Question

The hypothesis is that the mean BMI of the students is lower than 24.

A. What is the right set of null and alternative hypotheses?

B. What's the p-value for this test to FOUR decimals ? (Note: check if the above test is one-sided or two-sided first)

C. At significance level 5%, we can reject the null hypothesis and claim that the mean BMI is less than 24 for the student population of interest. True or False?

Age	BMI
35	24
23	20
23	18.2
24	22.3
.	.
28	.
32	25.8
24	22.8
27	19.1
24	.
22	18.5
22	22
23	18.6
49	.
41	25
21	27.5
24	20.4
22	24
25	21
45	25.8
26	22
.	27.2
32	21.1
.	25
42	27
28	20
47	24.8
29	17
31	20.9
28	19.8
26	.
21	19.9
22	29
30	0.2
26	22.3
24	19.9
25	.
28	23
23	22
27	24.6
30	20.5
22	.
24	23
29	20.8
23	21.1
25	17.8
22	21.8
24	21.9
24	23.7
22	21.5
33	18.9
40	.
26	21.9
24	.
32	21
26	19.91
30	19
27	28
27	29
49	.
48	39.5
29	35
50	23.6
33	33
38	25.6
26	.
40	28
33	22.6
37	.
28	19
24	19.9
24	24.4
26	19.5
30	19.7
30	24.5
50	27.3
27	27.9
23	19
28	24.3
25	25.6
25	18.7
23	.
22	21.3
27	23.1
28	26.8
36	34.9
50	27.4
24	22
21	26.4
24	24.1
26	26.6
25	23
31	22.2
50	22.8
24	21.6
27	19.2
22	.

Answer 1

1 Tailed Z test, Single Mean

Given: = 24, = 23, s = 8.1, n = 83, = 0.05

(A) The Hypothesis:

H0: = 24: The mean BMI of the students is equal to 24.

Ha: < 24: The mean BMI of the students is lesser than 24..

This is a 1 tailed test (Left tailed)

______________________________________________________

(B) For the p value, we need to find the test statistic.

The Test Statistic: The test statistic is given by the equation:

Z observed = -1.125

The p Value: The p value(Left Tailed) for Z = -1.125, p value = 0.1303

______________________________________________________________

(C) The Decision Rule:   If P value is < , Then Reject H0.

The Decision: Since P value (0.1303) is > (0.05) , We Fail Reject H0.

Therefore the statement: At the significance level 5%, we can reject the null hypothesis.....is FALSE.

_______________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where

SS = SUM(X - Mean)².

#	BMI	Mean	(X - Mean)2
1	24	23	1
2	20	23	9
3	18.2	23	23.04
4	22.3	23	0.49
5	25.8	23	7.84
6	22.8	23	0.04
7	19.1	23	15.21
8	18.5	23	20.25
9	22	23	1
10	18.6	23	19.36
11	25	23	4
12	27.5	23	20.25
13	20.4	23	6.76
14	24	23	1
15	21	23	4
16	25.8	23	7.84
17	22	23	1
18	27.2	23	17.64
19	21.1	23	3.61
20	25	23	4
21	27	23	16
22	20	23	9
23	24.8	23	3.24
24	17	23	36
25	20.9	23	4.41
26	19.8	23	10.24
27	19.9	23	9.61
28	29	23	36
29	0.2	23	519.84
30	22.3	23	0.49
31	19.9	23	9.61
32	23	23	0
33	22	23	1
34	24.6	23	2.56
35	20.5	23	6.25
36	23	23	0
37	20.8	23	4.84
38	21.1	23	3.61
39	17.8	23	27.04
40	21.8	23	1.44
41	21.9	23	1.21
42	23.7	23	0.49
43	21.5	23	2.25
44	18.9	23	16.81
45	21.9	23	1.21
46	21	23	4
47	19.91	23	9.5481
48	19	23	16
49	28	23	25
50	29	23	36
51	39.5	23	272.25
52	35	23	144
53	23.6	23	0.36
54	33	23	100
55	25.6	23	6.76
56	28	23	25
57	22.6	23	0.16
58	19	23	16
59	19.9	23	9.61
60	24.4	23	1.96
61	19.5	23	12.25
62	19.7	23	10.89
63	24.5	23	2.25
64	27.3	23	18.49
65	27.9	23	24.01
66	19	23	16
67	24.3	23	1.69
68	25.6	23	6.76
69	18.7	23	18.49
70	21.3	23	2.89
71	23.1	23	0.01
72	26.8	23	14.44
73	34.9	23	141.61
74	27.4	23	19.36
75	22	23	1
76	26.4	23	11.56
77	24.1	23	1.21
78	26.6	23	12.96
79	23	23	0
80	22.2	23	0.64
81	22.8	23	0.04
82	21.6	23	1.96
83	19.2	23	14.44
Total	1909.01		1890.0781

n	83
Sum	1909.01
Average	23.000
SS(Sum of squares)	1890.0781
Variance = SS/n-1	23.050
Std Dev=Sqrt(Variance)	4.80