A textbook of Computer Based Numerical and Statiscal Techniques part 56 potx

H0: µ = 47.5 i.e., there is no significant difference between the sample and population mean.. Conclusion: Since calculated t is less than tabulated t0.05 for 9 d.f., H0 may be accepted

Similarly, we will get

b – E(b) = −ad bc

N

–

= c – E(c); d – E(d) = ad bc

N

–

Substituting in (2), we get

χ2 = ad bc

N

–

2

E aa f a f a f a f+ E b +E c + E d

L

= ad bc

N

–

b g2

a+b a+c + a+b b+d a c c d b d c d

RS|

RS|

L N

N

–

b g2

b d a c

a b a c b d

b d a c

a c c d b d

+ + +

+ + +

L NMM b gb gb g b gb gb g O QPP

= (ad – bc)2 c d a b

a b a c b d c d

+ + +

L NMM b gb gb gb g O QPP

a b a c b d c d

–

b gb gb gb g

2

Example 11 From the following table regarding the colour of eyes of father and son test if the colour

of son’s eye is associated with that of the father.

Eye colour of son

Light Not light

Eye colour of father Light

Not light

Sol Null Hypothesis H 0 : The colour of son’s eye is not associated with that of the father i.e.,

they are independent

Under H0, we calculate the expected frequency in each cell as

= Product of column total and row total

whole total

Expected frequencies are:

Eye colour

Eye colour

of father

900 = 359.02

289 522×

900 = 167.62 522

900

×378

= 259.98 289

900

×378

= 121.38 378

χ2 = 471 359 02

359 02

51 167 62

167 62

148 259 98

259 98

230 121 38

121 38

–

–

–

–

= 261.498

Conclusion: At 5% level for 1 d.f., χ2 is 3.841 (tabulated value)

Since tabulated value of χ2 < calculated value of χ2 Hence H0 is rejected

Example 12 The following table gives the number of good and bad parts produced by each of the three shifts in a factory:

Test whether or not the production of bad parts is independent of the shift on which they were produced.

Sol Null Hypothesis H 0 : The production of bad parts is independent of the shift on which they were produced

i.e., the two attributes, production and shifts are independent.

i

i j i j

i j j

A B

N

MM MM

O Q

PP PP

1

2

0

2

0 1

e j

–

Calculation of expected frequencies

Let A and B be the two attributes namely production and shifts A is divided into two classes

A1, A2 and B is divided into three classes B1, B2, B3

(A1B1)0 = A B

N

1 2

2985

= 954.77;

(A1B2)0 = A B

N

1 2

b gb g = 2850 990

2985

= 945.226

(A1B3)0 = A B

N

1 3

b gb g = 2850 995

2985

= 950;

(A2B1)0 = A B

N

2 1

b gb g = 135 1000

2985

= 45.27

(A2B2)0 = A B

N

2 2

b gb g = 135 990

2985

= 44.773;

(A2B3)0 = A B

N

2 3

b gb g = 135 995

2985

To calculate the value of χ2

1.28126 Conclusion: The tabulated value of χ2 at 5% level of significance for 2 degrees of freedom

(r – 1)(s –1) is 5.991 Since the calculated value of χ2 is less than the tabulated value, we accept

H0 i.e., the production of bad parts is independent of the shift on which they were produced.

12.7.2 Student’s t-distribution

The t-distribution is used when sample size is less than equal to 30 (≤ 30) and the population

standard deviation is unknown

Let X i , i = 1, 2, , n be a random sample of size n from a normal population with mean

µ and variance σ2 Then student’s t is defined by

t = X

S n

– /

µ

~ t (n –1 d.f.)

1

n i X i

n

=

∑ is the sample mean

S2 = 1

1

n – i

n

=

∑

1

X i–X

d i2

is an unbiased estimate of the population variance σ2

The t-distribution has different values for each d.f and when the d.f are infinitely large, the t-distribution is equivalent to normal distribution.

Example 13 The 9 items of a sample have the following values 45, 47, 50, 52, 48, 47, 49, 53, 51 Does the mean of these values differ significantly from the assumed mean 47.5 ?

Sol H0: µ = 47.5

i.e., there is no significant difference between the sample and population mean.

