Question

1. In order to evaluate a spectrophotometric method for the determination of titanium, the method was applied to alloy samples containing difference certified amounts of titanium. The results (%Ti) are shown below.

Sample                        Certified Value           Mean               Standard Deviation

1                      0.496                           0.482               0.0257

2                      0.995                           1.009               0.0248

3                      1.493                           1.505               0.0287

4                      1.990                           2.002               0.0212

For each alloy, eight replicate determinations were made. For each alloy, test whether the mean value differs significantly from the certified value.

Answer 1

Sample 1: Data:     n = 8    μ = 0.496    s = 0.0257    x-bar = 0.482    Hypotheses:    Ho: μ = 0.496    Ha: μ ≠ 0.496    Decision Rule:    α = 0.05    Degrees of freedom = 8 - 1 = 7 Lower Critical t- score = -2.364624251 Upper Critical t- score = 2.364624251 Reject Ho if |t| > 2.364624251 Test Statistic:    SE = s/√n = 0.0257/√8 = 0.009086322 t = (x-bar - μ)/SE = (0.482 - 0.496)/0.00908632213824714 = -1.54078 p- value = 0.167269951    Decision (in terms of the hypotheses): Since 1.540777422 < 2.364624 we fail to reject Ho Conclusion (in terms of the problem): This sample meets the certified value Sample 2: Data:     n = 8    μ = 0.995    s = 0.0248    x-bar = 1.009    Hypotheses:    Ho: μ = 0.995    Ha: μ ≠ 0.995    Decision Rule:    α = 0.05    Degrees of freedom = 8 - 1 = 7 Lower Critical t- score = -2.364624251 Upper Critical t- score = 2.364624251 Reject Ho if |t| > 2.364624251 Test Statistic:    SE = s/√n = 0.0248/√8 = 0.008768124 t = (x-bar - μ)/SE = (1.009 - 0.995)/0.00876812408671319 = 1.596693 p- value = 0.15436533    Decision (in terms of the hypotheses): Since 1.596692732 < 2.364624 we fail to reject Ho Conclusion (in terms of the problem): This sample meets the certified value Sample 3: Data:     n = 8    μ = 1.493    s = 0.0287    x-bar = 1.505    Hypotheses:    Ho: μ = 1.493    Ha: μ ≠ 1.493    Decision Rule:    α = 0.05    Degrees of freedom = 8 - 1 = 7 Lower Critical t- score = -2.364624251 Upper Critical t- score = 2.364624251 Reject Ho if |t| > 2.364624251 Test Statistic:    SE = s/√n = 0.0287/√8 = 0.010146982 t = (x-bar - μ)/SE = (1.505 - 1.493)/0.010146982310027 = 1.182618 p- value = 0.275567805    Decision (in terms of the hypotheses): Since 1.182617613 < 2.364624 we fail to reject Ho Conclusion (in terms of the problem): This sample meets the certified value Sample 4: Data:     n = 8    μ = 1.99    s = 0.0212    x-bar = 2.002    Hypotheses:    Ho: μ = 1.99    Ha: μ ≠ 1.99    Decision Rule:    α = 0.05    Degrees of freedom = 8 - 1 = 7 Lower Critical t- score = -2.364624251 Upper Critical t- score = 2.364624251 Reject Ho if |t| > 2.364624251 Test Statistic:    SE = s/√n = 0.0212/√8 = 0.007495332 t = (x-bar - μ)/SE = (2.002 - 1.99)/0.0074953318805774 = 1.600996 p- value = 0.153411302    Decision (in terms of the hypotheses): Since 1.600996486 < 2.364624 we fail to reject Ho Conclusion (in terms of the problem): This sample meets the certified value.