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 is no difference between the mean values of the populations.
The technique is based on the following theorem:
If x1and x2are the means of random samples of size N1 and N2drawn from populations having means ofà1and à2and standard deviations ofσ1andσ2, then the sampling distribution of the differences of the means, (x1−x2), is a close approximation to a normal distribution, having a mean of zero and a standard deviation of
σ12 N1+ σ22
N2
.
For large samples, when comparing the mean val- ues of two samples, the variate is the difference in the means of the two samples, x1−x2; the mean of sampling distribution (and hence the difference in population means) is zero and the standard error of the sampling distributionσxis
56 67
σ21 N1+σ22
N2
. Hence, the z-value is
(x1−x2)−0 56
67 σ12
N1 + σ22 N2
= x1−x2 56
67 σ12
N1 + σ22 N2
(9)
For small samples, Student’s t-distribution values are used and in this case:
|t| = x1−x2 56
67
σ12 N1 + σ22
N2
(10)
where|t| means the modulus of t, i.e. the positive value of t.
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, σ2=s2
N N−1
(see page 598). For large popu- lations, the factor
N N−1
is small and may be neglected. However, when N<30, this correction factor should be included. Also, since estimates of bothσ1 andσ2 are being made, the k factor in the
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SIGNIFICANCE TESTING 603
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degrees of freedom in Student’s t-distribution tables becomes 2 and ν is given by (N1+N2−2). 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, the t-value is given by:
|t| = x1−x2 σ
1 N1 + 1
N2
(11)
An estimate of the standard deviationσ is 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 of σ12= N1s21
N1−1 and another estimate is based on a second sample, giving σ22= N2s22
N2−1, then a better estimate of the population variance,σ2, is given by:
σ2 = N1s21+N2s22 (N1−1)+(N2−1)
i.e. σ = 56 67
N1s12+N2s22 N1+N2−2
(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,σ1=σ2=σ=0.0 2 cm. Also N1=N2=40 and x1=1.51 cm, x2=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. x1=x2, hence it is H0: x1−x2=0.
The alternative hypothesis is that the size of the components has altered, i.e. that x1=x2, hence it is H1: x1−x2=0.
For a large sample having a known standard devi- ation of the population, the z-value of the difference of means of two samples is given by equation (9), i.e.,
z= x1−x2 56
67 σ12
N1 + σ22 N2
Since N1=N2=say, N, andσ1=σ2=σ, this equa- tion becomes
z= x1−x2 σ
2 N
= 1.51−1.52 0.02
2 40
= −2.236
Since the difference between x1and x2has 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 the z-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 be à1, and that of product 2 beà2.
The null hypothesis is that the means are the same, i.e. H0:à1−à2=0.
The alternative hypothesis is that the means are not the same, i.e. H1:à1−à2=0.
The population standard deviations are known, i.e.
σ1=8 ohms and σ2=7 ohms, the sample means are known, i.e. x1=74 ohms and x2=78 ohms.
Also the sample sizes are known, i.e. N1=40 and
604 STATISTICS AND PROBABILITY
N2=50. Hence, equation (9) can be used to deter- mine the z-value of the difference of the sample means. From equation (9),
z= x1−x2 56
67
σ12 N1 + σ22
N2
= 74−78 82
40+ 72 50
= −4
1.606 = −2.49
(a) For a two-tailed test, the results are probably significant at a 0.05 level of significance when z lies between −1.96 and +1.96. Hence the z-value of the difference of means shows there is ‘no significance’, i.e. that product 1 is signif- icantly different from product 2 at a level of significance of 0.05.
(b) For a two-tailed test, the results are highly signif- icant at a 0.01 level of significance when z lies between −2.58 and +2.58. Hence there is no significant difference between product 1 and product 2 at a level of significance of 0.01.
Problem 9. The reaction time in seconds of two people, A and B, are measured by electrodermal responses and the results of the tests are as shown below.
