Higher Engineering Mathematics

600 STATISTICS AND PROBABILITY

Problem 5. The mean lifetime of a random

sample of 50 similar torch bulbs drawn from

a batch of 500 bulbs is 72 hours. The stan-

dard deviation of the lifetime of the sample is

10.4 hours. The batch is classed as inferior if the

mean lifetime of the batch is less than the popu-

lation mean of 75 hours. Determine whether, as

a result of the sample data, the batch is consid-

ered to be inferior at a level of significance of

(a) 0.05 and (b) 0.01.

Population size,Np=500, population mean,μ= 75
hours, mean of sample,x=72 hours, standard devi-
ation of sample, s= 10 .4 hours, size of sample,
N=50.
The null hypothesis is that the mean of the sample
is equal to the mean of the population, i.e.H 0 :x=μ.
The alternative hypothesis is that the mean of the
sample is less than the mean of the population, i.e.
H 1 :x<μ.
(The fact thatx=72 should not lead to the con-
clusion that the batch is necessarily inferior. At a
level of significance of 0.05, the result is ‘probably
significant’, but since this corresponds to a confi-
dence level of 95%, there are still 5 times in every 100
when the result can be significantly different, that
is, be outside of the range ofz-values for this data.
This particular sample result may be one of these
5 times.)
The decision rules associated with the hypo-
theses are:

(i) rejectH 0 if thez-value (ort-value) of the sample

mean is less than thez-value (ort-value) corre-

sponding to a level of significance of (a) 0.05

and (b) 0.01, i.e. the batch is inferior,

(ii) accept H 0 otherwise, i.e. the batch is not

inferior.

The data given isN,Np,x,sandμ. The alterna-
tive hypothesis indicates a one-tailed distribution
and sinceN>30 the ‘large sample’ theory applies.

From equation (6),

z=

x−μ

s

√

N

√(

Np−N

Np− 1

)=

72 − 75

10. 4

√

50

√(

500 − 50

500 − 1

)

=

− 3

(1.471)(0.9496)

=− 2. 15

(a) For a level of significance of 0.05 and a one-

tailed test, all values to the left of thez-ordinate

at−1.645 (see Table 62.1 on page 594) indi-

cate that the results are ‘not significant’, that is,

they differ significantly from the null hypoth-

esis. Since thez-value of the sample mean is

−2.15, i.e. less than−1.645,the batch is con-

sidered to be inferior at a level of significance

of 0.05.

(b) Thez-value for a level of significance of 0.01

for a one-tailed test is−2.33 and in this case,

z-values of sample means lying to the left of the

z-ordinate at−2.33 are ‘not significant’. Since

thez-value of the sample lies to the right of this

ordinate, it does not differ significantly from the

null hypothesis andthe batch is not considered

to be inferior at a level of significance of 0.01.

(At first sight, for a mean value to be signifi-

cant at a level of significance of 0.05, but not

at 0.01, appears to be incorrect. However, it is

stated earlier in the chapter that for a result to

be probably significant, i.e. at a level of signifi-

cance of between 0.01 and 0.05, the range of

z-values is less than the range for the result to

be highly significant, that is, having a level of

significance of 0.01 or better. Hence the results

of the problem are logical.)

Problem 6. An analysis of the mass of carbon

in six similar specimens of cast iron, each of

mass 425.0 g, yielded the following results:

17 .1g, 17.3g, 16.8g, 16.9g,

17 .8 g, and 17.4g

Test the hypothesis that the percentage of carbon

is 4.00% assuming an arbitrary level of signifi-

cance of (a) 0.2 and (b) 0.1.

The sample mean,

x=

17. 1 + 17. 3 + 16. 8 + 16. 9 + 17. 8 + 17. 4

6

= 17. 22

The sample standard deviation,

s=

√ √ √ √ √ √ √

⎧

⎪⎪

⎨

⎪⎪

⎩

(17. 1 − 17 .22)^2 +(17. 3 − 17 .22)^2

+(16. 8 − 17 .22)^2 +···+(17. 4 − 17 .22)^2

6

⎫

⎪⎪

⎬

⎪⎪

⎭

= 0. 334