CHAPTER 93 SIGNIFICANCE TESTING

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1 CHAPTER 93 SIGNIFICANCE TESTING EXERCISE 342 Page Random samples of 1000 rings are drawn from the output of a machine periodically for inspection purposes. A defect rate of 5% is acceptable to the manufacturer, but if the defect rate is believed to have exceeded this value, the machine producing the rings is stopped and adjusted. Determine the type I errors which occur for the following decision rules: Stop production and adjust the machine if a sample contains (a) 54, (b) 62 and (c) 70 or more defective rings. (a) N = 1000, p = 0.05, q = 0.95, mean of normal distribution = Np = 50, standard deviation of the normal distribution = ( Npq ) = ( 1000)( 0.05)( 0.95) = A type I error is the probability of stopping production when getting more than 54 defective rings in the sample, even though defect rate is 5% z-value = variate mean standard deviation = = 0.58 and from Table 89.1, page 944 of the textbook, the area between the mean and a z-value of 0.58 is Thus, the probability of more than 54 defective rings = = Hence, the type I error is 28.1% (b) z-value = is = 1.74 and from Table 89.1, the area between the mean and a z-value of 1.74 Thus, the probability of more than 62 defective rings = = Hence, the type I error is 4.09% (c) z-value = = 2.90 and from Table 89.1, the area between the mean and a z-value of 2.90 is

2 Thus, the probability of more than 70 defective rings = = Hence, the type I error is 0.19% 2. Random samples of 1000 rings are drawn from the output of a machine periodically for inspection purposes. A defect rate of 5% is acceptable to the manufacturer, but if the defect rate is believed to have exceeded this value, the machine producing the rings is stopped and adjusted. Determine the type II errors which are made if the decision rule is to stop production if there are more than 60 defective components in the sample when the actual defect rate has risen to (a) 6%, (b) 7.5% and (c) 9%. (a) N = 1000, p = 0.06, q = 0.94 Mean = Np = 60, standard deviation = ( Npq ) = ( 1000)( 0.06)( 0.94) = 7.51 z-value = = 0.13 (note that more than 60 components defective means 61 or more) and from Table 89.1, page 944 of textbook, the area between the mean and a z-value of 0.13 is Thus, the probability of more than 60 defective components = = Hence, the type II error is 55.2% (b) N = 1000, p = 0.075, q = Mean = Np = 75, standard deviation = ( Npq ) = ( 1000)( 0.075)( 0.925) = z-value = = 1.68 and from Table 89.1, the area between the mean and a z-value of 1.68 is Thus, the probability of more than 75 defective components = = Hence, the type II error is 4.65% (c) N = 1000, p = 0.09, q = 0.91 Mean = Np = 90, standard deviation = ( Npq ) = ( 1000)( 0.09)( 0.91) =

3 z-value = = 3.20 and from Table 89.1, the area between the mean and a z-value of 3.20 is Thus, the probability of more than 90 defective components = = Hence, the type II error is 0.07% 3. A random sample of 100 components is drawn from the output of a machine whose defect rate is 3%. Determine the type I error if the decision rule is to stop production when the sample contains: (a) four or more defective components, (b) five or more defective components, and (c) six or more defective components. N = 100, p = 0.03 and Np = 3 Since N 50 and Np 5 the Poisson distribution is used. λ = Np = 3 The probability of 0, 1, 2, 3, 4, 5, defective components are given by the terms: = =, λ e λ = 3e 3 = , e λ e λ 2 32 λ 3 e = e = , 2! 2 λ 3 33 λ 3 e = e = , 3! 6 λ 4 34 λ 3 e = e = , 4! 24 λ 5 35 λ 3 e = e = ! 120 (a) The probability of a sample containing four or more defective components is: 1 ( ) = = i.e. the type I error is 35.3% (b) The probability of a sample containing five or more defective components is: 1 ( ) = = i.e. the type I error is 18.5% (c) The probability of a sample containing six or more defective components is: 1 ( ) = =

4 i.e. the type I error is 8.4% 4. If there are four or more defective components in a sample drawn from the machine given in Problem 3, determine the type II error when the actual defect rate is: (a) 5%, (b) 6% and (c) 7%. A type II error is the probability of accepting a hypothesis when it should be rejected. The type II error in this problem is the probability of a sample containing four or more defective components (a) N = 100, p = 0.05, q = 0.95 Mean = Np = 5, standard deviation = ( Npq ) = ( 100)( 0.05)( 0.95) = 2.18 z-value = = 0.46 and from Table 89.1, page 944 of textbook, the area between the mean and a z-value of 0.46 is Thus, the probability of four or more defective components = = Hence, the type II error is 32.3% (b) N = 100, p = 0.06, q = 0.94 Mean = Np = 6, standard deviation = ( Npq ) = ( 100)( 0.06)( 0.94) = z-value = = 0.84 and from Table 89.1, the area between the mean and a z-value of 0.84 is Thus, the probability of four or more defective components = = Hence, the type II error is 20.1% (c) N = 100, p = 0.07, q = 0.93 Mean = Np = 7, standard deviation = ( Npq ) = ( 100)( 0.07)( 0.93) = z-value = = 1.18 and from Table 89.1, the area between the mean and a z-value of 1.18 is Thus, the probability of four or more defective components = =