Person A (s) 0.243 0.243 0.239 Person B (s) 0.238 0.239 0.225 Person A (s) 0.232 0.229 0.241 Person B (s) 0.236 0.235 0.234 Find if there is any significant difference between the reaction times of the two people at a level of significance of 0.1.
The mean, x, and standard deviation, s, of the response times of the two people are determined.
xA=
0.243+0.243+0.239+0.232 +0.229+0.241
6
=0.2378 s
xB=
0.238+0.239+0.225+0.236 +0.235+0.234
6
=0.2345 s
sA = 56 66 66 7
⎡
⎢⎢
⎣
(0.243−0.2378)2+(0.243−0.2378)2 + ã ã ã +(0.241−0.2378)2
6
⎤
⎥⎥
⎦
=0.00543 s
sB = 56 66 66 7
⎡
⎢⎢
⎣
(0.238−0.2345)2+(0.239−0.2345)2 + ã ã ã +(0.234−0.2345)2
6
⎤
⎥⎥
⎦
=0.00457 s
The null hypothesis is that there is no difference between the reaction times of the two people, i.e.
H0: xA−xB=0.
The alternative hypothesis is that the reaction times are different, i.e. H1: xA−xB=0 indicating a two-tailed test.
The sample numbers (combined) are less than 30 and a t-distribution is used. The standard deviation of all the reaction times of the two people is not known, so an estimate based on the standard deviations of the samples is used. Applying Bessel’s correction, the estimate of the standard deviation of the population,
σ2=s2 N
N−1
gives σA =(0.00543) 6
5
=0.00595
and σB =(0.00457) 6
5
=0.00501
From equation (10), the t-value of the difference of the means is given by:
|t| = xA−xB 56
67 σA2
NA + σB2 NB
= 0.2378−0.2345 0.005952
6 +0.005012 6
=1.039
For a two-tailed test and a level of significance of 0.1, the column heading in the t-distribution of Table 61.2 (on page 587) is t0.95(refer to Problem 6).
The degrees of freedom due to k being 2 is ν = N1+N2−2, i.e. 6+6−2=10. The corresponding t-value from Table 61.2 is 1.81. Since the t-value
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SIGNIFICANCE TESTING 605
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of the difference of the means is within the range
±1.81, there is no significant difference between the reaction times at a level of significance of 0.1.
Problem 10. An analyst carries out 10 analyses on equal masses of a substance which is found to contain a mean of 49.20 g of a metal, with a standard deviation of 0.41 g. A trainee operator carries out 12 analyses on equal masses of the same substance which is found to contain a mean of 49.30 g, with a standard deviation of 0.32 g.
Is there any significance between the results of the operators?
Letà1andà2be the mean values of the amounts of metal found by the two operators.
The null hypothesis is that there is no difference between the results obtained by the two operators, i.e. H0:à1=à2.
The alternative hypothesis is that there is a differ- ence between the results of the two operators, i.e.
H1:à1=à2.
Under the hypothesis H0 the standard deviations of the amount of metal,σ, will be the same, and from equation (12)
σ = 56 67
N1s12+N2s22 N1+N2−2
=
(10)(0.41)2+(12)(0.32)2 10+12−2
=0.3814
The t-value of the results obtained is given by equation (11), i.e.,
|t| = x1−x2 σ
1 N1 + 1
N2
= 49.20−49.30 (0.3814)
1 10+ 1
12
=−0.612
For the results to be probably significant, a two-tailed test and a level of significance of 0.05 is taken. H0is rejected outside of the range t−0.975and t0.975. The number of degrees of freedom is N1+N2−2. For t0.975,ν=20, from Table 61.2 on page 587, the range is from−2.09 to+2.09. Since the t-value based on
the sample data is within this range, there is no sig- nificant difference between the results of the two operators at a level of significance of 0.05.
Now try the following exercise.
