Question

Subject	Vitamin D
1	55.994
2	47.346
3	51.744
4	51.171
5	57.201
6	56.836
7	53.378
8	43.720
9	49.885
10	49.398
11	50.763
12	58.868
13	53.006
14	55.206
15	61.534
16	52.371
17	54.461
18	53.017
19	48.578
20	49.415
21	51.758
22	57.314
23	57.444
24	48.401
25	48.328
26	49.784
27	48.893
28	55.942
29	46.437
30	55.696
31	51.958
32	51.040
33	56.077
34	50.727
35	57.855
36	57.818
37	55.631
38	59.430
39	53.673
40	45.377
41	56.994

What is the estimated 95% confidence interval (CI) of the average blood vitamin D level of US landscapers in ng/mL?

Please note the following: 1) you might calculate a CI that is different from any of the multiple choice options listed below due to rounding differences, therefore select the closest match; 2) ensure you use either the large or small sample CI formula as appropriate; and 3) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. 51.7 to 54.2 ng/mL

b. 45.5 to 62.3 ng/mL

c. 57.6 to 61.0 ng/mL

d. 57.3 to 62.9 ng/mL

Answer 1

$\alpha$ = 1-0.95= 0.05

z_0.05/2 = 1.960

confidence interval =

Lower bound = 52.938 - 1.960 * 4.175 / $\sqrt{41}$ = 51.66 = 51.7

Lower bound = 52.938 + 1.960 * 4.175 / $\sqrt{41}$ = 54.216

So answer is a.