5 Hence, the type II error is 11.9% 1425

6 EXERCISE 343 Page 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 are found to have a mean breaking strength of 2050 kn. Test the hypothesis that the breaking strength of the sample is greater than the breaking strength of the population from which it is drawn at a level of significance of 0.01 µ = 2000 kn, σ = 100 kn, N = 50, x = 2050kN The null hypothesis, H0 : x > µ The alternative hypothesis, H1 : x = µ z = µ σ x N = =± =± The value for a two-tailed test is given in Table 93.1, and at a significance level of 0.01 is ± 2.58 Since the z-value of the sample is outside of this range, the hypothesis is rejected 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 normally 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 0.05 µ = 80.8% = 0.808, N = 9, x = 80% = 0.80 and s = 1.2% = The null hypothesis is: H 0 : µ = 8000 t ( x µ ) ( N 1) ( ) ( 9 1) = = = 1.89 s From Table 92.2, page 971, t0.95, ν = 8 has a value of 1.86 Since 1.89 > 1.86, the null hypothesis is rejected 1426

7 3. The internal diameter of a pipe has a mean diameter of cm with a standard deviation of cm. A random sample of 30 measurements are taken and the mean of the samples is cm. Test the hypothesis that the mean diameter of the pipe is cm at a level of significance of 0.01 µ = cm, σ = cm, N = 30, x = cm The null hypothesis, H 0 : mean diameter = cm The alternative hypothesis, H: 1 mean diameter cm z = x µ σ N = =± =± The value for a two-tailed test is given in Table 93.1, page 978, and at a significance level of 0.01 is ±2.58 Since the z-value of the sample is outside of this range, the hypothesis is rejected 4. A fishing line has a mean breaking strength of kn. Following a special treatment on the line, the following results are obtained for 20 specimens taken from the line. Breaking strength (kn) Frequency Test the hypothesis that the special treatment has improved the breaking strength at a level of significance of 0.05 µ = kn and N = 20 Sample mean, ( ) ( ) ( ) ( ) x = = = Sample standard deviation, s = ( ) ( ) ( ) ( ) = =

8 The null hypothesis is that the sample breaking strength is better than the mean breaking strength. N < 30, therefore a t-distribution is used t ( x µ ) ( N 1) ( ) ( 20 1) = = = 1.72 s 0.33 At a level of significance of 0.1, the t value is 0.1 t i.e. t 0.95 and ν = N 1 = 20 1 = 19, and from Table 92.2, page 971, t 0.95 ν = 19 has a value of 1.73 Since 1.72 is within this range, the hypothesis is accepted 5. A machine produces ball bearings having a mean diameter of 0.50 cm. A sample of ten ball bearings is drawn at random and the sample mean is 0.53 cm with a standard deviation of 0.03 mm. Test the hypothesis that the mean diameter is 0.50 cm at a level of significance of (a) 0.05 and (b) 0.01 µ = 0.50 cm, N = 10, x = 0.53 cm and s = 0.03 cm The null hypothesis is: H: 0 µ = 0.50 cm t ( x µ ) ( N 1) ( ) ( 10 1) = = = 3.00 s 0.03 (a) From Table 92.2, page 971, t = t0.975, ν = 9 has a value of Since 3 is outside of this range, the hypothesis is rejected (b) From Table 92.2, t = t0.995, ν = 9 has a value of Since 3 is within this range, the hypothesis is accepted 6. Six similar switches are tested to destruction at an overload of 20% of their normal maximum current rating. The mean number of operations before failure is 8200, with a standard deviation of 145. The manufacturer of the switches claims 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 1428

9 (a) 0.1 and (b) 0.02? µ = 8000, N = 6, x = 8200 and s = 145 The null hypothesis is: H: 0 µ = 8000 t ( x µ ) ( N 1) ( ) ( 6 1) = = = 3.08 s 145 (a) From Table 92.2, page 971, t = t0.95, ν = 5 has a value of Since 3.08 > 2.02, the hypothesis is supported (b) From Table 92.2, t = t0.99, ν = 5 has a value of Since 3.08 < 3.36, the hypothesis is not supported 1429