H1: µ ≠ 47.5 (two tailed test): Given: n = 9, µ = 47.5

X–X – 4.1 – 2.1 0.9 2.9 –1.1 –2.1 –0.1 3.9 1.9

X–X

16.81 4.41 0.81 8.41 1.21 4.41 0.01 15.21 3.61

X = Σx

n =

442

9 = 49.11; Σd iX–X 2

= 54.89;

s2 = Σ X X

n

– –

d i

b g

2

1 = 6.86 ∴ s = 2.619

Applying t-test t = X

s n

– /

µ = 49 1 47 5

1 6 8

2 619

a f = 1.7279

t0.05 = 2.31 for γ = 8

Conclusion: Since t < t0.05, the hypothesis is accepted i.e., there is no significant difference

between their mean

Example 14 A random sample of 10 boys had the following I Q’ s: 70, 120, 110, 101, 88, 83, 95,

98, 107, 100 Do these data support the assumption of a population mean I.Q of 100 ? Find a reasonable range in which most of the mean I.Q values of samples of 10 boys lie.

Sol Null hypothesis, H0: The data are consistent with the assumption of a mean I.Q of 100

in the population, i.e., µ = 100

Alternative hypothesis: H1 :µ ≠ 100

Test Statistic Under H0, the test statistic is:

t = x

– /

µ

d i

2 ∼ t(n –1) where x and S2 are to be computed from the sample values of I.Q.’s

Calculation for Sample Mean and S.D.

Hence n = 10, x = 972

10 = 97.2 and S2 =

1833 60 9

= 203.73

203 73 10

– / =

2 8

20 37

2 8

4 514

= 0.62

Tabulated t0.05 for (10 – 1) i.e., 9 d.f for two-tailed test is 2.262.

Conclusion: Since calculated t is less than tabulated t0.05 for 9 d.f., H0 may be accepted at 5% level of significance and we may conclude that the data are consistent with the assumption

of mean I.Q of 100 in the population

The 95% confidence limits within which the mean I.Q values of samples of 10 boys will lie are given by

x ± t0.05 S/ n = 97.2 ± 2.262 × 4.514 = 97.2 ± 10.21 = 107.41 and 86.99 Hence the required 95% confidence intervals is [86.99, 107.41]

Example 15 The mean weekly sales of soap bars in departmental stores was 146.3 bars per store After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2 Was the advertising campaign successful?

Sol We are given: n = 22, x = 153.7, s = 17.2.

Null Hypothesis: The advertising campaign is not successful, i.e.,

H0: µ = 146.3

Alternative Hypothesis: H1: µ > 146.3 (Right-tail)

Test Statistic: Under the null hypothesis, the test statistic is:

s n

–

µ

2 b g1 ~ t22 – 1 = t21

Now t = 153 7 146 3

17 2 2 21

−

a f = 7 4 21

17 2

× = 9.03

Conclusion: Tabulated value of t for 21 d.f at 5% level of significance for single-tailed test

is 1.72 Since calculated value is much greater than the tabulated value, therefore it is highly significant Hence we reject the null hypothesis

Example 16 A machinist is making engine parts with axle diameters of 0.700 inch A random sample of 10 parts shows a mean diameter of 0.742 inch with a standard deviation of 0.040 inch Compute the statistic you would use to test whether the work is meeting the specifications Also state how you would proceed further.

Sol Here we are given:

µ = 0.700 inches, x = 0.742 inches, s = 0.040 inches and n = 10 Null Hypothesis, H0: µ = 0.700, i.e., the product is conforming to specifications

Alternative Hypothesis, H1: µ ≠ 0.700

Test Statistic : Under H0, the test statistic is:

t = x

s n

– /

µ

s n

–

µ

2 b g1 ∼ t(n – 1)

Now, t = 9 0 742 0 700

0 040

–

Here the test statistic ‘t’ follows Student’s t-distribution with 10 – 1 = 9 d.f We will now compare this calculated value with the tabulated value of t for 9 d.f and at certain level of significance, say 5% Let this tabulated value be denoted by t0

(i) If calculated ‘t’ viz., 3.15 > t0, we say that the value of t is significant This implies that

x differs significantly from µ and H0 is rejected at this level of significance and we conclude that the product is not meeting the specifications

(ii) If calculated t < t0, we say that the value of t is not significant, i.e., there is no significant

difference between x and µ In other words, the deviation d ix –µ is just due to

fluctuations of sampling and null hypothesis H0 may be retained at 5% level of

significance, i.e., we may take the product conforming to specifications.

