A consultant is asked to evaluate some bolt failures found during testing. He is informed that the yield point of a high-strength steel (HSS) bolt is known to be normally distributed with a variance of 1.6. He decides to test the hypothesis that [latex]σ^{2} = 1.6[/latex] against the alternative that [latex]σ^{2} ≠ 1.6[/latex]. He manages to “pull” a random sample of 5 of these fasteners for testing, and he finds that they have a standard deviation S = 2.1. He decides to use a 0.01 level of significance for testing. He utilizes the following approach:

Question 4.DS.14: A consultant is asked to evaluate some bolt failures found d......

Blake’s Design of Mechanical Joints [1062706]

A consultant is asked to evaluate some bolt failures found during testing. He is informed that the yield point of a high-strength steel (HSS) bolt is known to be normally distributed with a variance of 1.6. He decides to test the hypothesis that $σ^{2} = 1.6$ against the alternative that $σ^{2} ≠ 1.6$ . He manages to “pull” a random sample of 5 of these fasteners for testing, and he finds that they have a standard deviation S = 2.1. He decides to use a 0.01 level of significance for testing.

He utilizes the following approach:

Step-by-Step

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1. $H_{0} :\sigma^{2}=H_{a} :\sigma^{2} ≠1.6$ (two-tailed)
2. $\alpha= 0.01, n = 5$

3. $\chi ^{2}=\left(n-1\right)S^{2}/ \left(\sigma _{0}\right)^{2}$ where S = 2.1 and $\sigma^{2} _{0}=1.6$

which yields : $\chi ^{2}=\frac{\left(n-1\right)5^{2}}{\sigma^{2} _{0}} = \left(5-1\right)\frac{\left(2.1\right)^{2} }{\left(1.6\right)^{2} } =6.89$

4. From Table 4.6

For $\alpha /2=\frac{0.01}{2}=0.005 ∂=n-1 =5-1 =4$

\left(\chi_{\left(\alpha/2\right) } \right)^{2} =\chi ^{2}_{0.005}=0.207 \\ \left(\chi_{\left(1-\alpha/2\right) } \right)^{2} =\chi ^{2}_{0.995}=14.86

5. Analysis

Therefore, since $\chi^{2} = 6.89$ , which is greater than $\chi ^{2}_{0.005}= 0.207$ and less than $\chi ^{2}_{0.995}=14.86$ , the consultant concludes that at a significance level of 0.01 the null hypothesis is accepted (accept $H_{0}$ ) and he, therefore, concludes that the bolt variance is $σ^{2} = 1.6$ .

TABLE 4.6
Critical Values of the Chi-Square Distribution [9] Values of 2 P Corresponding to P

Df	χ^{2}_{.005}	χ^{2}_{.01}	χ^{2}_{.025}	χ^{2}_{.05}	χ^{2}_{.10}	χ^{2}_{.90}	χ^{2}_{.95}	χ^{2}_{.975}	χ^{2}_{.99}	χ^{2}_{.995}
1	.000039	.00016	.00098	.0039	.0158	2.71	3.84	5.02	6.63	7.88
2	.0100	.0201	.0506	.1026	.2107	4.61	5.99	7.38	9.21	10.60
3	.0717	.115	.216	.352	.584	6.25	7.81	9.35	11.34	12.84
4	.207	.297	.484	.711	1.064	7.78	9.49	11.14	13.28	14.86
5	.412	.554	.831	1.15	1.61	9.24	11.07	12.83	15.09	16.75
6	.676	.872	1.24	1.64	2.20	10.64	12.59	14.45	16.81	18.55
7	.989	1.24	1.69	2.17	2.83	12.02	14.07	16.01	18.48	20.28
8	1.34	1.65	2.18	2.73	3.49	13.36	15.51	17.53	20.09	21.96
9	1.73	2.09	2.70	3.33	4.17	14.68	16.92	19.02	21.67	23.59
10	2.16	2.56	3.25	3.94	4.87	15.99	18.31	20.48	23.21	25.19
11	2.60	3.05	3.82	4.57	5.58	17.28	19.68	21.92	24.73	26.76
12	3.07	3.57	4.4	5.23	6.30	18.55	21.03	23.34	26.22	28.30
13	3.57	4.11	5.01	5.89	7.04	19.81	22.36	24.74	27.69	29.82
14	4.07	4.66	5.63	6.57	7.79	21.06	23.68	26.12	29.14	31.32
15	4.60	5.23	6.26	7.26	8.55	22.31	25.00	27.49	30.58	32.8
16	5.14	5.81	6.91	7.96	9.31	23.54	26.30	28.85	32.00	34.27
18	6.26	7.01	8.23	9.39	10.86	25.99	28.87	31.53	34.81	37.16
20	7.43	8.26	9.59	10.85	12.44	28.41	31.41	34.17	37.57	40.00
24	9.89	10.86	12.40	13.85	15.66	33.2	36.42	39.36	42.98	45.56
30	13.79	14.95	16.79	18.49	20.6	40.26	43.77	46.98	50.89	53.67
40	20.71	22.16	24.43	25.51	29.05	51.81	55.76	59.34	63.69	66.77
60	35.53	37.48	40.48	43.19	46.46	74.40	79.08	83.30	88.38	91.95
120	83.85	86.92	91.58	95.70	100.62	40.23	146.57	152.21	158.95	163.64