10 EXERCISE 344 Page A comparison is being made between batteries used in calculators. Batteries of type A have a mean lifetime of 24 hours with a standard deviation of 4 hours, this data being calculated from a sample of 100 of the batteries. A sample of 80 of the type B batteries have a mean lifetime of 40 hours with a standard 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 Battery A: x A = 24, σ A = 4 and N A = 100 Battery B: x B = 40, σ B = 6 and N B = 80 The hypothesis is: H: x = x A + 15 Let x = = 39 z = x xb = = = σ A σb NA NB From Table 93.1, for α = 0.05, one-tailed test, z = Since the z-value is within this range, the hypothesis is accepted 2. Two randomly selected groups of 50 operatives in a factory are timed during an assembly 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 deviation 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 Group 1: x 1 = 112min, σ 1 = 12 min and N 1 = 50 Group 2: x 2 = 117 min, σ 2 = 9 min and N 2 = 50 The hypothesis is: H: x 1 = x

11 Let x = = 39 z = x σ x 2 1 σ = = = N1 N From Table 93.1, for α = 0.05, two-tailed test, z = ±1.96 Since the z-value is outside of this range, the hypothesis is rejected 3. Capacitors having a nominal capacitance of 24 µf but produced by two different companies are tested. The values of actual capacitance are: Company X Company Y Test the hypothesis that the mean capacitance of capacitors produced by company Y are higher than those produced by company X at a level of significance of 0.01 (Bessel s correction is ˆ 2 σ = sn 2 N 1 ) N = 5, x1 = = ( ) + ( ) + ( ) + ( ) + ( ) s1 = = N 5 σ1 = s1 = 1.73 = 1.93 N 1 4 x = = s 2 ( ) + ( ) + ( ) + ( ) + ( ) = = N 5 σ 2 = s2 = 2.50 = 2.80 N

12 t x2 x = = = = σ x σ y + + Nx Ny 5 5 t 0.01 = t and ν = N1+ N2 2= = From Table 92.2, page 971, t0.995, ν = 8 has a value of 3.36 Since the t value of the difference of the means, i.e. 1.32, is within the range ±3.36, the hypothesis is accepted 4. A sample of 100 relays produced by manufacturer A operated on average 1190 times before failure occurred, with a standard deviation of 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 manufacturers at a level of significance of (a) 0.05 and (b) 0.01 Manufacturer A: x A = 1190, σ A = and N A = 100 Manufacturer B: x B = 1220, σ B = 120 and N B = 100 The hypothesis is: H: x 1 = x 2 z = x σ x 2 1 σ = = = N1 N (a) From Table 93.1, for α = 0.05, two-tailed test, z = ±1.96 Since the z-value is outside of this range, the number of operations before failure is not significantly different (b) From Table 93.1, for α = 0.01, two-tailed test, z = ±2.58 Since the z-value is within this range, the number of operations before failure are significantly different 1432

13 5. A sample of 12 car engines produced by manufacturer A showed that the mean petrol consumption over a measured distance was 4.8 litres with a standard deviation of 0.40 litres. Twelve similar engines for manufacturer B were tested over the same distance and the mean petrol consumption 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) 0.05 N = x = σ = A 12, A 4.8 litres, A 0.40 litres N = x = σ = B 12, B 5.1 litres, B 0.36 litres The hypothesis is: H: manufacturer A is more economical than manufacturer B ( ) + ( ) NA sa + NBsB σ = = = = N + N A B t xa xb = = = = ( 0.397)( ) σ N N A B (a) The t value is 0.01 t 2 i.e. t and ν = = 22, hence, from Table 92.2, t0.995, ν = 22 has a value of 2.82 Since the t value of the difference of the means is outside the range ±1.85, the hypothesis is rejected (b) From Table 92.1, 0.1 t i.e. t 0.95, ν = 22 has a value of 1.72 Since the t value of the difference of the means is within the range ±1.85, the hypothesis is accepted 6. Unleaded and premium petrol is tested in five similar cars under identical conditions. For unleaded 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 premium petrol, the mean 1433

14 distance covered was 22.6 kilometres with a standard deivation of 0.48 km. Test the hypothesis that premium petrol gives more kilometres per litre than unleaded petrol at a level of significance of 0.05 N = 5, x = 21.4 km, s = 0.54 km UN UN x PREM = 22.6 km, s = 0.48 km PREM The hypothesis is: H: x PREM > x UN ( ) + ( ) NUN sun + NPREM sprem σ = = = = N + N UN PREM t x x PREM UN = = = = σ N N 5 5 PREM UN ( 0.571)( ) 3.32 The t value is 0.1 t i.e. t 0.95 and ν = = 8, hence, from Table 92.2, page 971, t0.95, ν = 8 has a value of 1.86 Since the t value of the difference of the means, i.e. 3.32, is outside of the range ±3.32, the hypothesis is rejected 1434