Question

Recall in our discussion of the normal distribution the research study that examined the blood vitamin D levels of the entire US population of landscape gardeners. The intent of this large-scale and comprehensive study was to characterize fully this population of landscapers as normally distributed with a corresponding population mean and standard deviation, which were determined from the data collection of the entire population.

Suppose you are now in a different reality in which this study never took place though you are still interested in studying the average vitamin D levels of US landscapers. In other words, the underlying population mean and standard deviation are now unknown to you. Furthermore, you would like to examine if wearing tank tops instead of short sleeve shirts significantly effects vitamin D levels. To accomplish this, you propose to collect data from the landscapers at two different points in time. Specifically, the landscapers are to wear short sleeve shirts while outside working during a period of three weeks. After three weeks, you collect blood specimens and the landscapers are then to wear tank tops for the next three weeks under the same working conditions, after which you collect blood draws a second time. You obtain research funding to randomly sample 44 landscapers, collect blood samples at two different time points as described above, and send these samples to your collaborating lab in order to quantify the amount of vitamin D in the landscapers' blood. After anxiously awaiting your colleagues to complete their lab quantification protocol, they email you the following vitamin D level data as shown in the following table.

Subject	Time Point 1, Shirts Vitamin D (ng/mL)	Time Point 2, Tank Tops Vitamin D (ng/mL)
1	44.265	22.441
2	41.681	29.863
3	38.418	28.554
4	53.197	30.752
5	51.202	25.736
6	33.329	30.924
7	33.558	35.181
8	51.820	23.336
9	44.105	35.307
10	46.219	30.449
11	48.724	23.699
12	43.675	33.445
13	50.605	32.659
14	56.323	29.940
15	54.620	34.779
16	43.244	31.221
17	47.470	28.504
18	45.657	22.623
19	48.714	34.379
20	43.183	30.759
21	46.505	28.025
22	46.643	30.560
23	42.883	28.374
24	38.949	32.660
25	42.552	30.668
26	51.432	28.308
27	48.329	32.815
28	45.195	27.508
29	56.078	27.174
30	43.361	29.104
31	46.869	28.384
32	46.076	25.247
33	41.810	34.027
34	47.533	29.059
35	47.970	35.541
36	46.101	35.743
37	43.185	32.460
38	46.863	35.466
39	43.693	37.052
40	46.926	38.258
41	43.997	31.581
42	41.986	25.363
43	55.691	31.229
44	40.977	30.938

What is the estimated 95% confidence interval (CI) of the average difference in blood vitamin D levels between short sleeve shirt and tank top attire amongst US landscapers in ng/mL?

Please note the following: 1) in practice, you as the analyst decide how to calculate the difference in vitamin D levels between time points for a given study participant, and subsequently interpret the aggregated results appropriately in the context of the data, though for the purposes of this exercise the difference is assigned for you as follows. Define the difference as the second minus the first time points, which is common practice, since the plus or minus sign of the resulting difference reflects any change over sequential time; 2) 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; 3) ensure you use either the large or small sample CI formula as appropriate; and 4) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. -17.53 to -13.45 ng/mL

b. -18.75 to -11.84 ng/mL

c. -18.84 to -12.04 ng/mL

d. -19.80 to -14.66 ng/mL

Answer 1

taking difference as 2nd -1st observations for each individual observation:

sample mean 'x̄=	-15.489
sample size   n=	44.00
sample std deviation s=	6.895
std error 'sx=s/√n=6.895/sqrt(44)=	1.039

for 95 % CI value of z=		1.960
margin of error E=z*std error =		2.04
lower bound=sample mean-E=		-17.53
Upper bound=sample mean+E=		-13.45
from above 95% confidence interval for population mean =(-17.53,-13.45)

option A is correct

a. -17.53 to -13.45 ng/mL

answered by: anonymous