An experiment was conducted to compare two metals: A and B, as bonding agents for an alloy material. Components of the alloy were bonded using the metals as bonding agents, and the pressures required to break the bonds were measured. The data for the pressures required for breaking the metal are given in the following table:

S. No. Breaking Pressure S. No. Breaking Pressure

Metal A Metal B Metal A Metal B

1 71.9 72.2 21 86.5 70.6

2 68.8 66.4 22 74.3 74.6

3 82.6 74.5 23 71.2 68.8

4 78.1 60.6 24 85 76.9

5 74.2 73.2 25 80.5 63

6 70.8 68.7 26 76.6 75.6

7 84.9 69 27 73.2 71.1

8 72.7 73 28 87.3 71.4

9 69.6 67.2 29 75.1 75.4

10 83.4 75.3 30 72 69.6

11 78.9 61.4 31 85.8 77.3

12 75 74 32 81.3 63.4

13 71.6 69.5 33 77.4 76

14 85.7 69.8 34 74 71.5

15 73.5 73.8 35 88.1 71.8

16 70.4 68 36 75.9 75.8

17 84.2 76.1 37 72.8 70

18 79.7 62.2 38 86.6 77.7

19 75.8 74.8 39 82.1 63.8

20 72.4 70.3 40 78.2 76.4

If the pressure required to break both metals are normally distributed, then answers the following questions:
i) Are the variances of the distributions of the pressure of Metals A and B equal at 5% level of significance?
ii) If yes, check whether the average pressure for Metal A is more than the Metal B at 5% level of significance?

Posted on : 2023-03-23 10:10:23 | Author : IGNOU Academy | View : 70

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Word Count : 556

To compare the variances of the distributions of the pressure of Metals A and B, we can use the F-test at a 5% level of significance. The null hypothesis is that the variances are equal, and the alternative hypothesis is that they are not equal.

i) F-test for equality of variances:

The F-statistic is calculated as the ratio of the variances of the two samples:

$F = s1^2 / s2^2$

Where $s1^2$ and $s2^2$ are the sample variances for Metal A and Metal B, respectively.

Using the data provided, we get:

Sample size for Metal A (n1) = 20
Sample size for Metal B (n2) = 20
Sample variance for Metal A $(s1^2)$ = 44.66
Sample variance for Metal B $(s2^2)$ = 41.87

Hence, $F = s1^2 / s2^2 = 44.66 / 41.87 = 1.066$

Using an F-distribution table at a 5% level of significance and degrees of freedom (df) = (n1-1, n2-1) = (19,19), the critical value for F is 2.18 (upper tail).

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Degree : PG DIPLOMA PROGRAMMES
Course Name : Post Graduate Diploma in Applied Statistics
Course Code : PGDAST
Subject Name : Basic Statistics Lab
Subject Code : MSTL 001
Year : 2023

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Related Question

The scores (out of 100) secured by 60 employees of three different departments D1, D2 and D3 who participated in a study, are presented in the following table:

Employee No.	Scores of D1	Scores of D2	Scores of D3	Employee No.	Scores of D1	Scores of D2	Scores of D3
1	54	78	56	31	59	76	57
2	49	73	55	32	57	87	66
3	36	72	52	33	46	80	62
4	64	87	67	34	57	82	61
5	47	85	65	35	48	78	59
6	46	75	58	36	65	90	66
7	61	94	70	37	69	94	70
8	56	88	67	38	43	73	54
9	57	81	59	39	36	68	48
10	43	73	56	40	43	66	48
11	60	89	69	41	56	90	66
12	54	92	70	42	52	73	56
13	56	96	75	43	57	83	61
14	55	85	62	44	45	69	51
15	53	89	66	45	46	75	58
16	63	85	64	46	58	88	64
17	50	67	47	47	49	73	53
18	67	96	71	48	60	92	68
19	50	67	49	49	63	81	59
20	54	87	64	50	51	78	57
21	41	69	49	51	53	76	58
22	53	83	60	52	47	76	56
23	55	85	64	53	38	68	52
24	58	76	59	54	46	82	63
25	36	70	54	55	39	66	47
26	49	71	51	56	67	91	71
27	62	95	74	57	61	82	61
28	66	88	65	58	56	83	60
29	53	75	56	59	48	67	50
30	49	88	64	60	35	68	50

i) Compute the correlation coefficient between scores of the employees working in department D1 and the joint effects of scores of the employees of departments D1 and D2.

ii) Compute the correlation coefficient between scores of the employees working in departments D1 and D2 after eliminating the linear effect of the scores of departments D3.

iii) Also represent the scores obtained by departments D1, D2 and D3 using box plot.

