Question

Conduct a formal hypothesis test of the claim that the mean longevity is less than 57 days. Test at significance α=0.05. Your written summary of this test must include the following: Your null and alternate hypotheses in the proper format. The type of distribution you used to construct the interval (t or normal). The P-value and its logical relationship to α (≤ or >). Your decision regarding the null hypothesis: reject or fail to reject. A statement regarding the sufficiency/insufficiency of the evidence to reject/support the original claim.

Pct Time Asleep	Longevity
22	35
9	37
49	49
1	46
23	63
83	39
23	46
15	56
9	63
81	65
12	56
15	65
37	70
24	63
26	65
17	70
14	77
14	81
6	86
25	70
18	70
26	77
24	77
29	81
27	77
18	40
6	37
19	44
7	47
16	47
13	47
35	68
2	47
35	54
6	61
15	71
14	75
18	89
50	58
25	59
10	62
33	79
43	96
35	58
17	62
27	70
22	72
16	75
20	96
37	75
23	46
4	42
20	65
42	46
9	58
32	42
66	48
28	58
10	50
4	80
12	63
17	65
12	70
23	70
40	72
18	97
10	46
38	56
7	70
23	70
36	72
9	76
21	90
62	76
36	92
23	21
62	40
28	44
18	54
10	36
28	40
22	56
29	60
15	48
73	53
10	60
5	60
13	65
27	68
20	60
21	81
12	81
49	48
17	48
22	56
71	68
17	75
10	81
24	48
18	68
34	16
6	19
4	19
22	32
28	33
31	33
16	30
27	42
8	42
32	33
20	26
35	30
12	40
14	54
17	34
29	34
31	47
6	47
30	42
27	47
40	54
19	54
8	56
8	60
15	44

Answer 1

Answer Date: 31/07/2020 To test the claim is that the mean longevity is less than 57 days at 5% significance level. The null and alternative hypothesis is, H: = 57 H:H< 57 The t-test statistics is, By using TI-84 calculator, find t-test statistics with the help of following steps: 1) Click ON. 2) Select 1:Edit from EDIT. 3) Enter Data 4) Select TESTS from Stat menu. 5) Select T Test from TESTS variables. 6) Select Data from Input:Data and L1 from List 1. 7) Enter Hypothesized mean (57), and select Alternative from T Test. 8) Select Alternative from Hypotheses. 9) Press Calculate.

T-Test <57 t=.2800831939 P=.610059775 X=57.44 Sx=17.56389258 n=125 From TI-84 calculator output, the t-test statistics is 0.280 The p-value for this test is, From TI-84 calculator output, the p-value for this test is 0.6101 Decision The p-value is higher than 0.05 which is 0.6101, so fail to reject the null hypothesis at 5% level of significance. There is insufficient evidence to support that the mean longevity is less than 57 days. The result is not statistically significant.