Example 17 A random sample of size 16 has 53 as mean The sum of squares of the derivation from mean is 135 Can this sample be regarded as taken from the population having 56 as mean ? Obtain 95% and 99% confidence limits of the mean of the population.

Sol H0: There is no significant difference between the sample mean and hypothetical population mean

H0: µ = 56; H1: µ ≠ 56 (Two tailed test)

t : X

s n

– /

µ ∼ t(n – 1 d.f.)

Given: X = 53, µ = 56, n = 16, Σd iX–X 2

= 135

s = ΣX X

n

– –

d i2

1 =

135

15 = 3; t =

53 56

3 16

–

t = 4, d.f = 16 – 1 = 15.

Conclusion: t0.05 = 1.753.

Since t = 4 > t0.05 = 1.753 i.e., the calculated value of t is more than the table value The

hypothesis is rejected Hence, the sample mean has not come from a population having 56 as mean

95% confidence limits of the population mean

=X ± s

n t0.05 = 53 ±

3

16 (1.725) = 51.706; 54.293

99% confidence limits of the population mean

=X ± s

n t0.01, = 53 ±

3

16 (2.602) = 51.048; 54.951.

(i ) t-Test of Significance for Mean of a Random Sample: To test whether the mean of a sample drawn from a normal population deviates significantly from a stated value when variance of the population is unknown

H0: There is no significant difference between the sample mean x and the population mean

µ i.e., we use the statistic

t = X

s n

– /

µ where X is mean of the sample

s2 = 1

1

n – i X i X

n

–

d i2 1

=

∑ with degrees of freedom (n – 1).

At given level of significance α1 and degrees of freedom (n – 1) We refer to t-table tα (two

tailed or one tailed) If calculated t value is such that t < tα the null hypothesis is accepted and for t > tα, H0 is rejected

(ii ) t-Test For Difference of Means of Two Samples: This test is used to test whether the

two samples x1, x2, x n

1, y1, y2, , y n

2 of sizes n1, n2 have been drawn from two normal populations with mean µ1 and µ2 respectively under the assumption that the population variance are equal (σ1 = σ2 = σ)

H0: The samples have been drawn from the normal population with means µ1 and µ2 i.e.,

H0: µ1 ≠ µ2

Let X, Y be their means of the two samples

Under this H0 the test of statistic t is given by t = X Y

s

–

d i

1 1

1 2 +

– t(n1 + n2 – 2 d.f.)

Also, if the two sample’s standard deviations s1, s2 are given then we have s2 = n s n s

n n

1 12 2 22

1 2 2

+

And, if n1 = n2 = n, t = X Y

s s n

–

–

12 22

1

+ can be used as a test statistic.

If the pairs of values are in some way associated (correlated) we can’t use the test statistic

as given in Note 2 In this case, we find the differences of the associated pairs of values and apply

for single mean i.e., t = X

s n

– /

µ

with degrees of freedom n – 1.

The test statistic is t = d

s/ n or t =

d

s/ n– 1 , where d is the mean of paired difference.

i.e., d i = x i – y i

d i = X–Y , where (x i , y i ) are the paired data i = 1, 2, , n Example 18 Samples of sizes 10 and 14 were taken from two normal populations with S.D 3.5 and 5.2 The sample means were found to be 20.3 and 18.6 Test whether the means of the two populations are the same at 5% level.

Sol H0: µ1 = µ2, i.e., the means of the two populations are the same.

H1: µ1 ≠ µ2 Given X = 20.3, X2 = 18.6; n1 = 10, n2 = 14, s1 = 3.5, s2 = 5.2

s2 = n s n s

1 12 2 22

1 2 2

+ + – =

10 3 5 14 5 2

10 14 2

–

+ = 22.775 ∴ s = 4.772

s

1 1

− +

= 20 3 18 6

1 10

1

14 4 772

–

+

F

= 0.8604

The value of t at 5% level for 22 d.f is t0.05 = 2.0739

Conclusion: Since t = 0.8604 < t0.05 the hypothesis is accepted i.e., there is no significant

difference between their means

Example 19 Two samples of sodium vapour bulbs were tested for length of life and the following results were got:

Sample

Is the difference in the means significant to generalise that Type I is superior to Type II regarding length of life ?

