Higher Engineering Mathematics

SIGNIFICANCE TESTING 601

J

The null hypothesis is that the sample and population
means are equal, i.e.H 0 :x=μ.
The alternative hypothesis is that the sample and
population means are not equal, i.e.H 1 :x=μ.
The decision rules are:

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

is outside of the range of thez-ort-value corre-

sponding to a level of significance of (a) 0.2 and

(b) 0.1, i.e. the mass of carbon is not 4.00%,

(ii) acceptH 0 otherwise, i.e. the mass of carbon is

4.00%.

The number of tests taken,N, is 6 and an infi-
nite number of tests could have been taken, hence
the population is considered to be infinite. Because
N<30, at-distribution is used.
If the mean mass of carbon in the bulk of the metal
is 4.00%, the mean mass of carbon in a specimen is
4.00% of 425.0, i.e. 17.00 g, thusμ= 17 .00.

From equation (7),

|t|=

(x−μ)

√

(N−1)

s

=

(17. 22 − 17 .00)

√

(6−1)

0. 334

= 1. 473

In general, for any two-tailed distribution there is
a critical region both to the left and to the right of the
mean of the distribution. For a level of significance
of 0.2, 0.1 of the percentile value of at-distribution
lies to the left of the mean and 0.1 of the percentile
value lies to the right of the mean. Thus, for a level
of significance ofα, a valuet(
1 −α 2

), is required for

a two-tailed distribution when using Table 61.2 on
page 587. This conversion is necessary because the
t-distribution is given in terms of levels of confidence
and for a one-tailed distribution. The rowt-value
for a value ofαof 0.2 ist( 1 − 0. 2
2

), i.e.t 0. 90. The

degrees of freedomνareN−1, that is 5. From
Table 61.2 on page 587, the percentile value corre-
sponding to (t 0. 90 ,ν=5) is 1.48, and for a two-tailed
test,±1.48. Since the mean value of the sample is
within this range, the hypothesis is accepted at a level
of significance of 0.2.
Thet-value forα= 0 .1ist(
1 −^02.^1

), i.e.t 0. 95. The

percentile value corresponding tot 0. 95 ,ν=5 is 2.02
and since the mean value of the sample is within

the range±2.02, the hypothesis is also accepted at

this level of significance.Thus, it is probable that

the mass of metal contains 4% carbon at levels of

significance of 0.2 and 0.1.

Now try the following exercise.

Exercise 224 Further problems on signifi-

cance tests for population means

1. A batch of cables produced by a manufacturer
   have a mean breaking strength of 2000 kN
   and a standard deviation of 100 kN. A sample
   of 50 cables is found to have a mean break-
   ing strength of 2050 kN. Test the hypothesis
   that the breaking strength of the sample is
   greater than the breaking strength of the pop-
   ulation from which it is drawn at a level of
   significance of 0.01.
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎣
   z(sample)= 3 .54,zα= 2 .58,
   hence hypothesis is rejected,
   wherezαis thez-value
   corresponding to a level of
   significance ofα

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

2. Nine estimations of the percentage of copper
   in a bronze alloy have a mean of 80.8% and
   standard deviation of 1.2%. Assuming that
   the percentage of copper in samples is nor-
   mally distributed, test the null hypothesis that
   the true percentage of copper is 80% against
   an alternative hypothesis that it exceeds 80%,
   at a level of significance of 0.1.
   [
   t 0. 95 ,ν 8 = 1 .86,|t|= 1 .88, hence
   null hypothesis rejected

]

3. The internal diameter of a pipe has a mean
   diameter of 3.0000 cm with a standard devi-
   ation of 0.015 cm. A random sample of 30
   measurements are taken and the mean of the
   samples is 3.0078 cm. Test the hypothesis that
   the mean diameter of the pipe is 3.0000 cm at
   a level of significance of 0.01.
   [
   z(sample)= 2 .85,zα=± 2 .58,
   hence hypothesis is rejected

]

4. A fishing line has a mean breaking strength
   of 10.25 kN. Following a special treatment on
   the line, the following results are obtained for
   20 specimens taken from the line.