Higher Engineering Mathematics

SIGNIFICANCE TESTING 603

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degrees of freedom in Student’st-distribution tables
becomes 2 andνis given by (N 1 +N 2 −2). With
these factors taken into account, when testing the
hypotheses that samples come from the same popu-
lation, or that there is no difference between the mean
values of two populations, thet-value is given by:

|t|=

x 1 −x 2

σ

√(

1

N 1

+

1

N 2

) (11)

An estimate of the standard deviationσis based on
a concept called ‘pooling’. This states that if one
estimate of the variance of a population is based

on a sample, giving a result ofσ^21 =

N 1 s^21

N 1 − 1

and

another estimate is based on a second sample, giving

σ 22 =

N 2 s^22

N 2 − 1

, then a better estimate of the population

variance,σ^2 , is given by:

σ^2 =

N 1 s^21 +N 2 s^22

(N 1 −1)+(N 2 −1)

i.e. σ=

√

√

√

√

(

N 1 s^21 +N 2 s^22

N 1 +N 2 − 2

)

(12)

Problem 7. An automatic machine is produc-

ing components, and as a result of many tests the

standard deviation of their size is 0.02 cm. Two

samples of 40 components are taken, the mean

size of the first sample being 1.51 cm and the

second 1.52 cm. Determine whether the size has

altered appreciably if a level of significance of

0.05 is adopted, i.e. that the results are probably

significant.

Since both samples are drawn from the same pop-
ulation,σ 1 =σ 2 =σ= 0 .0 2 cm. AlsoN 1 =N 2 = 40
andx 1 = 1 .51 cm,x 2 = 1 .52 cm.
The level of significance,α= 0 .05.
The null hypothesis is that the size of the com-
ponent has not altered, i.e. x 1 =x 2 , hence it is
H 0 :x 1 −x 2 =0.
The alternative hypothesis is that the size of the
components has altered, i.e. thatx 1 =x 2 , hence it is
H 1 :x 1 −x 2 =0.

For a large sample having a known standard devi-

ation of the population, thez-value of the difference

of means of two samples is given by equation (9), i.e.,

z=

x 1 −x 2

√

√

√

√

(

σ 12

N 1

+

σ^22

N 2

)

SinceN 1 =N 2 =say,N, andσ 1 =σ 2 =σ, this equa-

tion becomes

z=

x 1 −x 2

σ

√(

2

N

)=

1. 51 − 1. 52

0. 02

√(

2

40

)=−^2.^236

Since the difference betweenx 1 andx 2 has no spec-

ified direction, a two-tailed test is indicated. The

z-value corresponding to a level of significance of

0.05 and a two-tailed test is+1.96 (see Table 62.1,

page 594). The result for thez-value for the differ-

ence of means is outside of the range+1.96, that is,

it is probable that the size has altered appreciably

at a level of significance of 0.05.

Problem 8. The electrical resistances of two

products are being compared. The parameters

of product 1 are:

sample size 40, mean value of sample

74 ohms, standard deviation of whole of

product 1 batch is 8 ohms

Those of product 2 are:

sample size 50, mean value of sample

78 ohms, standard deviation of whole of

product 2 batch is 7 ohms

Determine if there is any significant differ-

ence between the two products at a level of

significance of (a) 0.05 and (b) 0.01.

Let the mean of the batch of product 1 beμ 1 , and

that of product 2 beμ 2.

The null hypothesis is that the means are the same,

i.e.H 0 :μ 1 −μ 2 =0.

The alternative hypothesis is that the means are not

the same, i.e.H 1 :μ 1 −μ 2 =0.

The population standard deviations are known, i.e.

σ 1 =8 ohms andσ 2 =7 ohms, the sample means

are known, i.e. x 1 =74 ohms and x 2 =78 ohms.

Also the sample sizes are known, i.e.N 1 =40 and