602 STATISTICS AND PROBABILITYBreaking strength Frequency
(kN)
9.8 1
10 1
10.1 4
10.2 5
10.5 3
10.7 2
10.8 2
10.9 1
11.0 1Test the hypothesis that the special treatment
has improved the breaking strength at a level
of significance of 0.1.
⎡⎣x= 10 .38,s= 0 .33,
t 0. 95 ν 19 = 1 .73,|t|= 1 .72,
hence hypothesis is accepted⎤⎦- A machine produces ball bearings having
a mean diameter of 0.50 cm. A sample of
10 ball bearings is drawn at random and the
sample mean is 0.53 cm with a standard devi-
ation of 0.03 cm. Test the hypothesis that
the mean diameter is 0.50 cm at a level of
significance of (a) 0.05 and (b) 0.01.
⎡
⎢
⎢
⎢
⎣|t|= 3 .00,
(a)t 0. 975 ν 9 = 2 .26, hence
hypothesis rejected,
(b)t 0. 995 ν 9 = 3 .25, hence
hypothesis is accepted⎤⎥
⎥
⎥
⎦- Six similar switches are tested to destruction
at an overload of 20% of their normal max-
imum current rating. The mean number of
operations before failure is 8200 with a stan-
dard deviation of 145. The manufacturer of
the switches claims that they can be operated
at least 8000 times at a 20% overload current.
Can the manufacturer’s claim be supported at
a level of significance of (a) 0.1 and (b) 0.2?
⎡
⎢
⎢
⎢
⎣|t|= 3 .08,
(a)t 0. 95 ν 5 = 2 .02, hence claim
supported,
(b)t 0. 99 ν 5 = 3 .36, hence claim
not supported⎤⎥
⎥
⎥
⎦62.4 Comparing two sample means
The techniques introduced in Section 62.3 can be
used for comparison purposes. For example, it may
be necessary to compare the performance of, say, two
similar lamps produced by different manufacturers
or different operators carrying out a test or tests on
the same items using different equipment. The null
hypothesis adopted for tests involving two different
populations is that there isno differencebetween
the mean values of the populations.
The technique is based on the following theorem:Ifx 1 andx 2 are the means of random samples of sizeN 1
andN 2 drawn from populations having means ofμ 1 and
μ 2 and standard deviations ofσ 1 andσ 2 , then the sampling
distribution of the differences of the means,(x 1 −x 2 ),is
a close approximation to a normal distribution, having amean of zero and a standard deviation of√(
σ 12
N 1+σ^22
N 2)
.For large samples, when comparing the mean val-
ues of two samples, the variate is the difference in
the means of the two samples,x 1 −x 2 ; the mean of
sampling distribution (and hence the difference in
population means) is zero and the standard error ofthe sampling distributionσxis√
√
√
√(
σ^21
N 1+σ 22
N 2).Hence, thez-value is(x 1 −x 2 )− 0
√
√
√
√(
σ 12
N 1+σ^22
N 2)=x 1 −x 2
√
√
√
√(
σ^21
N 1+σ^22
N 2) (9)For small samples, Student’st-distribution values
are used and in this case:|t|=x 1 −x 2
√√
√
√(
σ 12
N 1+σ 22
N 2) (10)where|t|means the modulus oft, i.e. the positive
value oft.
When the standard deviation of the population
is not known, then Bessel’s correction is applied
to estimate it from the sample standard devia-
tion (i.e. the estimate of the population variance,σ^2 =s^2(
N
N− 1)
(see page 598). For large popu-lations, the factor(
N
N− 1)
is small and may beneglected. However, whenN<30, this correction
factor should be included. Also, since estimates of
bothσ 1 andσ 2 are being made, thekfactor in the