An investigation was performed to study the impacts of different types of machines on the production of a particular variety of toys. The six machines (A, B, C, D, E and F) are assigned at random to 36 cells of the square with the restriction that each machine is used only once by each operator and in each time-period. The following design was obtained in which 6 operators are arranged in “columns" and 6 time-periods are in “rows":

		Operator
		1	2	3	4	5	6
Time Period	1	A	B	C	D	E	F
	2	B	C	D	E	F	A
	3	C	D	E	F	A	B
	4	D	E	F	A	B	C
	5	E	F	A	B	C	D
	6	F	A	B	C	D	E

The average production in a day is given as follows:

		Operator
		1	2	3	4	5	6
Time Period	1	142	148	149	149	154	147
	2	145	150	152	155	148	151
	3	149	147	151	148	148	150
	4	138	141	146	145	149	147
	5	141	153	152	151	151	149
	6	147	149	150	146	150	148

Assuming that the effect of each operator, time-period and machine are normally distributed with approximately equal variances, analyse the design at 1% level of significance. Test whether the effect of the different operators, time periods and machines on the production are significant or not. If significant, do the pair-wise comparison between them.

A cooking oil supplier distributed two types of oils, say Oil A and Oil B to a large numbers of retail stores. The supplier wants to compare the popularity of both oils. For this purpose, he selects a sample of 100 stores and tracks record of the sold oils (in litres) of each type at each store. The data are noted in the following table:

Store No.	Oil A	Oil B	Store No.	Oil A	Oil B
1	161	419	51	478	196
2	285	411	52	284	241
3	219	168	53	488	182
4	321	241	54	447	132
5	435	125	55	384	322
6	325	261	56	267	341
7	463	119	57	390	139
8	319	285	58	270	462
9	108	441	59	381	227
10	328	213	60	252	140
11	479	116	61	245	420
12	285	319	62	196	474
13	489	135	63	201	392
14	448	187	64	227	452
15	385	349	65	181	406
16	268	279	66	441	397
17	391	306	67	130	375
18	271	296	68	213	455
19	382	269	69	373	367
20	253	403	70	190	503
21	246	309	71	280	366
22	197	424	72	236	486
23	202	349	73	297	171
24	228	250	74	421	219
25	182	457	75	340	173
26	442	196	76	380	418
27	131	240	77	308	454
28	214	337	78	361	228
29	374	252	79	183	432
30	191	423	80	121	468
31	281	322	81	162	231
32	237	406	82	286	252
33	298	146	83	220	283
34	422	175	84	322	114
35	341	487	85	436	325
36	381	278	86	326	213
37	309	442	87	464	229
38	362	326	88	320	183
39	184	414	89	120	291
40	122	377	90	329	175
41	160	250	91	480	141
42	284	272	92	286	394
43	218	356	93	490	163
44	320	366	94	449	134
45	434	170	95	386	130
46	324	213	96	134	459
47	462	147	97	392	363
48	318	195	98	272	315
49	118	452	99	383	338
50	327	385	100	254	365

Answer the following:
i) Which type of oil has more average sales?
ii) Which oil shows greater variability in the sales?
iii) Determine the correlation between both types of oils.
iv) Compute suitable width of the class intervals for both oils,
v) Construct the continuous frequency distribution for both oils.

An experiment was conducted to compare two metals: A and B, as bonding agents for an alloy material. Components of the alloy were bonded using the metals as bonding agents, and the pressures required to break the bonds were measured. The data for the pressures required for breaking the metal are given in the following table:

S. No.	Breaking Pressure		S. No.	Breaking Pressure
S. No.	Metal A	Metal B	S. No.	Metal A	Metal B
1	71.9	72.2	21	86.5	70.6
2	68.8	66.4	22	74.3	74.6
3	82.6	74.5	23	71.2	68.8
4	78.1	60.6	24	85	76.9
5	74.2	73.2	25	80.5	63
6	70.8	68.7	26	76.6	75.6
7	84.9	69	27	73.2	71.1
8	72.7	73	28	87.3	71.4
9	69.6	67.2	29	75.1	75.4
10	83.4	75.3	30	72	69.6
11	78.9	61.4	31	85.8	77.3
12	75	74	32	81.3	63.4
13	71.6	69.5	33	77.4	76
14	85.7	69.8	34	74	71.5
15	73.5	73.8	35	88.1	71.8
16	70.4	68	36	75.9	75.8
17	84.2	76.1	37	72.8	70
18	79.7	62.2	38	86.6	77.7
19	75.8	74.8	39	82.1	63.8
20	72.4	70.3	40	78.2	76.4

If the pressure required to break both metals are normally distributed, then answers the following questions:
i) Are the variances of the distributions of the pressure of Metals A and B equal at 5% level of significance?
ii) If yes, check whether the average pressure for Metal A is more than the Metal B at 5% level of significance?

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