100 (O - E)^2/E 11.63636 9.467532 0.454545 0.666667 3.857143 1.666667 16.33333 7.714286 3.333333 55.12987

From the above table, we have

χ = ∑∑
2 j =1 i =1

n

m

(O

ij

− Eij ) Eij

2

= 55.12987

Given: Critical x2 for df = 4 is 13.28 at α=1% The decision Rule is:  Reject the null hypothesis if the chi square test statistic is greater than 13.28  Fail to reject null hypothesis if the chi square test statistic is less than or equal to 13.28 Conclusion: Since the value of chi – square test statistic fall in the rejection region, we reject the null hypothesis at 1% level of significance. Therefore, there is statistical evidence to conclude that party affiliation influences the reaction to the tax plan
8. Given two examples having the following observation values establish the sample belonging

to the sample population or from population using Mann- Whitney U-test. X Y Solution: Null Hypothesis: There is no difference between the median values of the two groups. Alternative Hypothesis: There is a difference in the median values between the two groups. Let n1 be the size of the variable X and n2 be the size of the variable Y. Here, n1 = 10 and n2 = 10 Rank all the values for both samples from the smallest (=1) to the largest. Set them up as shown in the table below Sl. No 1 2 3 4 5 6 7 8 9 10 X 450 750 925 820 770 650 575 975 1050 1450 Rank (X) 1.5 9 15 12.5 10 8 5 16 17.5 20 Y 775 475 580 900 625 525 1050 450 820 1160 Rank (Y) 11 3 6 14 7 4 17.5 1.5 12.5 19 450 775 750 475 925 580 820 900 770 625 650 525 575 1050 975 450 1050 820 1450 1160

Total

114.5

95.5

Level of Significance: Let the level of significance be α=5% or 0.05 Test Statistic: The test statistic is:

U 1 = n1 n2 + U 2 = n1 n 2 +

n1 * ( n1 + 1) 10 * (10 + 1) − 114.5 = 40.5 − R1 = 10 * 10 + 2 2 n2 * ( n 2 + 1) 10 * (10 + 1) − R2 = 10 * 10 + − 95.5 = 59.5 2 2

Critical Value: The critical value of U corresponding to (10, 10) degree of freedom and at 5% level of significance is: U(10,10), 0.05 = 23 (By referring Mann Whitney U test table) The decision Rule is:  Reject the Null Hypothesis if the smallest value of U1 or U2 is below the table value of U  Fail to reject the Null Hypothesis if the smallest value of U1 or U2 is greater than the table value of U Conclusion: Since the value of Mann Whitney U test statistic does not fall in the rejection region, we fail to reject the null hypothesis at 5% level of significance. Therefore, we conclude that the sample belonging to the population

9. Compute the trend values for the following data using method of least squares: Year : 1982 Y Solution: The regression equation to predict Y using Year (X) as Independent variable is: Y=a+b*X Where a and b are the intercept and slope of the regression line In order to find a and b the following two equations are solved: : 83 1983 60 1984 54 1985 21 1986 22 1987 13 1988 23

∑Y = na + b ∑x ------------------------- (1) 2 ∑XY = a ∑x + b∑x ------------------------- (2) Where n is the total number of observations in a series. These equations are called normal equations The table given below shows the workings of least square method Sl. No 2 3 4 5 6 7 Total From the above table, we have 7a + 13895b = 276 Multiplying (3) by 13895, we have 13895 * 7 a + 13895 * 13895 b = 13895 * 276 97265 a + 193071025 b = 3835020 Multiplying (4) by 7 7 * 13895 a + 7 * 15028 b = 7 * 547554 97265 a + 193071221b = 3832878 ------------------------- (6) (5) – (6) 97265 a + 193071025 b = 3835020 97265 a + 193071221 b = 3832878 (-) (-) (-) __________________________ - 196 b = 2142 b = 2142 / (- 196) = - 10.9286 Thus, the value of b is – 10.9286 ------------------------- (5) ------------------------- (3) 13895 a + 193071025 b = 547554 ------------------------- (4) Year (X) 1982 1983 1984 1985 1986 1987 1988 13895 Y 83 60 54 21 22 13 23 276 XY 164506 118980 107136 41685 43692 25831 45724 547554 X^2 3928324 3932289 3936256 3940225 3944196 3948169 3952144 193071025

On substituting the value of b in (3) 7a + 13895 * (- 10.9286) = 276 7 a – 151853 = 276 7 a = 276 + 151853  a = 152128.5/7 = 21732.64 Thus, the value of a is 21732.64 The required trend line equation is Y = 21732.64 – 10.9286 * Year When Year is 1982, the trend value is Y = 21732.64 – 10.9286 * 1982 = 72.21428571 Similarly, the trend values for the remaining period are computed and are given below: Sl. No 1 2 3 4 5 6 7 Year X 1982 1983 1984 1985 1986 1987 1988 Y 83 60 54 21 22 13 23 Trend Values 72.21428571 61.28571429 50.35714286 39.42857143 28.5 17.57142857 6.642857143

10. Explain the various sampling techniques.

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