Sol H0: µ1 = µ2, i.e., two types of bulbs have same lifetime.

H1: µ1 > µ2 i.e., type I is superior to type II.

s2 = n s n s

1 12 2 22

1 2 2

+

= 8 36 7 40

8 7 2

a f a f+ + – = 1659.076 ∴ s = 40.7317

s

1 1

− +

8

1 7

–

= 18.1480 ~ t(n1 + n2 –2d.f)

t0.05 at d.f 13 is 1.77 (one tailed test).

Conclusion: Since calculated t > t0.05, H0 is rejected i.e., H1 is accepted

∴ Type I is definitely superior to Type II

where X =

i

n

=

∑

1

1 X

n i

i , Y =

j

n

=

∑

1

2 Y n j

2

; s2 = 1

2

n +n – E Xd i –Xi e2 Y j –Yj2

+

estimate of the population variance σ2

t follows t-distribution with n1 + n2 – 2 degrees of freedom

Example 20 The following figures refer to observations in live independent samples:

Sample I

Sample II

Analyse whether the samples have been drawn from the populations of equal means.

Sol H0: The two samples have been drawn from the population of equal means i.e., there

is no significant difference between their means

H1: µ1 ≠ µ2 (Two tailed test) Given n1 = Sample I size = 10; n2 = Sample II size = 10

To calculate the two sample mean and sum of squares of deviation from mean Let X1 be

the Sample I and X2 be the Sample II

X–X1 – 1.6 3.4 1.4 7.4 –2.6 –6.6 –13.6 5.4 4.6 11.4

X1 X1

2

–

d i 2.56 11.56 1.96 54.76 6.76 43.56 184.96 29.16 21.16 129.96

X2 –X2 10.7 4.7 – 7.3 – 9.3 1.7 10.7 0.7 – 6.3 6.7 –12.3

X2 X2

2

–

d i 114.49 22.09 53.29 86.49 2.89 114.49 0.49 39.67 44.89 151.29

X1 =

i=

∑

1

10

X n

1 1

= 26.6 X2=

i=

∑

1

10 X n

2 2

= 293

10 = 29.3

Σ X1 X1

2

–

d i = 486.4 Σ X2 X2

2

–

d i = 630.08

2

n +n – Σ X1 X1 ΣX X

2

2

10 10 2+ – [486.4 + 630.08] = 62.026 ∴ S = 7.875

Under H0 the test statistic is given by

s

1 1

–

+

= 26 6 29 3

10

1 10

= – 0.7666 –t(n1 + n2 – 2 d.f)

t = 0.7666

Conclusion: The tabulated value of t at 5% level of significance for 18 d.f is 2.1 Since the

calculated value t = 0.7666 < t0.05 H0 is accepted

i.e., there is no significant difference between their means.

i.e., the two samples have been drawn from the populations of equal means.

Applications of t-Distribution: The t-distribution has a wide number of applications in

statistics, some of them are:

1 To test if the sample mean d iX differs significantly from the hypothetical value µ of the population mean

2 To test the significance between two sample means

3 To test the significance of observed partial and mutiple correlation coefficients

4 To test the significance of an observed sample correlation co-efficient and sample regression

coefficient Also, the critical value or significant value of t at level of significance α and

degree of freedom ν for two tailed test are given by

P[J > tν (α)] = α

⇒ P[J ≤ tν(α)] = 1– α

The significant value of t at level of significance ‘α’ for a single tailed test can be obtained

from those of two tailed test by considering the values at level of significance ‘2α’

12.7.3 Snedecor’s Variance Ratio Test or F-test

Suppose we want to test (i) whether two independent samples x i and y j For i = 1, 2 , n1 and j

= 1, 2, , n2 have been drawn from the normal populations with the same variance σ2, (say) or

(ii) whether two independent estimates of the population variance are homogenous or not.