Exercise 225 Further problems on compar- ing two sample means
1. A comparison is being made between batter- ies used in calculators. Batteries of type A have a mean lifetime of 24 hours with a stan- dard deviation of 4 hours, this data being calculated from a sample of 100 of the bat- teries. A sample of 80 of the type B batteries has a mean lifetime of 40 hours with a stan- dard deviation of 6 hours. Test the hypothesis that the type B batteries have a mean lifetime of at least 15 hours more than those of type A, at a level of significance of 0.05.
⎡
⎢⎢
⎢⎢
⎣
Take x as 24+15,
i.e. 39 hours, z=1.28, z0.05, one-tailed test=1.645, hence hypothesis is accepted
⎤
⎥⎥
⎥⎥
⎦
2. Two randomly selected groups of 50 opera- tives in a factory are timed during an assem- bly operation. The first group take a mean time of 112 minutes with a standard deviation of 12 minutes. The second group take a mean time of 117 minutes with a standard devia- tion of 9 minutes. Test the hypothesis that the mean time for the assembly operation is the same for both groups of employees at a level of significance of 0.05.
⎡
⎢⎢
⎣
z=2.357, z0.05,
two-tailed test= ±1.96, hence hypothesis is rejected
⎤
⎥⎥
⎦
3. Capacitors having a nominal capacitance of 24àF but produced by two different compa- nies are tested. The values of actual capaci- tance are:
Company 1 21.4 23.6 24.8 22.4 26.3 Company 2 22.4 27.7 23.5 29.1 25.8 Test the hypothesis that the mean capacitance of capacitors produced by company 2 are higher than those produced by company 1 at
606 STATISTICS AND PROBABILITY
a level of significance of 0.01.
Bessel’s correction isσˆ2= s2N N−1
.
⎡
⎢⎢
⎢⎣
x1 =23.7, s1=1.73, σ1 =1.93, x2=25.7, s2 =2.50,σ2=2.80,
|t| =1.62, t0.995ν8=3.36, hence hypothesis is accepted
⎤
⎥⎥
⎥⎦
4. A sample of 100 relays produced by manufac- turer A operated on average 1190 times before failure occurred, with a standard deviation of 90.75. Relays produced by manufacturer B, operated on average 1220 times before failure with a standard deviation of 120. Determine if the number of operations before failure are significantly different for the two manufac- turers at a level of significance of (a) 0.05 and (b) 0.1.
⎡
⎢⎢
⎢⎣
z (sample)=1.99,
(a) z0.05, two-tailed test= ±1.96, no significance, (b) z0.1, two-tailed test= ±1.645,
significant difference
⎤
⎥⎥
⎥⎦
5. A sample of 12 car engines produced by manufacturer A showed that the mean petrol consumption over a measured dis- tance was 4.8 litres with a standard devia- tion of 0.40 litres. Twelve similar engines
for manufacturer B were tested over the same distance and the mean petrol consump- tion was 5.1 litres with a standard deviation of 0.36 litres. Test the hypothesis that the engines produced by manufacturer A are more economical than those produced by manufacturer B at a level of significance of (a) 0.01 and (b) 0.1.
⎡
⎢⎢
⎢⎢
⎣
Assuming null hypothesis of no difference,σ=0.397,|t| =1.85,
(a) t0.995,ν22=2.82, hypothesis rejected, (b) t0.95,ν22=1.72, hypothesis accepted
⎤
⎥⎥
⎥⎥
⎦
6. Four-star and unleaded petrol is tested in 5 similar cars under identical conditions. For four-star petrol, the cars covered a mean distance of 21.4 kilometres with a standard deviation of 0.54 kilometres for a given mass of petrol. For the same mass of unleaded petrol, the mean distance covered was 22.6 kilometres with a standard devia- tion of 0.48 kilometres. Test the hypothesis that unleaded petrol gives more kilometres per litre than four-star petrol at a level of significance of 0.1.
σ=0.571,|t| =3.32, t0.95, ν8=1.86, hence hypothesis
is rejected
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Statistics and probability
63
Chi-square and distribution-free tests